$\lim_{j\to\infty}d_{p}(P_{\theta}[\psi](u_{0}^{j}),P_{\theta}[\psi](u_{1}^{j}))=d_{p}(P_{\theta}[\psi](u_{0}),P_{\theta}[\psi](u_{1}))$.
Hence from the result in the previous paragraph we have

$\displaystyle d_{p}(u_{0},u_{1})$
$\displaystyle=\lim_{j\to\infty}d_{p}(u_{0}^{j},u_{1}^{j})$
$\displaystyle\geq\lim_{j\to\infty}d_{p}(P_{\theta}[\psi](u_{0}^{j}),P_{\theta}[\psi](u_{1}^{j}))$
$\displaystyle=d_{p}(P_{\theta}[\psi](u_{0}),P_{\theta}[\psi](u_{1})).$

This proves Equation (20) in the full generality as desired. ∎

A consequence of this contraction formula is that the approximation formula
for $d_{p}$ on $\mathcal{H}_{\theta}\subset\mathcal{E}^{p}(X,\theta)$ from
potentials in analytic singularity type can be extended to any potentials in
$\mathcal{E}^{p}(X,\theta)$. More precisely, we can prove

######  Theorem 7.4.

Let $\psi_{k}\in\text{PSH}(X,\theta)$ have analytic singularities and
$\psi_{k}\nearrow V_{\theta}$ as described in the beginning of Section 4. If
$u_{0},u_{1}\in\mathcal{E}^{p}(X,\theta)$, then

(23)
$d_{p}(u_{0},u_{1})=\lim_{k\to\infty}d_{p}(P_{\theta}[\psi_{k}](u_{0}),P_{\theta}[\psi_{k}](u_{1})).$

###### Proof.

Equation (23) is the definition of $d_{p}$ when
$u_{0},u_{1}\in\mathcal{H}_{\theta}$. Here we want to prove it more generally.

Let $u_{0}^{j},u_{1}^{j}\in\mathcal{H}_{\theta}$ be such that
$u_{0}^{j}\searrow u_{0}$ and $u_{1}^{j}\searrow u_{1}$. Then by definition of
$d_{p}$ on $\mathcal{E}^{p}(X,\theta)$, we have

$d_{p}(u_{0},u_{1})=\lim_{j\to\infty}d_{p}(u_{0}^{j},u_{1}^{j})$

and

$d_{p}(u_{0}^{j},u_{1}^{j})=\lim_{k\to\infty}d_{p}(P_{\theta}[\psi_{k}](u_{0}^{j}),P_{\theta}[\psi_{k}](u_{1}^{j})).$

Combining the two we get

(24)
$d_{p}(u_{0},u_{1})=\lim_{j\to\infty}\lim_{k\to\infty}d_{p}(P_{\theta}[\psi_{k}](u_{0}^{j}),P_{\theta}[\psi_{k}](u_{1}^{j})).$

We want to exchange the limit. First, observe that as $j\to\infty$,
$P_{\theta}[\psi_{k}](u_{0}^{j})\searrow P_{\theta}[\psi_{k}](u_{0})$ and
$P_{\theta}[\psi_{k}](u_{1}^{j})\searrow P_{\theta}[\psi_{k}](u_{1})$. Thus
from Lemma 3.13,

(25)
$\lim_{j\to\infty}d_{p}(P_{\theta}[\psi_{k}](u_{0}^{j}),P_{\theta}[\psi_{k}](u_{1}^{j}))=d_{p}(P_{\theta}[\psi_{k}(u_{0}),P_{\theta}[\psi_{k}](u_{1})).$

Now we will show that the limit in Equation (25) is uniform in $k$. For that,
we observe by triangle inequality that

$\displaystyle|d_{p}(P_{\theta}[\psi_{k}](u_{0}^{j}),P_{\theta}[\psi_{k}](u_{1}^{j}))-d_{p}(P_{\theta}[\psi_{k}](u_{0}),P_{\theta}[\psi_{k}](u_{1}))|$
$\displaystyle\leq$ $\displaystyle
d_{p}(P_{\theta}[\psi_{k}](u_{0}^{j}),P_{\theta}[\psi_{k}](u_{0}))+d_{p}(P_{\theta}[\psi_{k}](u_{1}^{j}),P_{\theta}[\psi_{k}](u_{1}))$
$\displaystyle\leq$ $\displaystyle
d_{p}(u_{0}^{j},u_{0})+d_{p}(u_{1}^{j},u_{0}),$

where in the last line we used Theorem 7.3. Moreover,
$\lim_{j\to\infty}d_{p}(u_{0}^{j},u_{0})=0$ and
$\lim_{j\to\infty}d_{p}(u_{1}^{j},u_{1})=0$. Using the uniform convergence in
$k$, we obtain that we can exchange the limits in Equation (24). Thus

$\displaystyle d_{p}(u_{0},u_{1})$
$\displaystyle=\lim_{k\to\infty}\lim_{j\to\infty}d_{p}(P_{\theta}[\psi_{k}](u_{0}^{j}),P_{\theta}[\psi_{k}](u_{1}^{j}))$
$\displaystyle=\lim_{k\to\infty}d_{p}(P_{\theta}[\psi_{k}](u_{0}),P_{\theta}[\psi_{k}](u_{1})),$

as desired. ∎

##  8\. Uniform Convexity in the big case

With the help of Theorem 7.4, we can prove uniform convexity in the big case
as well. First, we see that the uniform convexity extends to the analytic
singularity setting as well.

######  Theorem 8.1.

If $\theta$ represents a big cohomology class and
$\psi\in\text{PSH}(X,\theta)$ has analytic singularities, then the metric
space $(\mathcal{E}^{p}(X,\theta,\psi),d_{p})$ for $p>1$, as described in
Section 3, is uniformly convex.

###### Proof.

Recall that we constructed the metric $d_{p}$ on
$\mathcal{E}^{p}(X,\theta,\psi)$ in Section 3.1 by resolving the singularities
of $\psi$. Let $\mu:\tilde{X}\to X$ be the resolution as described in Section
3. Recall that from Theorem 3.1 there is a smooth closed real $(1,1)$-form
$\tilde{\theta}$ on $\tilde{X}$ and a bounded function
$g\in\text{PSH}(\tilde{X},\tilde{\theta})$ such that the map
$\text{PSH}(X,\theta,\psi)\ni
u\mapsto\tilde{u}:=(u-\psi)\circ\mu+g\in\text{PSH}(\tilde{X},\tilde{\theta})$
is an order preserving bijection and from Corollary 3.4,
$\mathcal{E}^{p}(X,\theta,\psi)\ni
u\mapsto\tilde{u}\in\mathcal{E}^{p}(\tilde{X},\tilde{\theta})$ is a bijection
as well.

If $u,v_{0},v_{1}\in\mathcal{E}^{p}(X,\theta,\psi)$ and $v_{\lambda}$ is the
weak geodesic joining $v_{0}$ and $v_{1}$, then
$\tilde{u},\tilde{v}_{0},\tilde{v}_{1}\in\mathcal{E}^{p}(\tilde{X},\tilde{\theta})$
and by Theorem 3.6 $\tilde{v}_{\lambda}:=(v_{\lambda}-\psi)\circ\mu+g$ is the
weak geodesic joining $\tilde{v}_{0}$ and $\tilde{v}_{1}$. From Theorem 6.4,
$(\mathcal{E}^{p}(\tilde{X},\tilde{\theta}),d_{p})$ is uniformly convex, thus

$\displaystyle d_{p}(\tilde{u},\tilde{v}_{\lambda})^{2}$
$\displaystyle\leq(1-\lambda)d_{p}(\tilde{u},\tilde{v}_{0})^{2}+\lambda
d_{p}(\tilde{u},\tilde{v}_{1})^{2}-(p-1)\lambda(1-\lambda)d_{p}(\tilde{v}_{0},\tilde{v}_{1})^{2},\text{
if }1<p\leq 2\text{ and }$ $\displaystyle
d_{p}(\tilde{u},\tilde{v}_{\lambda})^{p}$
$\displaystyle\leq(1-\lambda)d_{p}(\tilde{u},\tilde{v}_{0})^{p}+\lambda
d_{p}(\tilde{u},\tilde{v}_{1})^{p}-\lambda^{p/2}(1-\lambda)^{p/2}d_{p}(\tilde{v}_{0},\tilde{v}_{1})^{p},\text{
if }p\leq 2.$

For $u_{0},u_{1}\in\mathcal{E}^{p}(X,\theta,\psi)$, we defined
$d_{p}(u_{0},u_{1}):=d_{p}(\tilde{u}_{0},\tilde{u}_{1})$ in Equation (6).
Applying this we get

$\displaystyle d_{p}(u,v_{\lambda})^{2}$
$\displaystyle\leq(1-\lambda)d_{p}(u,v_{0})^{2}+\lambda
d_{p}(u,v_{1})^{2}-(p-1)\lambda(1-\lambda)d_{p}(v_{0},v_{1})^{2},\text{ if
}1<p\leq 2\text{ and }$ $\displaystyle d_{p}(u,v_{\lambda})^{p}$
$\displaystyle\leq(1-\lambda)d_{p}(u,v_{0})^{p}+\lambda
d_{p}(u,v_{1})^{p}-\lambda^{p/2}(1-\lambda)^{p/2}d_{p}(v_{0},v_{1})^{p},\text{
if }p\leq 2$

implying uniform convexity of $(\mathcal{E}^{p}(X,\theta,\psi),d_{p})$ for
$p>1$. ∎

We would also need the analytic singularity version of [DL20, Proposition 3.6]
which holds true, because the proof in [DL20] only relies on the uniform
convexity of Theorem 8.1.

######  Theorem 8.2.

Let $\psi\in\text{PSH}(X,\theta)$ have analytic singularities. Let
$u_{0},u_{1}\in\mathcal{E}^{p}(X,\theta,\psi)$ for $p>1$, and $u_{t}$ be the
weak geodesic joining $u_{0}$ and $u_{1}$. If
$v\in\mathcal{E}^{p}(X,\theta,\psi)$ satisfies
$d_{p}(u_{0},v)\leq(t+\varepsilon)d_{p}(u_{0},u_{1})$ and
$d_{p}(u_{1},v)\leq(1-t+\varepsilon)d_{p}(u_{0},u_{1})$, for some
$\varepsilon>0$ and $t\in[0,1]$, then for some constant $C(p)>0$,

$d_{p}(v,u_{t})\leq\varepsilon^{1/r}Cd_{p}(u_{0},u_{1})$

where $r=\max\\{2,p\\}$.

Now we can prove one of our main results:

######  Theorem 8.3.

If $\theta$ represents a big cohomology class, then the metric space
$(\mathcal{E}^{p}(X,\theta),d_{p})$ for $p>1$ is uniformly convex. This means
for $u,v_{0},v_{1}\in\mathcal{E}^{p}(X,\theta)$, if $v_{\lambda}$ is the
geodesic joining $v_{0}$ and $v_{1}$, then

$\displaystyle d_{p}(u,v_{\lambda})^{2}$
$\displaystyle\leq(1-\lambda)d_{p}(u,v_{0})^{2}+\lambda
d_{p}(u,v_{1})^{2}-(p-1)\lambda(1-\lambda)d_{p}(v_{0},v_{1})^{2},\text{ if
}1<p\leq 2\text{ and }$ $\displaystyle d_{p}(u,v_{\lambda})^{p}$
$\displaystyle\leq(1-\lambda)d_{p}(u,v_{0})^{p}+\lambda
d_{p}(u,v_{1})^{p}-\lambda^{p/2}(1-\lambda)^{p/2}d_{p}(v_{0},v_{1})^{p},\text{
if }p\leq 2.$

###### Proof.

Let $\psi_{k}\nearrow V_{\theta}$ be the increasing sequence of $\theta$-psh
functions with analytic singularities. Let $u^{k}=P_{\theta}[\psi_{k}](u)$,
$v_{0}^{k}=P_{\theta}[\psi_{k}](v_{0})$, and
$v_{1}^{k}=P_{\theta}[\psi_{k}](v_{1})$. Let $v_{\lambda}^{k}$ be the weak
geodesic joining $v_{0}^{k}$ and $v_{1}^{k}$. From Theorem 8.1, we know that

$\displaystyle d_{p}(u^{k},v^{k}_{\lambda})^{2}$
$\displaystyle\leq(1-\lambda)d_{p}(u^{k},v^{k}_{0})^{2}+\lambda
d_{p}(u^{k},v^{k}_{1})^{2}-(p-1)\lambda(1-\lambda)d_{p}(v^{k}_{0},v^{k}_{1})^{2},\text{
if }1<p\leq 2\text{ and }$ $\displaystyle d_{p}(u^{k},v^{k}_{\lambda})^{p}$
$\displaystyle\leq(1-\lambda)d_{p}(u^{k},v^{k}_{0})^{p}+\lambda
d_{p}(u^{k},v^{k}_{1})^{p}-\lambda^{p/2}(1-\lambda)^{p/2}d_{p}(v^{k}_{0},v^{k}_{1})^{p},\text{
if }p\leq 2.$

From Theorem 7.4 we know that
$\lim_{k\to\infty}d_{p}(u^{k},v_{0}^{k})=d_{p}(u,v_{0})$,
$\lim_{k\to\infty}d_{p}(u^{k},v_{1}^{k})=d_{p}(u,v_{1})$, and
$d_{p}(v_{0}^{k},v_{1}^{k})=d_{p}(v_{0},v_{1})$. Thus to finish the proof by
taking the limit $k\to\infty$, we need to show that
$d_{p}(u^{k},v_{\lambda}^{k})\to d_{p}(u,v_{\lambda})$. Unfortunately, it may
not be true that $P_{\theta}[\psi_{k}](v_{\lambda})=v_{\lambda}^{k}$. But by
using Theorem 7.3, and Theorem 8.2, we can show that $v_{\lambda}^{k}$ and
$P_{\theta}[\psi_{k}](v_{\lambda})$ are $d_{p}$-close.

From Theorem 7.3, and the fact that
$(\mathcal{E}^{p}(X,\theta,\psi_{k}),d_{p})$ is a geodesic metric space, we
know that

$d_{p}(v_{0}^{k},P_{\theta}[\psi_{k}](v_{\lambda}))\leq
d_{p}(v_{0},v_{\lambda})=\lambda d_{p}(v_{0},v_{1})$

and

$d_{p}(v_{1}^{k},P_{\theta}[\psi_{k}](v_{\lambda}))\leq
d_{p}(v_{1},v_{\lambda})=(1-\lambda)d_{p}(v_{0},v_{1}).$

Again by the contraction theorem $d_{p}(v_{0}^{k},v_{1}^{k})\leq
d_{p}(v_{0},v_{1})$, moreover by Theorem 7.4,
$\lim_{k\to\infty}d_{p}(v_{0}^{k},v_{1}^{k})=d_{p}(v_{0},v_{1})$. Thus we can
write

$\frac{d_{p}(v_{0},v_{1})}{d_{p}(v_{0}^{k},v_{1}^{k})}\leq 1+\varepsilon_{k}$

where $\varepsilon_{k}>0$ and $\varepsilon_{k}\to 0$ as $k\to\infty$. Thus we
have

$d_{p}(v_{0}^{k},P_{\theta}[\psi_{k}](v_{\lambda}))\leq(\lambda+\lambda\varepsilon_{k})d_{p}(v_{0}^{k},v_{1}^{k})\leq(\lambda+\varepsilon_{k})d_{p}(v_{0}^{k},v_{1}^{k})$

and

$d_{p}(v_{1}^{k},P_{\theta}[\psi_{k}](v_{\lambda}))\leq(1-\lambda)(1+\varepsilon_{k})d_{p}(v_{0}^{k},v_{1}^{k})\leq(1-\lambda+\varepsilon_{k})d_{p}(v_{0}^{k},v_{1}^{k}).$

Applying Theorem 8.2 we get that

$d_{p}(v_{\lambda}^{k},P_{\theta}[\psi_{k}](v_{\lambda}))\leq(\varepsilon_{k})^{1/r}Cd_{p}(v_{0}^{k},v_{1}^{k})\leq(\varepsilon_{k})^{1/r}Cd_{p}(v_{0},v_{1}).$

Taking the limit $k\to\infty$ and using that $\varepsilon_{k}\to 0$, we get

(26)
$\lim_{k\to\infty}d_{p}(v_{\lambda}^{k},P_{\theta}[\psi_{k}](v_{\lambda}))=0.$

Now we will show that $d_{p}(u^{k},v_{\lambda}^{k})\to d_{p}(u,v_{\lambda})$
as $k\to\infty$. By applying the triangle inequality twice we get

$\displaystyle|d_{p}(u^{k},v_{\lambda}^{k})-d_{p}(u,v_{\lambda})|$
$\displaystyle\leq|d_{p}(u^{k},v_{\lambda}^{k})-d_{p}(u^{k},P_{\theta}[\psi_{k}](v_{\lambda}))|+|d_{p}(u^{k},P_{\theta}[\psi_{k}](v_{\lambda}))-d_{p}(u,v_{\lambda})|$
$\displaystyle\leq
d_{p}(v_{\lambda}^{k},P_{\theta}[\psi_{k}](v_{\lambda}))+|d_{p}(u^{k},P_{\theta}[\psi_{k}](v_{\lambda}))-d_{p}(u,v_{\lambda})|$

As $k\to\infty$, the first term goes to 0 due to Equation (26), and the second
term goes to 0 due to Theorem 7.4. ∎

The same proofs as in [DL20, Theorem 3.5] gives

######  Corollary 8.4.

In the metric space $(\mathcal{E}^{p}(X,\theta),d_{p})$ for $p>1$, the weak
geodesics are the only metric geodesics.

The same proof as in [DL20, Theorem 3.6] proves that

######  Corollary 8.5.

Let $u,v_{0},v_{1}\in\mathcal{E}^{p}(X,\theta)$ for $p>1$. Let $t\in[0,1]$ and
$\varepsilon>0$ such that $d_{p}(u,v_{0})\leq(t+\varepsilon)d_{p}(v_{0},_{1})$
and $d_{p}(u,v_{1})\leq(1-t+\varepsilon)d_{p}(v_{0},v_{1})$. If $v_{s}$ is the
weak geodesic joining $v_{0}$ and $v_{1}$, then there exists $C(p)>0$ such
that

$d_{p}(u,v_{t})\leq\varepsilon^{\frac{1}{r}}d_{p}(v_{0},v_{1})$

where $r=\max\\{2,p\\}$.

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