# Probabilistic Trust Intervals for Out of Distribution Detection

Gagandeep Singh
<EMAIL_ADDRESS>Deepak Mishra
<EMAIL_ADDRESS>
###### Abstract

Building neural network classifiers with an ability to distinguish between in
and out-of distribution inputs is an important step towards faithful deep
learning systems. Some of the successful approaches for this, resort to
architectural novelties, such as ensembles, with increased complexities in
terms of the number of parameters and training procedures. Whereas some other
approaches make use of surrogate samples, which are easy to create and work as
proxies for actual out-of-distribution (OOD) samples, to train the networks
for OOD detection. In this paper, we propose a very simple approach for
enhancing the ability of a pretrained network to detect OOD inputs without
even altering the original parameter values. We define a probabilistic trust
interval for each weight parameter of the network and optimize its size
according to the in-distribution (ID) inputs. It allows the network to sample
additional weight values along with the original values at the time of
inference and use the observed disagreement among the corresponding outputs
for OOD detection. In order to capture the disagreement effectively, we also
propose a measure and establish its suitability using empirical evidence. Our
approach outperforms the existing state-of-the-art methods on various OOD
datasets by considerable margins without using any real or surrogate OOD
samples. We also analyze the performance of our approach on adversarial and
corrupted inputs such as CIFAR-10-C and demonstrate its ability to clearly
distinguish such inputs as well.

##  1 Introduction

Deep neural network classifiers match the human level accuracy in many
applications when test data is sampled from a distribution which is same or
approximately similar to the training data distribution, also known as in-
distribution (ID) data [19, vgg10, 10, 5, 35]. However, the networks fail in
presence of data which comes from a different distribution, known as out-of-
distribution (OOD) data [26, 13]. These networks perform classification with
an assumption of complete knowledge of the classes and therefore, wrongly
classify the OOD inputs into the known categories, often with high confidence
[27]. In real world scenarios, it is challenging to ensure the similarity of
training and test data distributions [1] which raises concerns on the
predictions made by deep neural networks due to their vulnerability against
OOD inputs.

A solution to the above problem is to design networks which can detect OOD
inputs and hence, produce reliable results. One successful design choice in
this direction is to use siblings of a network. These siblings are often the
copies of a single network with different weight parameter values. The
siblings with similar architecture but different weights are expected to
produce different outputs for the same input, therefore, are trained on ID
data to produce identical outputs for an ID input. As a result, disagreement
between the outputs of siblings for ID inputs reduces and it becomes an
indicator of OOD inputs whenever a large value observed. For a robust
detection, all the weights among the siblings are varied to explore a diverse
functional space, for example in deep ensembles [20, 8]. This naturally
increases computational burden in terms of training and testing along with
increased memory requirements. A trade-off to limit the memory expenses is to
allow only few of the weights to vary among the siblings while assigning equal
values to the rest or sharing the rest of the weights [33]. Another alternate
to create such siblings is to learn distributions over the weight parameters,
motivated from the Bayesian principles. However such networks are
computationally more complex and often pose a challenge in terms of
convergence for deep architectures [20].

In this paper, we propose a simple approach to create siblings of an already
trained network. We define a probabilistic trust interval for each weight
around its optimized value. It allows to sample multiple values for each
weight parameter and produce corresponding outputs for a single input. The
sizes of these intervals are optimized for ID data such that the disagreement
among multiple outputs for an ID input is minimized. Fig. 1 shows an
illustration of probabilistic trust intervals for some of the randomly picked
weights of a standard network where the observed interval sizes before (in
red) and after (in green) optimization are presented. Centers of the intervals
represent the original optimized values of the corresponding weight
parameters. For each parameter, values around the center are sampled with
higher probability as compared to the values near to the boundary of the
interval. It ensures that accuracy of the considered network on ID inputs is
maintained while enabling OOD detection mechanism.

Figure 1: Illustration of Probabilistic Trust Intervals for randomly picked
weights of neural classifier before (red) and after (green) training. Trust
intervals before training are of same size for each weight and get adjusted to
ID data after training for maintaining the generalization abilities of the
underlying neural network while enabling OOD detection.

Optimization of the probabilistic trust intervals’ sizes is the most critical
component of our proposed approach. These intervals should be small enough to
not allow the network siblings to go far off from the optimal point. At the
same time, the interval sizes should be large enough to produce
distinguishable output variations between ID and OOD data. For this purpose,
we design a loss function which tries to optimise the size of intervals around
each weight such that inside these intervals the output variations are small
for ID inputs. Subsequently, these variations are used to quantify the
similarity among the outputs of different sibling networks using a newly
designed measure of agreement. The deployment of probabilistic trust intervals
around weights with measure of agreement on the trained networks enables OOD
detection without using any kind of OOD validation samples or surrogate
proxies of OOD data.

We evaluate our method on different shallow and deep neural classifiers,
including VGG-16 [30], ResNet-20 [10] and DenseNet-100 [15]. We use MNIST
[21], CIFAR-10 [17], SVHN [25], CIFAR-100 [18], Fashion-MNIST [34], CMATERDB
[6], and Gaussian images, where each pixel is sampled from $\mathcal{N}(0,1)$.
Our approach outperforms previous works related to OOD input detection in most
of the experiments. For example, the False Positive Rate at 95% True Positive
Rate, for MNIST as ID and Fashion-MNIST as OOD data, is found to be 12.46% by
using just 2 sets of weights as compared to 27.48% which is the best among the
considered baselines. In addition, we also tested our approach on corrupted
CIFAR-10 (CIFAR-10-C) [11] and adversarial samples generated using Fast
Gradient Sign Method [9], and observed that the proposed approach is able to
comfortably highlight both corrupted and adversarial inputs. In summary our
main contributions are as follows,

  * •

We define probabilistic trust intervals for optimized weight parameters of
neural networks and provide a mechanism to optimize interval sizes according
to ID data. These trust intervals allow sampling of multiple weight values to
create siblings of a network.

  * •

We design a measure of agreement, which in combination with optimized
probabilistic trust intervals is useful to create OOD detectors.

  * •

The proposed approach provides a simple way of enabling OOD detection ability
in any neural network without using any OOD samples during training and also
without raising any major concern of memory requirement.

  * •

Our experiments show that the proposed approach is also suitable to flag
corrupted and adversarial inputs.

  * •

We conduct experiments with networks of various depths and with different
datasets, and observe considerably better performance as compared to state-of-
the-art OOD detection approaches in most cases.

##  2 Problem Formulation and Proposed Solution

Let $\operatorname{\mathbb{C}}$ be a deep neural network classifier and
$\operatorname{\mathbb{C}}^{(1)},\operatorname{\mathbb{C}}^{(2)},\operatorname{\mathbb{C}}^{(3)},...,\operatorname{\mathbb{C}}^{(n)}$
be its n siblings, $P_{r}$ and $P_{e}$ be two input data distributions which
are not similar, $x$ be an input sample,
$M\left(\operatorname{\mathbb{C}}(x),\operatorname{\mathbb{C}}^{(1)}(x),\operatorname{\mathbb{C}}^{(2)}(x),...,\operatorname{\mathbb{C}}^{(n)}(x)\right)$
be a real valued measure of agreement and $\delta\in\mathbb{R}^{+}$ be a small
number, then the problem of OOD input detection is defined as follows,

OOD Input Detection: If $\operatorname{\mathbb{C}}$ is trained on
$X=\\{x:x\sim P_{r}\\}$ and receives an input $x\sim P_{e}$ then
$M\left(\operatorname{\mathbb{C}}(x),\operatorname{\mathbb{C}}^{(1)}(x),\operatorname{\mathbb{C}}^{(2)}(x),...,\operatorname{\mathbb{C}}^{(n)}(x)\right)<\delta$
should hold.

In what follows, we define probabilistic trust intervals which will allow the
creation of siblings.

###  2.1 Probabilistic Trust Intervals

If the $i^{\text{th}}$ weight of an already trained neural network is denoted
by $\mu_{i}$, and $\gamma_{i}\in\mathbb{R}^{+}$ is a real positive number,
then probabilistic trust interval is defined as follows,

Probabilistic Trust Interval:
$I(\mu_{i})=[\mu_{i}-\gamma_{i},\mu_{i}+\gamma_{i}]$ is the probabilistic
trust interval for weight $\mu_{i}$ of size $2\times\gamma_{i}$ if the
following hold

  1. 1.

For a weight value ($\widetilde{\mu_{i}}$) sampled from $I(\mu_{i})$, the
error in the output of the considered network
($\mathcal{E}(\widetilde{\mu_{i}})$) is low.

  2. 2.

The weight values near to $\mu_{i}$ are sampled with higher probability as
compared to the values away.

In simple words, we trust that the weight values sampled from $I(\mu_{i})$
will generalize well for ID data. Probabilistic nature of the interval is
motivated from the training perspective, we explain this in the next section.
We drop the subscript $i$ in subsequent discussion for generalization and
simplicity of explanation. Let us now consider the errors in predictions from
the already trained network and its sibling created using
$\widetilde{\boldsymbol{\mu}}=\boldsymbol{\mu}+\Delta\boldsymbol{\mu}$ as
$\mathcal{E(\boldsymbol{\mu})}$ and
$\mathcal{E(\widetilde{\boldsymbol{\mu}})}$, respectively. Here
$\boldsymbol{\mu}$ represents the parameter vector. As shown in [2], by the
fundamental theorem of calculus applied on neural networks, one may find
bounds on the variation in errors as:

$\Delta\mathcal{E}=\mathcal{E}(\widetilde{\boldsymbol{\mu}})-\mathcal{E}(\boldsymbol{\mu})\leq\max_{t}\sum_{i=1}^{L}g(\mu_{i}+t\Delta\mu_{i})^{T}\Delta\mu_{i}$
(1)

Similarly,

$\min_{t}\sum_{i=1}^{L}g(\mu_{i}+t\Delta\mu_{i})^{T}\Delta\mu_{i}\leq\Delta\mathcal{E}$
(2)

where $g(\mu_{i})$ denotes the gradient of $\mathcal{E}(\boldsymbol{\mu})$
with respect to $\mu_{i}$ and $t\in[0,1]$. This shows that, for a small
$\Delta\boldsymbol{\mu}$, the variation in errors of siblings is bounded by
the corresponding gradients. This directly gets translated to the variation in
the outputs of the siblings as the ground truth used to measure the prediction
errors is identical. For a network well trained on ID data, the error is
expected be minimized with low gradient values which is not expected to be
true for OOD inputs. This leads to difference in output variations which can
be used for OOD detection which is facilitated by the probabilistic trust
intervals. To understand it better we consider the quadratic approximation of
$\Delta\mathcal{E}$ given by the equation (3)

$\Delta\mathcal{E}=g({\boldsymbol{\mu}})^{T}\Delta\boldsymbol{\mu}+\frac{1}{2}\Delta\boldsymbol{\mu}^{T}\mathbf{H}\Delta\boldsymbol{\mu}$
(3)

where $\mathbf{H}$ is the Hessian matrix. Fig. 2 shows the occurrences of
$\Delta\mathcal{E}$ values as per equation (3) for ID and OOD samples, where
$\Delta\boldsymbol{\mu}$ is in accordance with the optimized probabilistic
trust intervals of the weight parameters of a simple CNN
($\operatorname{\mathbb{C}}_{1}$) reported in [4]. As we see,
$\Delta\mathcal{E}$ values for most OOD samples is higher than the values
observed for ID samples.

Figure 2:  Distribution of quadratic approximation of error for
$\operatorname{\mathbb{C}}_{1}$ architecture with MNIST dataset as ID data and
Fashion-MNIST as OOD data. It can be seen that for ID samples the values are
close to zero whereas for OOD they are spread over a considerably larger
range.

###  2.2 Optimizing the size of Probabilistic Trust Intervals

We model $\Delta{\mu}=\widetilde{{\mu}}-{\mu}$ as a random variable to allow
sampling of multiple $\widetilde{{\mu}}$. This limits the memory expenses to a
constant overhead in comparison to storing multiple values of
$\widetilde{{\mu}}$ which is done in approaches like ensembles. To achieve the
desired probabilistic characteristics of the trust intervals, we can consider
sampling of $\widetilde{{\mu}}$ according to distributions such as Cauchy,
Laplace, Triangular or any bell-shaped distribution, however we consider
Gaussian distribution to take the advantage of rich literature on variational
inference [16, 3]. Accordingly, we write $\gamma=3\times\sigma$ and
$\Delta\mu=\sigma\circ\epsilon$ where $\epsilon$ is sampled from
$\mathcal{N}(0,1)$, which gives $\widetilde{\mu}={\mu}+\sigma\circ\epsilon$.

As shown by equation (1), to ensure small variations in the outputs of
siblings, it is necessary that each sibling generalizes well for the ID data
and $\Delta\mu$ is small that means $\sigma$ is small. Accordingly we define
following cost function to optimize the sizes of probabilistic trust intervals

$\mathcal{L}(D,\sigma)=\operatorname{\mathbb{E}}_{P(\widetilde{\boldsymbol{\mu}})}[-\log
P(D|\widetilde{\boldsymbol{\mu}})]+\pi_{1}s^{2}$ (4)

where $D$ represents ID data and $\operatorname{\mathbb{E}}$ is expectation.
The first term in the cost is the conventional negative log-likelihood of the
data which helps in retaining generalization ability of the considered network
on ID samples. The second term, $s^{2}$, is the sum of variances of the
softmax scores of each class in different outputs, obtained from multiple
samples of weights. $\pi_{1}$ is a hyperparameter which controls the
contribution of $s^{2}$ in total cost. Minimization of this term reduces the
interval size $\sigma$ to ensure that all siblings agree with each other on ID
data for all the classes.

One obvious solution which can be produced by the cost in equation (4) is
$\sigma=0$. In that case the trust interval will collapse to a single point
and $\widetilde{\boldsymbol{\mu}}=\boldsymbol{\mu}$ will make all siblings
identical to the original network. We, therefore, include $-\log(\sigma)$ as a
regularizer to prevent an undesired collapse and allow the discovery of a non-
zero value of $\sigma$ to facilitate OOD detection. Other suitable
regularizers such as negative $\ell_{1}$ or $\ell_{2}$ can also be used. The
updated cost function is as follows:

$\mathcal{L}(D,\sigma)=\operatorname{\mathbb{E}}_{P(\widetilde{\boldsymbol{\mu}})}[-\log
P(D|\widetilde{\boldsymbol{\mu}})]+\pi_{1}s^{2}-\pi_{2}\log(\sigma)$ (5)

where $\pi_{2}$ controls the effect of the regularizer. In short,
$\mathcal{L}$ in equation (5) ensures that the probabilistic trust interval
for each weight will be non-zero in size and will be optimized in such a way
that the disagreement between multiple samples of
$\widetilde{\boldsymbol{\mu}}$ is minimized for the given data without
compromising on the test accuracy.

It is interesting to analyze the effect of $\pi_{1}$ and $\pi_{2}$ on the
trust interval size and test accuracy. For this purpose, we consider a simple
CNN ($\operatorname{\mathbb{C}}_{1}$) reported in [4] with MNIST as ID data to
observe the variation in test accuracy of the resultant siblings. These
observations are plotted in Fig. 3. As can be seen small values of $\pi_{1}$
are unable to maintain the test accuracy of the resultant siblings. On the
other hand for small values of $\pi_{2}$ the test accuracy remains consistent,
however, as we increase the value of $\pi_{2}$, the generalization
capabilities of the network drop with increase in $\sigma$ and stabilizes
towards the end. Accordingly a trade-off between the values of $\pi_{1}$ and
$\pi_{2}$ is used in the experiments. The proposed approach provides a network
where sampling of multiple weight values from the probabilistic trust interval
effectively results in controlled perturbations on the optimized weights with
certain characteristics, therefore, we call this new network as Perturbed-NN
or PNN. Next we define a suitable measure of agreement to quantify the
variation in outputs of a single input and use it for OOD detection.

Figure 3: The change in test accuracy with updates. In the left plot different
values of $\pi_{1}$ is varied while $\pi_{2}$ is fixed to $10^{-6}$ and in the
right plot different values of $\pi_{2}$ is taken while $\pi_{1}$ is fixed to
1.

###  2.3 Measure of Agreement

PNN allows us to obtain multiple outputs for a single input by sampling
weights using $\mu$ and $\sigma$. Now, we need a measure which can help in
effectively quantifying the variations in the outputs of the PNN for flagging
OOD inputs. As opposed to many of the previous works, we consider both average
value and standard deviation of the outputs of PNN. Formally, suppose that $n$
Monte Carlo samples of weights are drawn using PNN’s parameters for
classifying the input to $c$ classes, thereby giving us $n$ softmax scores for
each of the $c$ classes. Further suppose that, the mean softmax score and
standard deviation for $k^{th}$ class is $\alpha_{k}$ and $\beta_{k}$
(slightly deviating from the standard notations for mean i.e., $\mu$ and
standard deviation i.e., $\sigma$, for avoiding confusion) respectively. We
can see that $\frac{\beta_{k}^{2}}{\alpha_{k}}$ is nothing but the index of
dispersion of $n$ softmax scores for the $k^{th}$ class. Thus, the quantity
$\sum_{k=1}^{c}\frac{\beta_{k}^{2}}{\alpha_{k}}$ will attain a low value only
if the output variation and index of dispersion for all the classes is low.
This quantity can now be combined with entropy of the average softmax scores
of the classes to provide the desired measure of agreement as,

$M=\frac{1}{\sum_{k=1}^{c}\frac{\beta_{k}^{2}}{\alpha_{k}}}+\frac{1}{\sum_{k=1}^{c}-\alpha_{k}\log{\alpha_{k}}}$
(6)

A large value of $M$ indicates high level of agreement between the outputs of
different siblings. A small constant in the denominator of $M$ can be added
for stability.

In order to explain why $M$ is better than other measures used in previous
works, we have made a conceptual comparison in the following points.

Shannon’s Entropy \- Usually, the entropy of the average softmax scores
obtained from different siblings, including ensembles and Bayesian neural
network (BNN), is used as a measure of uncertainty to identify OOD inputs. One
problem with this is that by using only the average softmax scores, the
variances are ignored. Hence, through the entropy values, only a partial
knowledge of the amount of agreement between the multiple sets of weights for
a given input is captured. In contrast $M$ in equation (6) takes into account
the variances of the output scores along with entropy of average scores to
extract more information.

Maximum of average softmax scores \- This measure has been used in [13, 24] to
detect OOD inputs. Similar to the entropy of average scores, the spread in
average scores for different classes is ignored if we only use the maximum of
the averages. Our measures give more weight to the variances of the classes
with lower softmax scores (observe the $\frac{1}{\alpha_{k}}$ multiplied with
$\beta_{k}^{2}$ in equation (6)) as compared to those with higher softmax
scores. This means if a sibling is selecting a particular class with higher
confidence then it should reject other classes in agreement with the other
siblings. Otherwise index of dispersion for the rejected classes will increase
resulting in PNN suspecting the input as OOD.

Sum of KL divergence between individual and the average softmax scores \- This
measure was used by [20] for quantifying uncertainty in the outputs of an
ensemble of neural networks. A conceptual limitation of this measure arises in
the scenario when the same class is accepted with high softmax scores by all
the networks in the ensemble. In this case, the KL divergence between the
individual and the average softmax scores will be dominated by the class with
maximum softmax score. Hence, the agreement between the networks with respect
to all the classes will not be reflected entirely.

##  3 Related Work

OOD detection has gained a lot of attention in recent years with an interest
of creating trustable deep learning solutions. Hendrycks & Gimpel [13]
proposed a baseline method for OOD detection using the low softmax scores
produced by a neural network for the OOD inputs. Their baseline was shown to
work with a variety of datasets and neural networks on diverse sets of tasks.
Subsequently, ODIN [24] was built on top of the baseline method [13] which
uses temperature scaling and adds small directed noise to the inputs, based on
the gradient of maximum softmax scores with respect to the input. The
perturbed samples can be considered as the proxies for OOD data to increase
gap between the softmax scores produced by a network for ID and OOD inputs.
Similar approaches where the proxies or auxiliary data is created using tools
such as generative adversarial networks are reported in [22, 14, 27, 31].
Another generative model based approach is presented in [28], which assumes
that an input is composed of two components, a semantic component and a
background component. Accordingly, it trains two models and proposes a log
likelihood ratio statistic on the outputs of the two models to flag OOD
inputs. To train the model on background components, the dataset was created
by applying directed noise on the ID samples with an assumption that it will
inhibit the semantic component. Lee et al. [23] also used the directed noise
but in different manner where it was applied on the test inputs, which allowed
the network to produce multiple outputs and select the best among them using
Mahalanobis distance based confidence scores. The approach works under the
assumption that class-conditional Gaussian distributions can be fitted to the
features of an already trained softmax neural classifier and subsequently can
be used to measure the Mahalanobis distance of the inputs.

A better approach to obtain multiple outputs for a single input is to use deep
ensembles [20] where multiple copies of a network are trained in parallel on
the ID data and variation in outputs of the ensembles during inference is used
an indicator of the uncertainty in the decision and flag OOD inputs. However,
this leads to a linearly increasing memory overhead with the number of
ensembles used and also limits the number of outputs one can obtain. An
alternate to create multiple copies of a network is BNN which learns
distribution over weights and samples multiple weight values to create Bayes
ensembles [3]. BNNs do not suffer from the problems of linearly increasing
memory overhead and limited number of outputs, however, face the challenge of
computational complexity and convergences of deep architectures [8]. There
have been attempts to create less complex and scalable BNNs [29, 7], however,
these work on the basis of several assumptions, and often require proxies of
OOD samples for uncertainty calibration and OOD detection. A PNN takes the
best of both worlds in a sense that it allows sampling of multiple weight
values from the trust intervals where the trust intervals can be interpreted
as the space of directed noise, although added to the optimized weights of an
already trained neural network, not the input samples. PNNs, therefore, do not
need any proxies of the OOD samples and offer a very simple training mechanism
with robust OOD detection.

##  4 Experiments and Results

We use the following combination of architectures and ID datasets in our
experiments and corresponding PNNs.

${\operatorname{\mathbb{C}}_{1}}$ \- A shallow CNN architecture containing
only two convolutional and two fully connected layers [4]. The first two
layers are convolution layers with $5\times 5$ kernels and number of channels
as 32 and 64, respectively. Each convolutional layer output is downsampled
using max pooling and the resultant output passes through the two fully
connected layers containing 1024 and 10 output units, respectively, to produce
the final output. All but last layer use ReLU activation. MNIST dataset is
used as ID data. We train the corresponding PNN for 1708 iterations with a
batch size of 256.

$\operatorname{\mathbb{C}}_{2}$ \- The second architecture is VGG-16 for which
we used CIFAR-10 as ID data to create the corresponding PNN in 1712 iterations
with a batch size of 64.

$\operatorname{\mathbb{C}}_{3}$ \- Third architecture considered is ResNet-20
on CIFAR-10 dataset. The corresponding PNN converges in 802 iterations with a
batch size of 128.

$\operatorname{\mathbb{C}}_{4}$ \- This is also a simple CNN architecture
trained on SVHN dataset. It contains a sequence of convolution-batch-
normalization-convolution-max-pool-dropout blocks. There are three such blocks
with 32, 64 and 128 channels in the convolutional layers. The kernel size of
all the convolutional layers is $3\times 3$. The window size for max-pool is
$2\times 2$. The dropout rate is 0.3. The last two layers are fully connected
layers with 128 and 10 output units. For PNN, we use 315 iterations with a
batch size of 128.

$\operatorname{\mathbb{C}}_{5}$ \- Final architecture is a comparatively
deeper architecture - DenseNet-100 with a growth rate of 12. It is trained on
CIFAR-10 dataset. For PNN model of this architecture, we use 120 iterations
with a batch size of 64.

To ensure neumarical stability during minimization of the cost in equation
(5), we parametrize $\sigma$ using $\rho$ as, $\sigma=\log(1+\exp(\rho))$, and
$\rho$ is initialized using $U(0,1)$ 222Source code for all the experiments is
provided in the supplementary material. The optimizer used for all the
architectures is $RMSprop$ [32] with a learning rate of 0.01. The values of
$\pi_{2}$ are kept less than $10^{-6}$, which allows $\pi_{1}$ to be fixed at
1 for all of our experiments without raising any concerns of convergence. We
stopped the training when either the negative log likelihood, cross entropy
for image classification tasks, started to increase or there wasn’t any
further decrease in $\pi_{1}s^{2}$. The intuition behind this is that
optimizing $\sigma$ in further iterations won’t be beneficial because,
increasing $\sigma$ will either fluctuate the outputs for ID data too much and
decreasing $\sigma$ will increase $-\log(\sigma)$.

To begin with, we first evaluate the obtained PNNs performances on ID data.
Table 1 shows the differences in mean classification accuracy of CNNs with
point estimate and the corresponding PNNs on ID test samples. For each PNN
only a single sample of weights is drawn from their trust intervals for
computing the test accuracy. It can be observed that the change in test
accuracy is negligible. This shows that when an already trained neural
classifier is converted into a PNN using the proposed approach, its
performance on ID test data is maintained. Next we show that while maintaining
the performance on ID data, PNNs facilitate the detection of OOD inputs.

Architecture | Dataset | CNN (in %) | PNN (in %)
---|---|---|---
$C_{1}$ | MNIST | 98.53 | 98.49
$C_{2}$ | CIFAR10 | 93.54 | 93.66
$C_{3}$ | CIFAR10 | 91.68 | 91.44
$C_{4}$ | SVHN | 95.93 | 94.43
$C_{5}$ | CIFAR10 | 93.51 | 93.55
Table 1: Test accuracy of CNNs with point estimate and the corresponding PNNs for various architectures. For all the architectures, PNN accuracy is very close to CNN which shows that PNNs maintain the performance of corresponding CNNs on ID data while enabling OOD detection. Model | ID | OOD | Baseline | ODIN | BayesAdapter | DeepEnsemble | MHB | PNN (ours)
---|---|---|---|---|---|---|---|---
C1 | MNIST | FMNIST | 27.48 | 27.09 | 70.61 | 36.86 | 30.31 | 12.46
C2 | CIFAR10 | CMATERDB | 61.48 | 56.80 | 58.63 | 88.32 | 100.0 | 54.80
CIFAR10 | CIFAR100 | 71.35 | 71.39 | 70.15 | 71.46 | 99.86 | 67.41
C3 | CIFAR10 | CMATERDB | 58.88 | 59.03 | 66.11 | 44.12 | 75.63 | 48.18
CIFAR10 | CIFAR100 | 70.74 | 71.6 | 76.73 | 82.97 | 99.71 | 64.37
C4 | SVHN | CIFAR10 | 43.80 | 43.18 | 71.00 | 66.12 | 88.50 | 37.00
SVHN | CMATERDB | 49.00 | 52.59 | 73.45 | 45.54 | 87.08 | 51.00
SVHN | GAUSSIAN | 19.95 | 19.97 | 45.36 | 60.73 | 95.08 | 1.61
C5 | CIFAR10 | CIFAR100 | 70.18 | 69.71 | 72.37 | 69.71 | 97.122 | 68.71
Table 2: Comparison of our proposed approach with the state-of-the-art methods with respect to FPR at 95 % TPR. Low FPR at high TPR is important when it comes to selecting a threshold for deployment of an OOD detection technique Model | ID | OOD | Baseline | ODIN | BayesAdapter | DeepEnsemble | MHB | PNN (ours)
---|---|---|---|---|---|---|---|---
C1 | MNIST | FMNIST | 97.00 | 96.92 | 51.60 | 55.72 | 96.49 | 99.44
C2 | CIFAR10 | CMATERDB | 92.58 | 93.11 | 78.21 | 60.00 | 62.01 | 95.98
CIFAR10 | CIFAR100 | 82.56 | 82.88 | 68.51 | 57.56 | 40.89 | 88.92
C3 | CIFAR10 | CMATERDB | 90.68 | 90.10 | 88.27 | 55.22 | 90.60 | 92.41
CIFAR10 | CIFAR100 | 85.89 | 85.63 | 82.88 | 66.49 | 50.74 | 98.82
C4 | SVHN | CIFAR10 | 93.12 | 93.02 | 96.99 | 56.59 | 92.70 | 94.85
SVHN | CMATERDB | 84.93 | 84.49 | 93.64 | 60.60 | 87.68 | 87.73
SVHN | GAUSSIAN | 98.00 | 97.92 | 96.91 | 54.78 | 91.41 | 98.51
C5 | CIFAR10 | CIFAR100 | 69.59 | 69.93 | 66.05 | 64.03 | 70.56 | 78.3
Table 3: Comparison of our proposed approach with the state-of-the-art
techniques with respect to AUPR. The proposed PNNs result in comparative
better performance in most experiments.

We consider different combination of ID and OOD datasets for different
architectures. We compare OOD detection performance of PNN with existing works
including state-of-the-art approaches such as Deep Ensembles [20] and
BayesAdapter [7] along with ODIN [24], MHB [23], and the Baseline OOD
detection approach [13]. For ODIN, we pick $T$ for temperature scaling from,
$\\{10,100,1000\\}$ and $\epsilon$ from $\\{0.0001,0.00625,0.025,0.05,0.1\\}$.
In Mahalanobis distance based OOD detection (MHB), we use the features from
the penultimate layer and the layer preceding it. Similarly for Deep
Ensembles, BayesAdapter, and the proposed PNNs, we considered two siblings of
each for a fair comparison. It should be noted that, no method including ours,
is adapted to OOD data or calibrated for uncertainty using proxies of OOD
data, as in a general scenario, the OOD examples can come from any
distribution which may or may not be known during training. We use False
Positive Rate at 95 % True Positive Rate (FPR at 95 % TPR), Area Under the
Precision Recall Curve (AUPR), and Area Under the Receiver Operating
Characteristic Curve (AUROC) as the performance metrics. The results for FPR
at 95 % TPR and AUPR are shown in Tables 2 and 3, respectively. AUROC results
are provided in the supplementary material. It can be seen that, PNNs
outperform the previous works in most of the experiments and achieve
comparable results in other, particularly in terms of AUPR. It is noteworthy
that since AUPR is a threshold independent measure it can be of great
significance when it comes to comparing different techniques for OOD
detection. Higher values of AUPR reflect better performance on a large number
of thresholds. To understand the capacity of the proposed PNNs better, we
further test them against corrupted and adversarial inputs, and present the
results as follows.

###  4.1 Experiments with Noisy Test Samples

We use noisy versions of ID CIFAR-10 samples, known as CIFAR-10-C [12]. We
consider two of the most commonly observed real world noises, Gaussian and
Speckle, and pass them to PNNs corresponding to
$\operatorname{\mathbb{C}}_{2}$, $\operatorname{\mathbb{C}}_{3}$ and
$\operatorname{\mathbb{C}}_{5}$ for which CIFAR-10 is used as the ID. We plot
distribution of the observed values of the measure of agreement ($M$) on a
logarithmic scale in Fig. 4. The figure also shows distribution of the values
of $M$ observed for the clean ID samples. A clear difference can be observed
for noisy and clean samples, particularly for the deep architecture
DenseNet-100. Notice the consistency in the observed distribution of the
values of $M$ for clean samples between the plots in the top and bottom row of
Fig. 4 for different architectures. Even though different values are sampled
for the two different types of noisy samples, a negligible variation is
observed while using these weights for clean samples. This shows the robust
behaviour of PNNs.

###  4.2 Experiments with Adversarial Test Samples

Next we consider adversarial images generated from the original ID samples
using FGSM attack [9]. FGSM is applied on each architecture with its point
estimates and adversarial samples are generated for the corresponding ID data
which was used during training for obtaining those point estimates. Similar to
the previous experiment, distributions of the observed values of $M$ for the
original and adversarial samples are plotted in Fig. 5. As the intensity of
the attack increases, quality of the samples degrades and PNNs highlight it by
bringing out clear separations in the peaks of the distribution corresponding
to the original and adversarial samples.

Figure 4: Distribution of values of $\log{(M)}$ for the original clean images
(in green) and their Gaussian noise corrupted versions (top row, in red) and
Speckle noise corrupted versions (bottom row, in red) obtained from CIFAR-10-C
dataset. A distinct nature of the plots for clean and noisy samples can be
observed.

Figure 5: Distribution of $\log{(M)}$ for original images (in green) and their
adversarial versions (in red). Rows from top to bottom show the observations
for $\operatorname{\mathbb{C}}_{1}$ to $\operatorname{\mathbb{C}}_{5}$,
respectively. Peeks of the distribution for original and adversarial images
can be easily distinguished from each other. Moreover, with the increase in
intensity of adversarial attack ($\epsilon$), the overlap between the
distributions decreases. Observe the shift in the red peak with increasing
$\epsilon$.

##  5 Discussion

As mentioned in section 3, proposed probabilistic trust intervals can be
interpreted as the space of directed noise. It is, therefore, important to
discuss the scenario where instead of learning $\sigma$, a trivial approach is
adopted in which small random perturbations are added to the weights of an
already trained neural network. We observed in our experiments that this
trivial approach results in degradation of test accuracy if the strength of
perturbations is too high and if it is too low then test accuracy is preserved
but the performance on the task of OOD input detection is suboptimal. For
example, when we directly added perturbations sampled from $\mathcal{N}(0,1)$
to the weights of $\operatorname{\mathbb{C}}_{1}$, the test accuracy on MNIST
dataset reduced from close to 98% to $\approx$ 10% and when we sampled the
perturbations from $\mathcal{N}(0,0.01)$, the FPR at 95% TPR for Fashion-MNIST
was $\approx$ 55% which is considerably poor as compared to what is observed
for the corresponding PNN, shown in Table 2. In fact, handcrafting the
strength of perturbations is nearly impossible because in complex
architectures it is very difficult to predict the effect of perturbing weights
on the outputs if done manually. Our proposed cost function automates this
process and $\sigma$ is learnt in such a way that both the requirements of OOD
input detection and test accuracy are satisfied simultaneously.

Further, it is also interesting to discuss the effect of number of siblings on
OOD detection. For this we consider PNN corresponding to
$\operatorname{\mathbb{C}}_{1}$ trained on MNIST dataset. We sample 2, 5, and
10 values for weight parameters and evaluate the created sets of siblings for
ID and OOD (Fashion-MNIST) samples. Similar to the experiments described in
previous section, distributions of values of $M$ are plotted in Fig. 6. As we
see, the distributions corresponding to OOD samples for different number of
weight samples are almost overlapping each other. There is a little
improvement with increasing number of weight samples in the performance on ID
data. Overall the plots show that 2 samples is a reasonable choice to get the
desired results while keeping the amount of computation during inference under
control.

Figure 6:  Distribution of $\log{(M)}$ values for
$\operatorname{\mathbb{C}}_{1}$ architecture with MNIST dataset as ID data
(solid lines) and Fashion-MNIST as OOD data (dashed lines) for different
number of weight samples. It can be seen that the distributions almost overlap
with each other for different number of samples taken.

##  6 Conclusion

In this paper, we propose a simple approach to enable OOD detection on already
trained deep neural network classifiers using the defined probabilistic trust
intervals and optimized for ID data around each weight parameter. We also
designed a measure of agreement for a robust detection using the resultant
PNNs. Once trained PNN can be safely deployed with a constant memory and
computation overhead. PNNs are able to distinguish adversarial and noisy
inputs as well. We have used Gaussian distribution to model the probabilistic
nature of the trust interval, in future other distributions can also be
considered. Another possible future direction is to explore the possibility of
removing redundant and very small sized intervals to reduce both, the time
consumed in sampling and memory required to store interval parameters. PNNs
can also be explored for uncertainty estimation in applications such as
regression and segmentation.

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# Inequalities for $L^{p}$-norms that sharpen the triangle inequality and
complement Hanner’s Inequality

Eric A. Carlen  Department of Mathematics, Hill Center, Rutgers University,
110 Frelinghuysen Road, Piscataway NJ 08854-8019, USA<EMAIL_ADDRESS>,  Rupert L. Frank  Mathematisches Institut, Ludwig-Maximilans Universität
München, Theresienstr. 39, 80333 München, Germany, and Mathematics 253-37,
Caltech, Pasadena, CA 91125, USA<EMAIL_ADDRESS>,  Paata Ivanisvili
Department of Mathematics, University of California, Irvine, CA 92617, USA
<EMAIL_ADDRESS>and  Elliott H. Lieb  Departments of Mathematics and Physics,
Princeton University, Princeton, NJ 08544, USA<EMAIL_ADDRESS>
###### Abstract.

In 2006 Carbery raised a question about an improvement on the naïve norm
inequality $\|f+g\|_{p}^{p}\leq 2^{p-1}(\|f\|_{p}^{p}+\|g\|_{p}^{p})$ for two
functions in $L^{p}$ of any measure space. When $f=g$ this is an equality, but
when the supports of $f$ and $g$ are disjoint the factor $2^{p-1}$ is not
needed. Carbery’s question concerns a proposed interpolation between the two
situations for $p>2$. The interpolation parameter measuring the overlap is
$\|fg\|_{p/2}$. We prove an inequality of this type that is stronger than the
one Carbery proposed. Moreover, our stronger inequalities are valid for all
$p$.

© 2018 by the authors. This paper may be reproduced, in its entirety, for non-
commercial purposes.
Work partially supported by NSF grants DMS–1501007 (E.A.C.), DMS–1363432
(R.L.F.), PHY–1265118 (E.H.L.)

December 6, 2018

##  1\. Introduction and main theorem

Since $|z|^{p}$ is a convex function of $z$ for $p\geq 1$, for any measure
space, the $L^{p}$ unit ball, $B_{p}:=\\{f\ :\ \int|f|^{p}\leq 1\\}$, is
convex. One way to express this is with Minkowski’s triangle inequality
$\|f+g\|_{p}\leq\|f\|_{p}+\|g\|_{p}$. Another is the inequality

$\|f+g\|_{p}^{p}\leq 2^{p-1}\left(\|f\|_{p}^{p}+\|g\|_{p}^{p}\right),$ (1.1)

valid for any functions $f$ and $g$ on any measure space. There is equality if
and only if $f=g$ and, in Theorem 1.1, we improve (1.1) substantially when $f$
and $g$ are far from equal.

In 2006 Carbery proposed [3] several plausible refinements of (1.1) for $p\geq
2$, of which the strongest was

$\int\left|f+g\right|^{p}\leq\left(1+\frac{\|fg\|_{p/2}}{\|f\|_{p}\|g\|_{p}}\right)^{p-1}\int\left(|f|^{p}+|g|^{p}\right).$
(1.2)

There is equality in (1.2) both when $f=g$ and when $fg=0$. Thus, (1.2), if
true, can be viewed as a refinement of (1.1) in which there is equality not
only when $f=g$ but also when $fg=0$.

The ratio ${\displaystyle\Gamma=\frac{\|fg\|_{p/2}}{\|f\|_{p}\|g\|_{p}}}$
varies between 0 and 1 and, therefore, the factor of $(1+\Gamma)^{p-1}$ varies
between 1 and $2^{p-1}$, interpolating between the two cases of equality in
(1.2).

We propose and prove a strengthening of (1.2) in which $\Gamma$ is replaced by
the quantity

$\widetilde{\Gamma}:=\|fg\|_{p/2}\left(\frac{\|f\|_{p}^{p}+\|g\|_{p}^{p}}{2}\right)^{-2/p}\
,$ (1.3)

which is smaller by virtue of the arithmetic-geometric mean inequality.

Our improved inequalities are not restricted to $p>2$, but are valid for all
$p\in\mathbb{R}$, as stated in Theorem 1.1. There we write

$\|f\|_{p}:=\left(\int|f|^{p}\right)^{1/p}\quad\text{for {\bf all}}\ p\neq 0.$

We note that inequality (1.2) involves three kinds of quantities on the right
side ($\|fg\|_{p/2},\ \|f\|_{p}^{p}+\|g\|_{p}^{p}$ and $\|f\|_{p}\|g\|_{p}$),
while our inequality involves only two ($\|fg\|_{p/2}$ and
$\|f\|_{p}^{p}+\|g\|_{p}^{p}$), a simplification that is essential for our
proof.

######  Theorem 1.1 (Main Theorem).

For all $p\in(0,1]\cup[2,\infty)$ and functions $f$ and $g$ on any measure
space,

$\boxed{\int\left|f+g\right|^{p}\leq\left(1+\frac{2^{2/p}\|fg\|_{p/2}}{\left(\|f\|_{p}^{p}+\|g\|_{p}^{p}\right)^{2/p}}\right)^{p-1}\int\left(\,|f|^{p}+|g|^{p}\,\right).}$
(1.4)

The inequality reverses if $p\in(-\infty,0)\cup(1,2)$, where, for $p\in(1,2)$,
it is assumed that $f$ and $g$ are positive almost everywhere.
For $p>2$, (resp. for $p\in(0,1)$ ) the inequality is false if
$\widetilde{\Gamma}$ is raised to any power $q>1$, (resp. $q<1$).
For $p<0$, (resp. for $p\in[1,2]$ ) the reversed inequality is false if
$\widetilde{\Gamma}$ is raised to any power $q>1$, (resp. for $q<1$).

For $p>0$, $p\neq 1,2$, $\|f\|_{p},\|g\|_{p}<\infty$, there is equality in
(1.4) if and only if $f$ and $g$ have disjoint supports, up to a null set, or
are equal almost everywhere. For $p<0$, $\|f\|_{p},\|g\|_{p}<\infty$, there is
equality in (1.4) if and only if $f$ and $g$ are equal almost everywhere.

We note that in proving the theorem, we may always assume that $f$ and $g$ are
non-negative. In fact, the right side of (1.4) only depends on $|f|$ and $|g|$
and the left side does not decrease for $p\geq 0$ and does not increase for
$p<0$ if $f$ and $g$ are replaced by $|f|$ and $|g|$. The latter follows since
$|f+g|\leq|f|+|g|$ implies $|f+g|^{p}\leq(|f|+|g|)^{p}$ for $p>0$ and
$|f+g|^{p}\geq(|f|+|g|)^{p}$ for $p<0$.

Carbery proved that his proposed inequality is valid when $f$ and $g$ are
characteristic functions. Our theorem can also be easily proved in this
special case.

Theorem 1.1 may be viewed as a refinement of Minkowski’s inequality. Since
(1.1), like Minkowski’s inequality, is a direct expression of the convexity of
$B_{p}$, it is equivalent to Minkowski’s inequality. We recall the simple
argument: For any unit vectors $u,v\in L^{p}$, (1.1) says that
$\|(u+v)/2\|_{p}\leq 1$, and then by continuity, $\|\lambda
u+(1-\lambda)v\|_{p}\leq 1$ for all $\lambda\in(0,1)$. Suppose
$0<\|f\|_{p},\|g\|_{p}<\infty$, and define
$\lambda=\|f\|_{p}/(\|f_{p}\|+\|g\|_{p})$, $u=\|f\|_{p}^{-1}f$, and
$v=\|g\|_{p}^{-1}g$. Then

$\|f+g\|_{p}^{p}=(\|f\|_{p}+\|g\|_{p})^{p}\|\lambda
u+(1-\lambda)v\|_{p}^{p}\leq(\|f\|_{p}+\|g\|_{p})^{p}\ ,$

which is Minkowski’s inequality.

When $p=1$ and $f,g\geq 0$, (1.1) is an identity; otherwise when $p>1$, there
is equality in (1.1) if and only if $f=g$. When the supports of $f$ and $g$
are disjoint, however, (1.1) is far from an equality and the factor $2^{p-1}$
is not needed. There is equality in Minkowski’s inequality whenever $f$ is a
multiple of $g$ or vice-versa. Hence although (1.1) is equivalent to
Minkowski’s inequality, it becomes an equality in fewer circumstances.

There is another well-known refinement of Minkowski’s inequality for
$1<p<\infty$, namely Hanner’s inequality, [4, 2, 6] which gives the exact
modulus of convexity of $B_{p}$, the unit ball in $L^{p}$. For $p\geq 2$, and
unit vectors $u$ and $v$, Hanner’s inequality says that

$\left\|\frac{u+v}{2}\right\|_{p}^{p}+\left\|\frac{u-v}{2}\right\|_{p}^{p}\leq
1,$ (1.5)

which is also a consequence of one of Clarkson’s inequalities [1]. When $u$
and $v$ have disjoint supports, $\|u+v\|_{p}^{p}=\|u-v\|_{p}^{p}=2$, and then
the left hand side is $2^{2-p}$, so that for unit vectors $u$ and $v$, the
condition $uv=0$, which yields equality in the inequality of Theorem 1.1, does
not yield equality in Hanner’s inequality. On the other hand, while one can
derive a bound on the modulus of convexity in $L^{p}$ from (1.4), one does not
obtain the sharp exact result provided by Hanner’s inequality. Both
inequalities express a quantitative strict convexity property of $B_{p}$, but
neither implies the other; they provide complimentary information, with the
information provided by Theorem 1.1 being especially strong when $f$ and $g$
have small overlap as measured by $\|fg\|_{p/2}$.

Our proof of Theorem 1.1 consists of three parts:

Part A: We show how to reduce the inequality to a simpler one involving only
one function, namely $\alpha:=f/(f+g)$ for $f,g>0$, which takes values in
$[0,1]$, and a reference measure that is a probability measure. This exploits
the fact that the only important quantity is the ratio of $f$ to $g$. This
part is very easy.

Part B: In the second part, which is more difficult than Part A, we show that
Theorem 1.1 is true if it is true when the function $\alpha$ is constant.
(This is the same as saying $f$ and $g$ are proportional to each other.) When
$\alpha:=f/(f+g)$ is constant and the reference measure is a probability
measure, (1.4) yields the inequality for _numbers_ $\alpha\in[0,1]$ and
$p\in[0,1]\cup[2,\infty]$,

$1\leq\left(1+\left(\frac{2\alpha^{p/2}(1-\alpha)^{p/2}}{\alpha^{p}+(1-\alpha)^{p}}\right)^{2/p}\right)^{p-1}\left(\alpha^{p}+(1-\alpha)^{p}\right)\
.$ (1.6)

with the reverse inequality for $p\notin[0,1]\cup[2,\infty]$.

######  Remark 1.2.

The quantity ${\displaystyle
R:=\frac{2\alpha^{p/2}(1-\alpha)^{p/2}}{\alpha^{p}+(1-\alpha)^{p}}}$ lies in
$[0,1]$ for all $\alpha$ and $p$. Therefore, $R^{q}$ decreases as $q$
increases. Thus for $p\geq 2$, the inequality

$1\leq\left(1+\left(\frac{2\alpha^{p/2}(1-\alpha)^{p/2}}{\alpha^{p}+(1-\alpha)^{p}}\right)^{q}\right)^{p-1}\left(\alpha^{p}+(1-\alpha)^{p}\right)\
,$ (1.7)

strengthens as $q$ increases, and for $p\in[0,1]$, it strengthens as $q$
decreases. Likewise, for $p\in[1,2]$ the reverse of (1.7) is stronger for
smaller $q$, and for $p<0$, it is stronger for larger $q$.

Part C: With Parts A and B complete, the proof reduces to a seemingly
elementary inequality, parametrized by $p$, for a number $\alpha\in[0,1]$. The
proof of this is Part C. While the validity of (1.6) appears to be a
consequence of Theorem 1.1, one can also view Theorem 1.1 as a consequence of
(1.6).

######  Theorem 1.3.

For all numbers $\alpha\in[0,1]$, inequality (1.6) is valid for all
$p\in(0,1]\cup[2,\infty)$, and the reverse inequality is valid for all
$p\in(-\infty,0)\cup(1,2)$.

For $p>2$, (resp. for $p\in(0,1)$ ) inequality (1.7) is false if $q>2/p$,
(resp. for $q<2/p$).

For $p<0$, (resp. for $p\in[1,2]$ ) the reverse inequality is false if
$q>2/p$, (resp. for $q<2/p$).

For $p>0$, $p\neq 1,2$, there is equality if and only if
$\alpha\in\\{0,1/2,1\\}$. For $p<0$, there is equality if and only if
$\alpha=1/2$.

###  1.1. Restatement of Theorem 1.3 in terms of means

Inequality (1.6) can be restated in terms of $q$th power means [5]: For
$x,y>0$, define

$M_{q}(x.y)=((x^{q}+y^{q})/2)^{1/q}\quad\text{if}\
q\in\mathbb{R}\setminus\\{0\\}\qquad\text{and}\qquad M_{0}(x,y)=\sqrt{xy}\,.$

Note that $M_{0}(x,y)$ is the _geometric mean_ of $x$ and $y$ and
$M_{-1}(x,y)$ is their _harmonic mean_.

######  Corollary 1.4.

For all $x,y>0$, and all $p\in[0,1]\cup[2,\infty]$

$M_{1}^{p}(x,y)\leq\left(\frac{M_{p}(x,y)+M_{-p}(x,y)}{2}\right)^{p-1}M_{p}(x,y)\
,$ (1.8)

while the reverse inequality is valid for all $p\in(-\infty,0)\cup(1,2)$.

###### Proof.

A simple calculation shows that for all $p>0$,
${\displaystyle\frac{M_{-p}(x,y)}{M_{p}(x,y)}=\frac{2^{2/p}xy}{(x^{p}+y^{p})^{2/p}}}$.
Thus, taking $x=\alpha$ and $y=1-\alpha$, the inequality (1.6) can be written
as

$\frac{1}{2}\leq\left(1+\frac{M_{-p}(\alpha,1-\alpha)}{M_{p}(\alpha,1-\alpha)}\right)^{p-1}M_{p}^{p}(\alpha,1-\alpha)\
,$

Then by homogeneity and the fact that $M_{1}(\alpha,1-\alpha)=1/2$, (1.6) is
equivalent to (1.8) ∎

The following way to write our inequality sharpens and complements the
arithmetic-geometric mean inequality for any two numbers $x,\,y>0$, provided
one has information on $M_{p}(x,y)$.

######  Corollary 1.5 (Improved and complemented AGM inequality).

For all $x,y>0$, and all $p>2$,

$1-\left(\frac{A}{M_{p}}\right)^{p^{\prime}}\,\geq\,\frac{1}{2}\left(1-\left(\frac{G}{M_{p}}\right)^{2}\right)\geq\frac{1}{2}\left(1-\left(\frac{G}{M_{p^{\prime}}}\right)^{2}\right)\geq
1-\left(\frac{A}{M_{p^{\prime}}}\right)^{p}$ (1.9)

where $p^{\prime}=p/(p-1)$, $A=(x+y)/2$ and $G=\sqrt{xy}$.

######  Remark 1.6.

Since $p,p^{\prime}\geq 1$, all of the quantities being compared in these
inequalities are non-negative.

Despite the classical appearance of (1.8), we have not been able to find it in
the literature, most of which concerns inequalities for means
$M_{q}(x_{1},\dots,x_{n})=(\frac{1}{n}\sum_{j=1}^{n}x_{j}^{p})^{1/p}$ of an
$n$-tuple of non-negative numbers, often with more general weights. The
obvious generalization of (1.8) from two to three non-negative numbers $x$,
$y$, and $z$ is false as one sees by taking $z=0$: Then there is no help from
$M_{-p}(x,y,z)$ on the right. A valid generalization to more variables
probably involves means over $M_{-p}(x_{j},x_{k})$ for the various pairs. In
any case, as far as we know, (1.8) is new.

A truly remarkable feature of the inequality (1.8) is that it is surprisingly
close to equality uniformly in the arguments. To see this, let $f(\alpha,p)$
denote the right hand side of (1.6). Contour plots of this function for
various ranges of $p$ are shown in Figs. 1, 2 and 3 below.

Fig. 1 Fig. 2 Fig. 3

Fig. 1 is a contour plot of this function in $[1/2,1]\times[2,4]$. The
contours shown in Fig. 1 range from $1.00001$ to $1.018$. Note that the
function $f$ is identically $1$ along three sides of plot: $\alpha=1/2,1$, and
$p=2$. The maximum value for $2\leq p\leq 4$, near $1.018$, occurs towards the
middle of the segment at $p=4$.

Fig. 2 is a contour plot of $f$ on $[1/2,1]\times[1,2]$. The contours range
from $0.9961$ (the small closed contour) to $0.99999999$ (close to the
boundary). _Amazingly, the function in ( 1.6) is quite close – within two
percent – to the constant 1 over the range $p\geq 1$ and $\alpha\in[0,1]$._
Moreover, the “landscape” is quite flat: The gradient has a small norm over
the whole domain.

Fig. 3 is a contour plot of $f$ in the domain $[0,1/2]\times[0,1]$. The
contours in Fig. 3 range from $1.0000001$ to $1.06$. Higher values are to the
right. For $p$ in this range, the maximum is not so large – about $1.06$ – but
the landscape gets very “steep” near $\alpha=1$ and $p=0$. The proof of the
inequality is especially delicate in this case.

For $p<0$, there is equality only at $\alpha=1/2$, and the inequality is not
so uniformly close to an identity. The contour plot is less informative, and
hence is not recorded here. This is the case in which the inequality is
easiest to prove.

It is possible to give a simple direct proof of the inequality for certain
integer values of $p$, as we discuss in Section 5. We also give a simple proof
that for $p>2$ and for $p<0$, validity of the inequality at $p$ implies
validity of the inequality at $2p$, and we briefly discuss an application of
this to the problem in which functions are replaced by operators and integrals
are replaced by traces.

######  Remark 1.7.

We close the introduction by briefly discussing one other way to write the
inequality (1.6). Introduce a new variable $s\in(0,1)$ through

$\alpha=\frac{1+\sqrt{s}}{2}$

Rewriting (1.6), and taking the $\frac{1}{p-1}$ root of both sides, we may
rearrange terms to obtain.

$2\leq\eta^{\frac{1}{p-1}}(s)\left(1+\frac{1-s}{\eta^{\frac{2}{p}}(s)}\right)=\eta^{\frac{1}{p-1}}(s)+(1-s)\eta^{\frac{2-p}{p(p-1)}}(s)$
(1.10)

for $0\leq s\leq 1$, where

$\eta(s):=\frac{(1+\sqrt{s})^{p}+(1-\sqrt{s})^{p}}{2}\ .$ (1.11)

Taking the $\frac{1}{p-1}$ eliminate the change of direction in the inequality
at $p=1$, and it now take on a non-trivial form at $p=1$: Define

$f_{p}(s):=\eta^{\frac{1}{p-1}}(s)+(1-s)\eta^{\frac{2-p}{p(p-1)}}(s)-2\ ,$
(1.12)

for $p\neq 1$, and one easily computes the limit at $p=1$:

$f_{1}(s):=(2-s)(1-\sqrt{s})^{\frac{1-\sqrt{s}}{2}}(1+\sqrt{s})^{\frac{1+\sqrt{s}}{2}}-2\
.$

Theorem 1.3 is equivalent to the assertion that for all $s\in(0,1)$

$f_{p}(s)\geq 0\ {\rm for}\ p\in(-\infty,0)\cup(2,\infty)\quad{\rm and}\quad
f_{p}(s)\leq 0\ {\rm for}\ p\in(0,2)\ .$ (1.13)

In this form, the inequality is easy to check for some values of $p$. For
example, for $p=-1$, $\eta(s)=\frac{1}{1-s}$ and
$f_{-1}(s)=(1-s)^{1/2}+(1-s)^{-1/2}-2$. which is clearly positive. One can
give simple proofs of (1.13) for other integer values of $p$, e.g., $p=3$ and
$p=4$ along these line, but this change of variables is not what we use to
prove the general inequality. It is, however, convenient for checking
optimality of of the power $2/p$ in (1.6).

##  2\. Part A. Reduction from two functions to one

While Theorem 1.1 involves two functions $f$ and $g$ one can use the
arbitrariness of the measure to reduce the question to a _single_ function
defined on a _probability space_ (that is, $\int 1=1$). We have already
observed that it suffices to prove the inequality in the case where $f$ and
$g$ are both non-negative. For non-negative functions $f$ and $g$, set

$\alpha=f/(f+g)\,,\qquad 1-\alpha=g/(f+g)\,.$

Replacing the underlying measure $dx$ by the new measure
$(f+g)^{p}\,dx/\|f+g\|_{p}^{p}$ we see that it suffices to prove the following
inequality for $p\in[0,1]\cup[2,\infty]$, and also to prove the reverse
inequalities for $p\notin[0,1]\cup[2,\infty]$ :

$1\leq\left(1+\frac{2^{2/p}\,\|\alpha(1-\alpha)\|_{p/2}}{\left(\,\|\alpha\|_{p}^{p}+\|1-\alpha\|_{p}^{p}\,\right)^{2/p}}\right)^{p-1}\left(\,\|\alpha\|_{p}^{p}+\|1-\alpha\|_{p}^{p}\,\right)$
(2.1)

for a single function $0\leq\alpha\leq 1$ on a _probability space_ , i.e.,
$\int 1=1$.

##  3\. Part B. Reduction to a constant function

In this section we prove the following.

######  Proposition 3.1.

If $p\in[0,1]\cup[2,\infty)$, then inequality (2.1) is true for all functions
$\alpha$ (which is equivalent to (1.2) for all $f,g$) if and only if it is
true for all constant functions, that is, for all numbers $\alpha\in[0,1]$,

$1\leq\left(1+\left(\frac{2\alpha^{p/2}(1-\alpha)^{p/2}}{\alpha^{p}+(1-\alpha)^{p}}\right)^{2/p}\right)^{p-1}\left(\alpha^{p}+(1-\alpha)^{p}\right).$
(3.1)

If $p\notin[0,1]\cup[2,\infty)$, then the reverse of inequality (2.1) is true
for all functions $\alpha$ (which is equivalent to the reverse of (1.2) for
all $f,g$) if and only if it is true for all constant functions, that is, for
all numbers $\alpha\in[0,1]$, the reverse of (3.1) holds.

Moreover, for $p\neq 0,1,2$, there is equality in (2.1) if and only if
$\max\\{\alpha(x),1-\alpha(x)\\}$ is constant almost everywhere.

To prove this Proposition we need a definition and a lemma.

######  Definition 3.2.

Fix $p\in\mathbb{R}$ and for $0\leq a\leq 1$, let $h(a):=a^{p/2}(1-a)^{p/2}$
and let $b(a):=a^{p}+(1-a)^{p}$. Clearly, $b$ determines the unordered pair
$a$ and $1-a$ and, therefore, $b$ determines $h$. Thus, we can consider the
function $b\mapsto H(b):=h(a^{-1}(b))$ (in which the dependence on $p$ is
suppressed in the notation).

######  Lemma 3.3 (convex/concave $H$).

The function $b\mapsto H(b)$ is strictly convex when $p\in(2,\infty)$ and
strictly concave when $p\in(-\infty,2)$, $p\neq 0,1$.

###### Proof.

To prove this lemma we use the chain rule to compute the second derivative of
$H$. As a first step we define a useful reparametrization as follows:
$e^{2x}:=a/(1-a)$. A quick computation shows that $h=(2\cosh x)^{-p}$ and
$b=2\cosh(px)(2\cosh x)^{-p}$. Thus, $h=b/(2\cosh(px))$. By symmetry, we can
restrict our attention to the half-line $x\geq 0$.

We now compute the first two derivatives:

$\displaystyle db/dx$ $\displaystyle=$ $\displaystyle
2^{1-p}p\frac{\sinh((p-1)x)}{(\cosh x)^{p+1}}$ (3.2) $\displaystyle dh/dx$
$\displaystyle=$ $\displaystyle-p\frac{\tanh x}{(\,2\cosh x\,)^{p}}$ (3.3)
$\displaystyle(dH/db)(x)$ $\displaystyle=$
$\displaystyle\frac{dh/dx}{db/dx}=-\frac{\sinh x}{2\sinh((p-1)x)}$ (3.4)
$\displaystyle(d/dx)(dH/db)(x)$ $\displaystyle=$
$\displaystyle\cosh(x)\frac{(p-1)\tanh x-\tanh((p-1)x)}{2\sinh((p-1)x)\
\tanh((p-1)x)}$ (3.5) $\displaystyle(d^{2}H/db^{2})(x)$ $\displaystyle=$
$\displaystyle\frac{(d/dx)(dH/db)(x)}{db/dx}$ (3.6)

Our goal is to show that (3.6) has the correct sign (depending on $p$) for all
$x\geq 0$.

Clearly, the quantity (3.2) is nonpositive for $p\in(0,1]$ and nonnegative
elsewhere. We claim that the quantity (3.5) is nonpositive for
$p\in(-\infty,0]\cup[1,2]$ and nonnegative elsewhere. In fact, the denominator
is always positive. For the numerator we write $t=p-1$ and use the fact that
for all $x>0$, $t\tanh(x)-\tanh(tx)>0$ for $t>1$ and for $-1<t<0$, while the
inequality reverses, and is strict for other values of $t$ except
$t\in\\{-1,0,1\\}$.

To see this, fix $x>0$, and define $f(t):=t\tanh(x)-\tanh(tx)$. Evidently
$f(t)=0$ for $t=-1,0,1$. Then, since
$f^{\prime\prime}(t)=2x^{2}\sinh(tx)/\cosh^{3}(tx)$, $f^{\prime\prime}(t)>0$
for $t>0$, and $f^{\prime\prime}(t)<0$ for $t<0$. It follows that $f(t)>0$ for
$-1<t<0$ and $t>1$, while $f(t)<0$ for $0<t<1$ and $t<-1$.

According to (3.6) the products of the signs of (3.2) and (3.5) yield the
strict convexity/concavity properties of $H(b)$ shown in rows $2$ to $4$ of
the table below. ∎

$p<0$$0<p<1$$1<p<2$$p>2$$\tfrac{db}{dx}$$\geq 0$$\leq 0$$\geq 0$$\geq
0$$\tfrac{d}{dx}\left(\tfrac{dH}{db}\right)$$\leq 0$$\geq 0$$\leq 0$$\geq
0$$H(b)$concaveconcaveconcaveconvex$p(p-1)$$\geq 0$$\leq 0$$\geq 0$$\geq
0$Direction$\geq$$\leq$$\geq$$\leq$

Fig. 6: Table of signs determining the direction of the main inequality (1.4).

######  Lemma 3.4 (constant $b(\alpha(x))$).

For $f,g\geq 0$, define $\alpha(x)=f(x)/(f(x)+g(x))$. Then with $b(\alpha(x))$
is almost everywhere constant if $\max\\{\alpha(x),1-\alpha(x)\\}$ is constant
almost everywhere, which is true if and only if only if either $f$ and $g$
have essentially disjoint support, or else $\tilde{f}(x):=\max\\{f(x),g(x)\\}$
and $\tilde{g}(x):=\min\\{f(x),g(x)\\}$ are proportional.

###### Proof.

Let $F=\\{x:\ f(x)>0\\}$ and $G=\\{x:\ g(x)>0\\}$. Then $\alpha(x)=1$ on
$F\backslash G$, and $1-\alpha(x)=1$ on $G\backslash F$. If the measure of
$F\cap G$ is zero, $b(\alpha(x))=1$ almost everywhere with respect to
$(f(x)+g(x)){\rm d}x$. Conversely, since $b(a)=1$ if and only if
$a\in\\{0,1\\}$, if $b(\alpha(x))=1$ almost everywhere, then almost everywhere
$\alpha(x)\in\\{0,1\\}$, which means that $f$ and $g$ have essentially
disjoint supports.

For $b\in[2^{1-p},1)$, there is a unique $a\in[1/2,1)$ such that $b(a)=b$.
Therefore, if $b(\alpha(x))=b\in[2^{1-p},1)$, there is a unique $a\in[1/2,1)$
such that $\alpha(x)\in\\{a,1-a\\}$ almost everywhere, and this is the case if
and only if
$\max\\{\alpha(x),1-\alpha(x)\\}=\tilde{f}(x)/(\tilde{f}(x)+\tilde{g}(x))=a$
almost everywhere. ∎

###### Proof of Proposition 3.1.

Consider the ratio in (2.1). The numerator is the integral $\int
H(b(\alpha(x)))$. By Jensen’s inequality (recalling that $\int 1=1$) and the
convexity/concavity of $H$ in Lemma 3.3, this integral is bounded from below
by $H(B)$ in the convex case and from above in the concave case, where

$B:=\int(\alpha^{p}+(1-\alpha)^{p})\ .$ (3.7)

That is,

$\frac{\int H(b(y))}{B}\geq\frac{H(B)}{B}$ (3.8)

for $p\geq 2$, while the reverse is true for $p\leq 2$. Moreover, by the
strict convexity/concavity of $H(b)$, the inequality in (3.8) is strict unless
$b(\alpha(x))$ is constant when $p\notin\\{0,1,2\\}$. By the first part of
Lemma 3.4, $b(\alpha(x))$ is constant if and only if
$\max\\{\alpha(x),1-\alpha(x)\\}$ is a constant, necessarily belonging to
$[1/2,1]$. Then, taking into account the signs of $2/p$ and $p-1$ in the
various ranges,

$\left(1+\left(\frac{2\int H(b(y)){\rm
d}y}{B}\right)^{2/p}\right)^{p-1}\geq\left(1+\left(\frac{2H(B)}{B}\right)^{2/p}\right)^{p-1}$

for $p\in(0,1]\cup[2,\infty)$, with the reverse in equality for
$p\in(-\infty,0)\cup[1,2]$. The last two rows in Fig. 6 summarize the
interaction of the convexity/concavity properties of $H(b)$ and the signs of
the exponents $p/2$ and $p-1$ in the direction of the inequality in (3.9) for
the different ranges of $p$, and taking into account the cases of equality
discussed above, this yields the result as stated.

Thus, it suffices for us to prove

$1\leq\left(1+\left(\frac{2H(B)}{B}\right)^{2/p}\right)^{p-1}B$ (3.9)

for $p\in[0,1]\cup[2,\infty]$ and the reverse inequality for
$p\notin[0,1]\cup[2,\infty]$. We do not know what the number $B$ is, but that
does not matter. In each case the range of $b$ is an interval and, therefore,
the average value $B$ lies in this same interval. Consequently, whatever $B$
might be, there is a number $\alpha$ such that $B=\alpha^{p}+(1-\alpha)^{p}$.
(Note that it is not claimed that this number $\alpha$ is related in any
particular way to the function $\alpha(x)$.) ∎

###### Proof of Theorem 1.1.

This is immediate from Proposition 3.1 and Theorem 1.3. ∎

##  4\. Part C. Proof of Theorem 1.3

###  4.1. Proof of the inequality

First we prove the inequality

$\displaystyle(\alpha^{p}+(1-\alpha)^{p})\left(1+\left(\frac{2\alpha^{p/2}(1-\alpha)^{p/2}}{\alpha^{p}+(1-\alpha)^{p}}\right)^{2/p}\right)^{p-1}\geq
1\qquad\text{for all}\ \alpha\in(0,1)$ (4.1)

if $p\in[0,1]\cup[2,\infty)$, and the reverse inequality if
$p\in(-\infty,0]\cup[1,2]$.

For $p>0$, there is evidently equality for $\alpha\in\\{0,\tfrac{1}{2},1\\}$,
and for $p<0$, there is equality for $\alpha=1/2$. Thus for the proof of (4.1)
it suffices to consider $\alpha\in(1/2,1)$ for $p>0$, and $\alpha\in(0,1/2)$
if $p<0$, and it is convenient to change variables

$t:=\left(\frac{1-\alpha}{\alpha}\right)^{p}\in(0,1]\qquad\text{and}\qquad
c:=1/p\,.$

Moreover, for fixed $c$ we introduce the function

$f(t):=-\frac{1}{c}\ln(1+t^{c})+\ln(1+t)+\frac{1-c}{c}\ln\left(1+\left(\frac{4t}{(t+1)^{2}}\right)^{c}\right).$

By taking logarithms we see that the claimed inequality (4.1) is equivalent to

$f(t)\geq 0\qquad\text{for}\ t\in(0,1)$

if $p\in[0,1]\cup[2,\infty)$ (that is, $c\in(0,1/2]\cup[1,\infty)$), and the
reverse inequality in (4.1) is equivalent to the reverse inequality if
$p\in(-\infty,0]\cup[1,2]$ (that is, $c\in(-\infty,0]\cup[1/2,1]$). We shall
show that for $c>0$ the derivative $f^{\prime}$ has a unique sign change in
$(0,1)$ and it changes sign from $+$ to $-$ if $c\in(0,1/2)\cup(1,\infty)$ and
from $-$ to $+$ if $c\in(1/2,1)$. Moreover, for $c<0$ we shall show that the
derivative $f^{\prime}$ is positive on $(0,1)$.

Since $f(0)=f(1)=0$ for $c>0$, this proves that $f\geq 0$ if
$c\in(0,1/2)\cup(1,\infty)$ and that $f\leq 0$ if $c\in(1/2,1)$. Moreover,
since $f(1)=0$ for $c<0$, this proves that $f\leq 0$ if $c<0$. Thus, we have
reduced the proof of Theorem 1.3 to proving the above sign change properties
of $f^{\prime}$.

In order to discuss the sign changes of $f^{\prime}$ we compute

$\displaystyle
f^{\prime}(t)=\frac{(1-c)(1-t)}{t(1+t)}\left(\frac{1}{\left(\frac{(1+t)^{2}}{4t}\right)^{c}+1}-\frac{t^{c}-t}{(1-c)(t^{c}+1)(1-t)}\right).$
(4.2)

Clearly, it suffices to consider the sign changes of the second factor and
therefore to consider the sign changes of

$\displaystyle
g(t):=\left(\frac{(1+t)^{2}}{4t}\right)^{c}-\left(\frac{(1-c)(t^{c}+1)(1-t)}{t^{c}-t}-1\right).$
(4.3)

We shall show that for $c>0$, $g$ has a unique sign change in $(0,1)$ and it
changes sign from $-$ to $+$ if $c\in(0,1/2)$ and from $+$ to $-$ if
$c\in(1/2,\infty)$. Moreover, for $c<0$ we shall show that $g$ is negative on
$(0,1)$. Clearly, these properties of $g$ imply the claimed properties of
$f^{\prime}$ and therefore will conclude the proof.

We next observe that the second term in (4.3) is positive.

######  Lemma 4.1.

For any $c\in\mathbb{R}\setminus\\{1\\}$ and $t\in(0,1)$,

$\frac{(1-c)(t^{c}+1)(1-t)}{t^{c}-t}>1\,.$

###### Proof.

First, consider the case $c\in[0,1)$. Then concavity of the map $t\mapsto
t^{c}$ implies $1-c+ct-t^{c}\geq 0$, therefore $\frac{(1-c)(1-t)}{t^{c}-t}\geq
1$, and the claim follows from $t^{c}+1>1$.

Next, for $c>1$ the argument is similar using convexity of the map $t\mapsto
t^{c}$.

Finally, for $c<0$ convexity of $t\mapsto t^{1-c}$ implies that

$\displaystyle\frac{(1-c)(t^{c}+1)(1-t)}{t^{c}-t}-1=\frac{(1-c)(1+t^{-c})(1-t)}{1-t^{1-c}}-1>\frac{(1-c)(1-t)}{1-t^{1-c}}-1\geq
0\,.$

This concludes the proof of the lemma. ∎

Because of Lemma 4.1, we can define

$\displaystyle
h(t):=c\ln\left(\frac{(1+t)^{2}}{4t}\right)-\ln\left(\frac{(1-c)(t^{c}+1)(1-t)}{t^{c}-t}-1\right).$
(4.4)

We shall show that for $c>0$, $h$ has a unique sign change in $(0,1)$ and it
changes sign from $-$ to $+$ if $c\in(0,1/2)$ and from $+$ to $-$ if
$c\in(1/2,\infty)$. Moreover, for $c<0$ we shall show that $h$ is negative on
$(0,1)$. Clearly, these properties of $h$ imply the claimed properties of $g$
and therefore will conclude the proof.

We will prove this by investigating sign changes of $h^{\prime}$. Namely, we
shall show that for $c>0$, $h^{\prime}$ has a unique sign change in $(0,1)$
and it changes sign from $+$ to $-$ if $c\in(0,1/2)$ and from $-$ to $+$ if
$c\in(1/2,\infty)$. Moreover, for $c<0$ we shall show that $h^{\prime}$ is
positive on $(0,1)$.

Let us show that this implies the claimed properties of $h$. Indeed, an
elementary limiting argument shows that

$h(0)=\begin{cases}-\infty&\text{if}\ c<0\,,\\\ -2c\ln 2-\ln(1-c)&\text{if}\
c\in(0,1)\,,\\\ +\infty&\text{if}\ c>1\,.\end{cases}$

and

$h(1)=0\qquad\text{for all}\ c\,.$

The function $-2c\ln 2-\ln(1-c)$ is convex on $(0,1)$ and vanishes at $c=0$
and $c=1/2$. From this we conclude that

$h(0)<0\ \text{if}\ c<1/2\,,\quad h(0)=0\ \text{if}\ c=1/2\,,\quad h(0)>0\
\text{if}\ c>1/2\,.$

Because of this behavior of $h(0)$ and $h(1)$, the claimed properties of
$h^{\prime}$ imply the claimed properties of $h$.

Therefore in order to complete the proof of Theorem 1.3 we need to discuss the
sign changes of $h^{\prime}$. We compute

$\displaystyle
h^{\prime}(t)=\frac{v(t)}{(1+t)(t^{c}-t)^{2}\left(\frac{(1-c)(t^{c}+1)(1-t)}{t^{c}-t}-1\right)}$

with

$v(t):=t(2c^{2}-1)-t^{2}c^{2}+2c(1-2c)(t^{c}-t^{c+1})+t^{2c}(1-2c^{2})+(t^{1+2c}-1)(1-c)^{2}+t^{2c-1}c^{2}\,.$

We shall show that for $c>0$, $v$ has a unique sign change in $(0,1)$ and it
changes sign from $+$ to $-$ if $c\in(0,1/2)$ and from $-$ to $+$ if
$c\in(1/2,\infty)$. Moreover, for $c<0$ we shall show that $v$ is positive on
$(0,1)$.

Since, by Lemma 4.1 the denominator in the above expression for $h^{\prime}$
is positive, these properties of $v$ clearly imply those of $h^{\prime}$ and
therefore complete the proof of the theorem.

In order to prove the claimed properties of $v$ we shall study the sign
changes of $v^{\prime\prime}$. We shall show that for $c>0$,
$v^{\prime\prime}$ has a unique sign change in $(0,1)$ and it changes sign
from $+$ to $-$ if $c\in(0,1/2)$ and from $-$ to $+$ if $c\in(1/2,\infty)$.
Moreover, for $c<0$ we shall show that $v^{\prime\prime}$ is positive.

Let us now argue that these properties of $v^{\prime\prime}$ indeed imply the
claimed properties of $v$. We compute

$\displaystyle v^{\prime}(t)$
$\displaystyle=2c^{2}-1-2c^{2}t+2c(1-2c)(ct^{c-1}-(c+1)t^{c})+2c(1-2c^{2})t^{2c-1}$
$\displaystyle\quad\quad\quad+(1-c)^{2}(1+2c)t^{2c}+c^{2}(2c-1)t^{2c-2},$
$\displaystyle v^{\prime\prime}(t)$
$\displaystyle=2c\cdot[-c+c(1-2c)(c-1)t^{c-2}-c(1-2c)(c+1)t^{c-1}+(2c-1)(1-2c^{2})t^{2c-2}$
$\displaystyle\quad\quad\quad+(1-c)^{2}(1+2c)t^{2c-1}+c(2c-1)(c-1)t^{2c-3}],$
(4.5)

and finally

$v^{\prime\prime\prime}(t)=2c(1-2c)(c-1)t^{c-3}w(t)$ (4.6)

with

$w(t):=c(c-2)-(c+1)ct-2t^{c}(1-2c^{2})+(1-c)(1+2c)t^{c+1}-c(2c-3)t^{c-1}\,.$

From these formulas we easily infer that

$v(1)=v^{\prime}(1)=v^{\prime\prime}(1)=0\,,\qquad
v^{\prime\prime\prime}(1)=2c(1-2c)(c-1)^{2}\,.$

In particular, $v^{\prime\prime\prime}(1)>0$ if $c\in(0,1/2)$ and
$v^{\prime\prime\prime}(1)<0$ if $c\in(-\infty,0)\cup(1/2,1)\cup(1,\infty)$.
This means that $v$ is convex near $t=1$ if $c\in(-\infty,0)\cup(1/2,\infty)$
and concave near $t=1$ if $c\in(0,1/2)$.

Let us discuss the behavior near $t=0$. If $c<1/2$, then $v(t)$ behaves like
$t^{2c-1}c^{2}$, so $v(0)=+\infty$, and $v^{\prime\prime}(0)>0$. If $c>1/2$,
then $v(0)=-(1-c)^{2}$ and $v^{\prime\prime}(0)<0$.

This behavior of $v$ near $0$ and $1$, together with the claimed sign change
properties of $v^{\prime\prime}$, imply the claimed sign change properties of
$v$ and will therefore complete the proof of Theorem 1.3. This is because, for
example, if $v$ is convex near $t=1$ with $v(1)=v^{\prime}(1)=0$, and $v$ has
a single inflection point $t_{0}\in(0,1)$, then $v$ is positive on
$[t_{0},0)$, and $v$ is concave on $(0,t_{0})$.

Thus, we are left with studying the sign changes of $v^{\prime\prime}$. In
order to do so, we need to distinguish several cases. For $c<1$ we will argue
via the sign changes of $v^{\prime\prime\prime}$, while for $c>1$ we will
argue directly.

_Case $c\in(0,1)$._ We want to show that $v^{\prime\prime}$ changes sign from
$+$ to $-$ if $c\in(0,1/2)$ and from $-$ to $+$ if $c\in(1/2,1)$.

Since $v^{\prime\prime}(0)>0$ if $c\in(0,1/2)$, $v^{\prime\prime}(0)<0$ if
$c\in(1/2,1)$, $v^{\prime\prime}(1)=0$, and $v^{\prime\prime\prime}(1)>0$, it
suffices to show that $v^{\prime\prime\prime}$ changes sign only once on
$(0,1)$. Because of (4.6) this is the same as showing that $w$ changes sign
only once on $(0,1)$. Notice that $w(0)=+\infty$, and $w(1)=c-1<0$. Moreover,

$\displaystyle w^{\prime\prime}(t)=c(1-c)t^{c-3}p(t)$

with

$p(t):=t^{2}(c+1)(1+2c)+2t(1-2c^{2})+2c^{2}-7c+6\,.$

The quadratic polynomial $p$ is positive. Indeed, when $c\in(0,1/2)$ this
follows from the fact that all its coefficients are positive. When
$c\in(1/2,1)$ we observe that the parabola $p$ is minimized on $\mathbb{R}$ at
$t=\frac{2c^{2}-1}{(c+1)(1+2c)}$, and its minimal value is
$\frac{(5-3c^{2})+c(11-8c^{2})}{(c+1)(1+2c)}$, which is positive for
$c\in(1/2,1)$.

The fact that $p$ is positive means that $w$ is convex. Since $w(0)=+\infty$
and $w(1)<0$, we conclude that $w$ has only one root.

_Case $c\in(-\infty,0)$._ We want to show that $v^{\prime\prime}$ is positive.

Since $v^{\prime\prime}(1)=0$, it suffices to show that
$v^{\prime\prime\prime}$ is negative which, by (4.6), is the same as showing
that $w$ is negative. Clearly, $w(0)=-\infty$, $w^{\prime\prime}(0)<0$, and
$w(1)=c-1<0$, and $w^{\prime}(1)=(3c-1)(c-1)>0$, so it suffices to show that
$w^{\prime\prime}<0$ on $(0,1)$. For this it suffices to show that $p>0$ on
$(0,1)$. We have $p(0)>0$, and $p(1)=9-4c>0$. Thus if $(1+c)(1+2c)\leq 0$ we
have proved the claim. Consider the case when $(1+c)(1+2c)>0$. The vertex of
the parabola is $t_{0}=\frac{2c^{2}-1}{(c+1)(1+2c)}$. If $c<-1$ then clearly
$\frac{2c^{2}-1}{(c+1)(1+2c)}>1$. If $c\in(-1/2,0)$, then clearly
$\frac{2c^{2}-1}{(c+1)(1+2c)}<0$.

_Case $c\in(1,\infty)$._ We want to show that $v^{\prime\prime}$ changes sign
from $-$ to $+$.

We begin with the case $c\in(1,2)$. We write (4.1) as
$v^{\prime\prime}(t)=2cq(t)$ with

$\displaystyle
q(t):=-c+c(1-2c)(c-1)t^{c-2}-c(1-2c)(c+1)t^{c-1}+(2c-1)(1-2c^{2})t^{2c-2}$
$\displaystyle+(1-c)^{2}(1+2c)t^{2c-1}+c(2c-1)(c-1)t^{2c-3}.$

Clearly $q(0)=-\infty$ and $q(1)=0$. It is enough to show that $q^{\prime}$
changes sign from $+$ to $-$. We have

$q^{\prime}(t)=t^{2c-4}(2c-1)(c-1)m(t)$

with

$m(t):=c(2-c)t^{1-c}+c(c+1)t^{2-c}+2(1-2c^{2})t+(c-1)(1+2c)t^{2}+c(2c-3)\,.$

We shall show that $m(t)$ changes sign only once from $+$ to $-$. Clearly
$m(0)=+\infty$ and $m^{\prime\prime}(0)>0$. Next, $m(1)=1-c<0$, and
$m^{\prime\prime}(1)=(c-1)(c^{2}+2c+2)>0$. Thus it suffices to show
$m^{\prime\prime}>0$ on $(0,1)$. Since
$m^{\prime\prime}(0)>0,m^{\prime\prime}(1)>0$, then $m^{\prime\prime}>0$ will
follow from $m^{\prime\prime\prime}$ having the constant sign. We have

$\displaystyle m^{\prime\prime\prime}(t)=t^{-c-2}c^{2}(c-1)(c-2)(c+1)(1-t)<0.$

This finishes the case $c\in(1,2)$.

If $c=2$, then $q(t)=(t-1)(5t^{2}-16t+8)$, and we see that it changes sign
only once.

In what follows we assume $c>2$. Let us rewrite (4.1) as
$v^{\prime\prime}(t)=2ct^{2c-3}u(t)$ with

$\displaystyle u(t)$
$\displaystyle:=-ct^{3-2c}+c(1-2c)(c-1)t^{1-c}-c(1-2c)(c+1)t^{2-c}+(2c-1)(1-2c^{2})t$
$\displaystyle\quad\ +(1-c)^{2}(1+2c)t^{2}+c(2c-1)(c-1)\,.$

We need to show that $u$ changes sign only once. We have $u(0)=-\infty$, and
$u^{\prime\prime}(0)<0$. At the point $t=1$, we have $u(1)=0$,
$u^{\prime}(1)=-(2c-1)(c-1)^{2}<0$, $u^{\prime\prime}(1)=-2(2c-1)(c-1)^{2}<0$.
It suffices to show that $u^{\prime\prime}<0$ on $(0,1)$. Since
$u^{\prime\prime}(0)<0,u^{\prime\prime}(1)<0$, the latter claim will follow
from showing that $u^{\prime\prime\prime}$ has a constant sign. We have

$\displaystyle u^{\prime\prime\prime}(t)=t^{-2-c}c(2c-1)(c-1)b(t)$

with

$b(t):=c^{3}-c-tc(c+1)(c-2)+2t^{2-c}(2c-3)\,.$

The factor $b$ has the property that $b(0)=+\infty$, $b(1)=(c+6)(c-1)>0$. On
the other hand,

$b^{\prime}(t)=-c(c+1)(c-2)t^{1-c}\left(t^{c-1}+\frac{2(2c-3)}{c(c+1)}\right)$

is negative, so $b$ is positive.

This concludes the proof of the inequality of Theorem 1.3.

###  4.2. Sharpness of the exponent $2/p$

The sharpnes of the exponent $2/p$ is easily checked using the variables
introduced in Remark 1.7. If one rpelaces the power of $2/p$ in (1.6) and
kames the transforations described there, one is led to the function

$g_{r,p}(s):=\eta^{\frac{1}{p-1}}(s)\left(1+\left(\frac{1-s}{\eta^{\frac{2}{p}}(s)}\right)^{r}\right)-2\
.$ (4.7)

instead of $f_{p}(s)$. A motivation for this reparametrization is that for
fixed $p$, the function on the right hand side of (1.6) is equal to $1$ up to
order ${\mathcal{O}}((\alpha-1/2)^{4})$ at $\alpha=1/2$. In the variable $s$,
the leading term in Taylor expansion in $s$ will be second order, and we
proves the sharpness by an expansion at this point.

###### Proof of the second paragraph of Theorem 1.3.

For fixed $r>0$, define the function $g_{r,p}(s)$ by (4.7). By the arithmetic-
geometric mean inequality, $(1-s)^{p/2}\leq\eta(s)$ for all $p$, and hence
$(1-s)/\eta^{2/p}(s)<1$ for $p>0$, while $(1-s)/\eta^{2/p}(s)>1$ for $p<0$.
Therefore, for fixed $s$ and $p$, $g_{r,p}(s)$ decreases as $r$ increases for
$p>0$, and does the opposite for $p<0$.

A Taylor expansion shows that

$g_{r,p}(s)=p(1-r)s+o(s)\ .$

It follows that $g_{r,p}(s)\geq 0$ on $[0,1]$ is false (near $s=0$) for $p>2$
and $r>1$, and for $p<0$ and $r<1$. Likewise, it follows that $g_{r,p}(s)\leq
0$ on $[0,1]$ is false for $p\in(0,2)$ and $r<1$. Since the exponent $q$ in
(1.7) corresponds to $r(2/p)$, this together with the reamrks leading to
(1.13) justifies the statements referring to $q$ in Theorem 1.3.

Consideration of the argument shows that for $p>0$, $p\neq 1,2$, there is
equality if and only if $\alpha\in\\{0,1/2,1\\}$ and for $p<0$, if and only if
$\alpha=1/2$. ∎

The proof of Theorem 1.3 is now complete. By what has been explained above,
the inequality of Theorem 1.1 is proved. Concerning the cases of equality, we
have seen in Section 3, that for all values of $p$ under consideration, if
there is equality then $\max\\{\alpha(a),1-\alpha(x)\\}$ is constant, and then
by what has just been proved here, and in Lemma 3.4 for $p>0$, this constant
is either $1$, in which case $f$ and $g$ have essentially disjoint support, or
$1/2$ in which case $=g$. For $p<0$, there is equality only in case $f=g$.
Finally, it is evident that there is equality in these cases.

##  5\. Doubling arguments and a generalization to Schatten norms.

###  5.1. Doubling arguments

We begin this section with a simple proof showing that if the inequality (1.4)
is valid for some $p\geq 2$ or some $p<0$, then it is also valid for $2$.
Since the inequality (1.4) holds as an identity for $p=2$, and is simple to
prove for $p=-1$ (see Remark 1.7), this yields a simple proof of infinitely
many cases of the inequality (1.4). The proof is not only simple and elegant;
it applies to certain non-commutative generalizations of (1.4) for which the
reductions in parts A and B of the proof we have just presented are not
applicable, as we discuss.

To introduce the doubling argument we present a direct proof of Theorem 1.1
for $p=4$.

###### Direct proof of Theorem 1.1 for $p=4$.

Suppose $f,g\geq 0$, By homogeneity, we may suppose that
$\|f\|_{4}^{4}+\|g\|_{4}^{4}=2$. Define

$X:=fg\ ,\quad Y:=f^{2}+g^{2}\ ,\quad\alpha:=\|X\|_{2}\quad{\rm
and}\quad\beta:=\|Y\|_{2}\ .$ (5.1)

By the arithmetic-geometric mean inequality, $X\leq\frac{1}{2}Y$, and hence

$\int X^{2}{\rm d}\mu\leq\frac{1}{4}\int Y^{2}{\rm
d}\mu=\frac{1}{4}\int(f^{4}+g^{4}+2f^{2}g^{2}){\rm
d}\mu=\frac{1}{2}+\frac{1}{2}\int X^{2}{\rm d}\mu\ .$

This yields $\alpha\leq 1$ and $\beta\leq 2$. Then $(f+g)^{2}=Y+2X$ and hence

$\|f+g\|_{4}^{2}=\|Y+2X\|_{2}\leq\|Y\|_{2}+2\|X\|_{2}=\beta+2\alpha\ .$ (5.2)

It suffices to prove that $\beta+2\alpha\leq 2^{1/2}(1+\alpha)^{3/2}$. Note
that $\beta^{2}=\int(f^{2}+g^{2})^{2}{\rm d}\mu=2+2\alpha^{2}$, and then since
$\alpha\in[0,1]$. Thus it suffices to show that

$(1+\alpha^{2})^{1/2}\leq(1+\alpha)^{3/2}-2^{1/2}\alpha\quad{\rm for\
all}\quad 0\leq\alpha\leq 1\ .$ (5.3)

Squaring both sides, this is equivalent to
$1+\alpha^{2}\leq(1+\alpha)^{3}+2\alpha^{2}-2^{3/2}\alpha(1+\alpha)^{3/2}$.
This reduces to $2^{3/2}(1+\alpha)^{3/2}\leq 3+4\alpha+\alpha^{2}$. Squaring
both sides again, this reduces to $(\alpha^{2}-1)^{2}\geq 0$, completing the
proof. ∎

What made this proof work is the fact that the inequality holds for $p=2$ – as
an identity, but that is unimportant. Then, using Minkowski’s inequality, as
in (5.2), together with the numerical inequality (5.3) we arrive at the
inequality for $p=4$. This is a first instance of the general doubling
proposition, to be proved next. The inequality (5.3) is s special case of the
general inequality (5.4) proved below.

This strategy can be adapted to give direct proof of the inequality for other
integer values of $p$; e.g., $p=3$. When $p$ is an integer, and $f$ and $g$
are non-negative, one has the binomial expansion of
$(f+g)^{p}=f^{p}+g^{p}+{\rm mixed\ terms}$. Under the assumption that
$\int(f^{p}+g^{p})=2$, one is left with estimating the mixed terms, and one
can use Hölder for this. When $p$ is not an integer, there is no useful
expression for $(f+g)^{p}-f^{p}-g^{p}$.

######  Proposition 5.1 (A “doubling” argument).

Suppose that for some $p\geq 2$, (1.4) is valid for all $f,g\geq 0$. Then
(1.4) is valid with $p$ replaced by $2p$ for all $f,g\geq 0$. Likewise, if for
some $p<0$ the reverse of (1.4) is valid for all $f,g>0$, then the reverse of
(1.4) is valid with $p$ replaced by $2p$ for all $f,g>0$.

The proof of Proposition 5.1 relies on the following lemma.

######  Lemma 5.2.

For $t\in\mathbb{R}$, define $\psi_{t}$ on $[0,\infty)$ by

$\psi_{t}(\alpha)=(1+\alpha)^{1+t}-(1+\alpha^{2})^{t}-2^{t}\alpha\ .$ (5.4)

Then for $t\in[0,1]$, $\psi_{t}(\alpha)\geq 0$ on $[0,\infty)$, while for
$t>1$, $\psi_{t}(\alpha)\leq 0$ on $[0,\infty)$.

###### Proof.

We write
$\psi_{t}(\alpha)=(1+\alpha)^{t}-(1+\alpha^{2})^{t}-(2^{t}-(1+\alpha)^{t})\alpha$.
Therefore,

$\frac{\psi_{t}(\alpha)}{\alpha(1-\alpha)}=\frac{(1+\alpha)^{t}-(1+\alpha^{2})^{t}}{\alpha(1-\alpha)}-\frac{2^{t}-(1+\alpha)^{t}}{1-\alpha}\
.$

Defining $a:=1+\alpha^{2}$, $b:=1+\alpha$ and $c:=2$, and defining
$\varphi(\alpha):=x^{t}$, the right hand side is the same as

$\frac{\varphi(b)-\varphi(a)}{b-a}-\frac{\varphi(c)-\varphi(b)}{c-b}\ .$

For $\alpha\in[0,1)$ we have $a<b<c$ and therefore this quantity is positive
when $\varphi$ is concave, and negative when $\varphi$ is convex. For
$\alpha\in(1,\infty)$ we have $a>b>c$ and therefore this quantity is negative
when $\varphi$ is concave, and positive when $\varphi$ is convex. ∎

###### Proof of Proposition 5.1.

Let $f,g\in L^{2p}$ with $\|f\|_{2p}^{2p}+\|g\|_{2p}^{2p}=2$. Define $X:=fg$
and $Y:=f^{2}+g^{2}$, and $\gamma:=\|X\|_{p}$ and $\beta:=\|Y\|_{p}$. By the
triangle inequality we have

$\|f+g\|_{2p}^{2}=\|Y+2X\|_{p}\begin{cases}\leq\|Y\|_{p}+2\|X\|_{p}=\beta+2\gamma&\text{if}\
p\geq 2\,,\\\ \geq\|Y\|_{p}+2\|X\|_{p}=\beta+2\gamma&\text{if}\
p<0\,.\end{cases}$

(Note that the triangle inequality reverses for $p<0$.) We now use the
assumption that the inequality (1.4) is valid for $p$. Applying the inequality
with exponent $p$ to the functions $f^{2}$ and $g^{2}$, which satisfy
$\|f^{2}\|_{p}^{p}+\|g^{2}\|_{p}^{p}=\|f\|_{2p}^{2p}+\|g\|_{2p}^{2p}=2$, we
obtain for $p\geq 2$,

$\beta^{p}=\|f^{2}+g^{2}\|_{p}^{p}\leq
2\left(1+\|f^{2}g^{2}\|_{p/2}\right)^{p-1}=2\left(1+\gamma^{2}\right)^{p-1}$

and similarly $\beta^{p}\geq 2\left(1+\gamma^{2}\right)^{p-1}$ for $p<0$. To
summarize, we have shown that

$\|f+g\|_{2p}^{2}\begin{cases}\leq
2^{1/p}(1+\gamma^{2})^{1-1/p}+2\gamma&\text{if}\ p\geq 2\,,\\\ \geq
2^{1/p}(1+\gamma^{2})^{1-1/p}+2\gamma&\text{if}\ p<0\,.\end{cases}$

According to Lemma 5.2 (with $t=1-1/p$ and $\alpha=\gamma$) this is bounded
from above for $p\geq 2$ and from below for $p<0$ by
$2^{1/p}(1+\gamma)^{2-1/p}=2^{1/p}(1+\|fg\|_{p})^{2-1/p}$, which is the
claimed inequality. ∎

###  5.2. A generalization to Schatten norms

For $p\in[1,\infty)$, an operator $A$ on some Hilbert space belongs the
Schatten $p$-class ${\mathcal{S}}_{p}$ in case $(A^{*}A)^{p/2}$ is trace
class, and the Schatten $p$ norm on ${\mathcal{S}}_{p}$ is defined by
$\|A\|_{p}=(\operatorname{Tr}[(A^{*}A)^{p/2}])^{1/p}$. One possible non-
commutative analog of (part of) Theorem 1.1 would assert that for positive
$A,B\in{\mathcal{S}}_{p}$, $p>2$.

$\operatorname{Tr}(A+B)^{p}\leq\left(1+\left(\frac{\operatorname{Tr}[B^{p/4}A^{p/2}B^{p/4}]}{\tfrac{1}{2}\|A\|_{p}^{p}+\tfrac{1}{2}\|B\|_{p}^{p}}\right)^{2/p}\right)^{p-1}\operatorname{Tr}\left(\,A^{p}+B^{p}\,\right).$
(5.5)

Note that for $p=2$, (5.5) holds as an identity.

In this setting, it is not clear how to implement analogs of Parts A and B of
our proof for functions. However, the direct proofs sketched at the beginning
of this section do allow us to prove the valididty of (5.5) for all $p=2^{k}$,
$k\in\mathbb{N}$.

######  Theorem 5.3.

If (5.5) is valid for some $p\geq 2$ and all positive
$A,B\in{\mathcal{S}}_{p}$, then it is valid for $2p$ and all
$A,B\in{\mathcal{S}}_{2p}$. In particular, since (5.5) holds as an identity
for $p=2$, it is valid for $p=2^{k}$ for all $k\in\mathbb{N}$.

###### Proof.

Let $A$ and $B$ be positive operators in ${\mathcal{S}}_{2p}$, and assume that
$\|A\|_{2p}^{2p}+\|B\|_{2p}^{2p}=2$, which, by homogeneity, entails no loss of
generality. Define

$X:=\frac{1}{2}(AB+BA)\qquad{\rm and}\qquad Y=A^{2}+B^{2}\ .$

Note that

$\|X\|_{p}\leq\frac{1}{2}(\|AB\|_{p}+\|BA\|_{p})\ .$

By definition, the Lieb–Thirring inequality [7], and cyclicity of the trace,

$\|AB\|_{p}^{p}=\operatorname{Tr}[(BA^{2}B)^{p/2}]\leq\operatorname{Tr}[B^{p/2}A^{p}B^{p/2}]=\operatorname{Tr}[A^{p/2}B^{p}A^{p/2}]\
.$

Define

$\beta:=\|Y\|_{p}\quad{\rm
and}\quad\gamma:=(\operatorname{Tr}[B^{p/2}A^{p}B^{p/2}])^{1/p}\ .$

Therefore,
$\|A+B\|_{2p}^{2}=\|Y+2X\|_{p}\leq\|Y\|_{p}+2\|X\|_{p}\leq\beta+2\gamma$.
Since $\|A^{2}\|_{p}^{p}+\|B^{2}\|_{p}^{p}=2$, we can apply (5.5) to deduce
that

$\beta^{p}=\|A^{2}+B^{2}\|_{p}^{p}\leq
2\left(1+(\operatorname{Tr}[B^{2p/4}A^{p}B^{2p/4}])^{2/p}\right)^{p-1}=2\left(1+\gamma^{2}\right)^{p-1}.$

Altogether

$\|A+B\|_{2p}^{2}\leq 2^{1/p}(1+\gamma^{2})^{1-1/p}+2\gamma$

and, by Lemma 5.2, the right side is bounded above by
$2^{1/p}(1+\gamma)^{2-1/p}$, which proves the inequality, ∎

Acknowledgement We thank Anthony Carbery for useful correspondence.

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