$\widehat{\mu}_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$, as
well as the following key estimate

$\lVert\mathcal{K}(\widehat{\mu}_{k}-\mu_{k})\rVert_{Y}\lesssim
r_{j}(\mu_{k})^{1/2}\,.$ (6.11)

This estimate relies on (6.9)-(6.10), and thus inherently requires Assumption
(B5).

  * iv)

In Section 6.2.9 we employ (6.11) to prove the estimate

$r_{j}(\mu_{k+1})\leq r_{j}(\mu_{k}^{s})\leq\zeta r_{j}(\mu_{k})\,,$ (6.12)

for some $\zeta\in[3/4,1)$ and a suitable step size $s\in(0,1)$, where
$\mu_{k}^{s}$ is defined in (6.8). From (6.12) we then obtain the local linear
convergence rate for PDAP, see Theorem 6.10.

  * v)

In Section 6.2.10 we finally prove Theorem 3.8. Thanks to the above analysis,
the proof is a simple consequence of the linear convergence rate for PDAP
established in Theorem 6.10, and of the link between PDAP and FC-GCG granted
by the lifting strategy, see Theorem 5.10.

######  Remark 6.4.

The proof strategy described above is inspired by [52], where the authors
propose a method in the spirit of Algorithm 2 for solving a minimization
problem in the space of measures supported on a compact set in
$\mathbb{R}^{n}$. However, the proofs in our setting often require novel
techniques compared to the ones in [52]. Loosely speaking, this can be
attributed to the fact that the metric space $(\mathcal{B},d_{\mathcal{B}})$
does not posses an obvious geometric structure. For example, in the euclidean
setting of [52], the estimates for the distance between $\bar{\mathcal{A}}$
and $\mathcal{A}_{k}$, or $\widehat{v}_{k}$, respectively, rely on the
convexity of euclidean balls, as well as on the higher order differentiability
of the dual variable. Such strategy does not extend to our setting, where
$d_{\mathcal{B}}$-neighborhoods are generally non-convex and no apparent
differentiable structure is available. The difference between both approaches
shows most prominently in the key estimate for
$\mathcal{K}(\widehat{\mu}_{k}-\mu_{k})$ anticipated in (6.11). In contrast,
the fast convergence result in [52] relies on

$\|\mathcal{K}(\widehat{\mu}_{k}-\mu_{k})\|_{Y}\lesssim(P_{k}(\widehat{v}_{k})-1)^{1/2}\,,$
(6.13)

see [52, Lemma 5.15]. With the notation of the current paper, estimate (6.13)
is obtained from a perturbed quadratic growth condition of the form

$\displaystyle\llangle
P_{k},\widehat{v}_{k}-u\rrangle\geq\frac{\kappa}{2}\,g(u,\widehat{v}_{k})^{2}$

in the vicinity of $\widehat{v}_{k}$, the derivation of which relies on higher
order differentiability of the dual variable. In general though, the quadratic
growth condition in Assumption (B5) is not stable with respect to
perturbations of the dual variable. Thus, such arguments cannot be applied in
our setting.

####  6.2.3. Properties of iterates

Since $\bar{\mu}\neq 0$ and $\lVert\cdot\rVert_{M(\mathcal{B})}$ is weak*
lower semicontinuous, from (6.6) we conclude the existence of $M\in\mathbb{N}$
such that

$\mu_{k}\neq 0\,,\quad\mathcal{A}_{k}\neq\emptyset\,,\quad\text{ for all
}\,k\geq M\,.$ (6.14)

Also, as a consequence of Theorem 6.2

$\\{\mu_{k}\\}_{k}\subset E_{\mu_{0}}\,,\qquad\sup_{k\in\mathbb{N}}\
\lVert\mu_{k}\rVert_{M(\mathcal{B})}\leq M_{0}\,,$ (6.15)

where $E_{\mu_{0}}$ is the weak* compact set in (6.3), and $M_{0}>0$ is an
upper bound on the norm of elements in $E_{\mu_{0}}$. In particular
$\bar{\mu}\in E_{\mu_{0}}$ and $\lVert\bar{\mu}\rVert_{M(\mathcal{B})}\leq
M_{0}$.

####  6.2.4. Properties of dual variables

Recall that $\bar{P}$ and $\bar{p}$ are related by
$\bar{P}(v)=\langle\bar{p},v\rangle$ for all $v\in\mathcal{B}$, thanks to
Proposition 5.5. In particular, Assumption (B2) reads

$\operatorname*{arg\,max}_{v\in\mathcal{B}}\bar{P}(v)=\\{v\in\mathcal{B}\;|\;\bar{P}(v)=\langle\bar{p},v\rangle=1\\}=\bar{\mathcal{A}}\,.$
(6.16)

Similarly, we can reformulate Assumption (B5) and Remark 3.7 in terms of
$\bar{P}$ to obtain

$\displaystyle\bar{P}(v)\leq 1-\sigma\quad\text{ for all
}\,\,v\in\mathcal{B}\setminus\bigcup^{N}_{i=1}\bar{U}_{i}\,,$ (6.17)
$\displaystyle\bar{P}(v)\leq 1-\kappa g(v,\bar{u}_{i})^{2}\quad\text{ for all
}\,\,v\in U_{i}\,,$ (6.18)

where we remind the reader that
$g\colon\operatorname{Ext}(B)\times\operatorname{Ext}(B)\to[0,\infty)$, and
$\kappa,\sigma>0$ are constants. The dual variables $P_{k}$ satisfy the
following optimality conditions. For a proof, see Appendix C.2.1.

######  Proposition 6.5.

Let $M$ be as in (6.14). Then, for all $k\geq M$ we have

$\displaystyle P_{k}=1\,\text{ on
}\,\mathcal{A}_{k}\,,\quad\leavevmode\nobreak\ \llangle
P_{k},\mu_{k}\rrangle=\|\mu_{k}\|_{M(\mathcal{B})}\,,\quad\max_{v\in\mathcal{B}}P_{k}(v)\geq
1\,.$

####  6.2.5. Convergence of dual variables and observations

Due to the strong convexity of $F$ around $\bar{y}$, see Assumption (B1), the
worst-case convergence guarantee of Theorem 6.2 also carries over to the
observations $y_{k}$ and the dual variables $P_{k}$, as stated in the
following proposition. For a proof, we refer the reader to Appendix C.2.2.

######  Proposition 6.6.

There exist $M\in\mathbb{N}$ and $c>0$ such that, for all $k\geq M$, there
holds

$\|y_{k}-\bar{y}\|_{Y}+\|\nabla F(y_{k})-\nabla
F(\bar{y})\|_{Y}+\|P_{k}-\bar{P}\|_{C(\mathcal{B})}\leq
c\sqrt{r_{j}(\mu_{k})}\,.$ (6.19)

In particular, we have $y_{k}\rightarrow\bar{y}$ in $Y$ and
$P_{k}\rightarrow\bar{P}$ in $C(\mathcal{B})$.

####  6.2.6. Asymptotic behavior of $\mathcal{A}_{k}$

In the following we show that active sets $\mathcal{A}_{k}$ cluster around
$\bar{\mathcal{A}}$. Specifically, for $k$ sufficiently large, we have
$\mathcal{A}_{k}\subset\cup_{i=1}^{N}U_{i}$, with $\mathcal{A}_{k}\cap
U_{i}\neq\emptyset$ for all $i=1,\ldots,N$.

######  Proposition 6.7.

There exists $M\in\mathbb{N}$ such that for all $k\geq M$ we have

$\displaystyle P_{k}(v)\leq 1-\sigma/2\quad\text{ for all
}\,\,v\in\mathcal{B}\setminus\bigcup^{N}_{i=1}\bar{U}_{i}\,,$ (6.20)
$\displaystyle\mathcal{A}_{k}\subset\bigcup^{N}_{i=1}U_{i}\,,$ (6.21)

where $\sigma>0$ is the constant in (6.17). Moreover, for all $k\geq M$ and
$i=1,\ldots,N$,

$\lim_{k\to\infty}\mu_{k}(\bar{U}_{i})=\bar{\lambda}_{i}\,,\qquad\mathcal{A}_{k}\cap
U_{i}\neq\emptyset\,.$ (6.22)

###### Proof.

As $P_{k}\to\bar{P}$ uniformly by Proposition 6.6, we deduce the existence of
$M\in\mathbb{N}$ such that, for all $k\geq M$, it holds
$\|P_{k}-\bar{P}\|_{C(\mathcal{B})}\leq\sigma/2$. As a consequence, for all
$v\in\mathcal{B}\setminus\bigcup^{N}_{i=1}\bar{U}_{i}$ and $k$ sufficiently
large we have

$\displaystyle P_{k}(v)$ $\displaystyle=\bar{P}(v)+P_{k}(v)-\bar{P}(v)$
$\displaystyle\leq 1-\sigma+\|P_{k}-\bar{P}\|_{C(\mathcal{B})}\leq
1-\sigma/2\,,$

where we used (6.17). This proves (6.20). Now, recall that $P_{k}=1$ on
$\mathcal{A}_{k}$ by Proposition 6.5. Therefore
$\mathcal{A}_{k}\subset\cup_{i=1}^{N}\bar{U}_{i}$, otherwise (6.20) would
yield a contradiction. Since by construction
$\mathcal{A}_{k}\subset\operatorname{Ext}(B)$ and
$U_{i}:=\bar{U}_{i}\cap\operatorname{Ext}(B)$, we conclude (6.21). Consider
now an arbitrary but fixed index $l$ in $\\{1,\dots,N\\}$. Recall that the
sets $\bar{U}_{i}$ are pairwise disjoint and $d_{\mathcal{B}}$-closed.
Therefore we can apply Urysohn’s lemma to obtain a
$d_{\mathcal{B}}$-continuous function $\varphi_{l}\colon\mathcal{B}\to[0,1]$
such that $\varphi_{l}=1$ in $\bar{U}_{l}$, and $\varphi_{l}=0$ in
$\bar{U}_{i}$, $i\neq l$. Recall that
$\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu$ along the whole
sequence by (6.6). Therefore, since
$\bar{\mu}=\sum_{i=1}^{N}\bar{\lambda}_{i}\delta_{\bar{u}_{i}}$, we get

$\displaystyle\bar{\lambda}_{l}=\llangle\varphi_{l},\bar{\mu}\rrangle=\lim_{k\rightarrow\infty}\llangle\varphi_{l},\mu_{k}\rrangle=\lim_{k\rightarrow\infty}\mu_{k}(\bar{U}_{l})\,,$

where in the last equality we used (6.21). Finally, since
$\bar{\lambda}_{l}>0$, from the above convergence we get
$\mu_{k}(\bar{U}_{l})>0$ for $k$ sufficiently large. Recalling that
$\mathcal{A}_{k}\subset\operatorname{Ext}(B)$ and that
$U_{l}:=\bar{U}_{l}\cap\operatorname{Ext}(B)$, we then infer
$\mu_{k}(U_{l})>0$ for $k$ sufficiently large, concluding $\mathcal{A}_{k}\cap
U_{l}\neq 0$. ∎

####  6.2.7. Distance between $\mathcal{A}_{k}^{+}$ and $\bar{\mathcal{A}}$

In the following proposition we use the previous results to quantify the
distance between the active set $\mathcal{A}_{k}$ and the set of optimal
extremal points $\bar{\mathcal{A}}$, see (6.23) below. We also provide an
estimate for the distance of $\widehat{v}_{k}$ to the closest element in
$\bar{\mathcal{A}}$, see (6.24). We remark that this is the first point in
which the estimates of Assumption (B5) are employed.

######  Proposition 6.8.

There exist $M\in\mathbb{N}$ and a constant $c>0$ such that for all $k\geq M$
and $i=1,\dots,N$ it holds

$\sum_{v\in\mathcal{A}_{k}\cap U_{i}}\mu_{k}(\\{v\\})g(v,\bar{u}_{i})\leq
c\sqrt{r_{j}(\mu_{k})}\,.$ (6.23)

Moreover, for each $k\geq M$ there is an index
$\hat{\imath}_{k}\in\\{1,\dots,N\\}$ such that $\widehat{v}_{k}\in
U_{\hat{\imath}_{k}}$ and

$\displaystyle g(\widehat{v}_{k},\bar{u}_{\hat{\imath}_{k}})\leq
c\sqrt{r_{j}(\mu_{k})}\,.$ (6.24)

###### Proof.

By minimality of $\bar{\mu}$ in ($\mathcal{P}_{M^{+}}$) and convexity of $F$
we obtain

$\displaystyle r_{j}(\mu_{k})$
$\displaystyle=F(\mathcal{K}\mu_{k})-F(\mathcal{K}\bar{\mu})+\left\|\mu_{k}\right\|_{M(\mathcal{B})}-\left\|\bar{\mu}\right\|_{M(\mathcal{B})}$
(6.25) $\displaystyle\geq(\nabla
F(\mathcal{K}\bar{\mu}),\mathcal{K}\mu_{k}-\mathcal{K}\bar{\mu})_{Y}+\left\|\mu_{k}\right\|_{M(\mathcal{B})}-\left\|\bar{\mu}\right\|_{M(\mathcal{B})}$
$\displaystyle=\llangle\bar{P},\bar{\mu}-\mu_{k}\rrangle+\left\|\mu_{k}\right\|_{M(\mathcal{B})}-\left\|\bar{\mu}\right\|_{M(\mathcal{B})}=\|\mu_{k}\|_{M(\mathcal{B})}-\llangle\bar{P},\mu_{k}\rrangle\,,$

where we used the optimality condition (5.11) in the last line. Fix
$i_{0}\in\\{1,\dots,N\\}$ and an arbitrary $k\in\mathbb{N}$ large enough such
that all previous results in this section hold. We will show (6.23) for the
index $i_{0}$. By (6.21) and definition of $\mu_{k}$ we obtain

$\displaystyle\left\|\mu_{k}\right\|_{M(\mathcal{B})}=\sum^{N}_{i=1}\sum_{v\in\mathcal{A}_{k}\cap
U_{i}}\mu_{k}(\\{v\\})\,,$ (6.26)
$\displaystyle\llangle\bar{P},\mu_{k}\rrangle=\sum^{N}_{i=1}\sum_{v\in\mathcal{A}_{k}\cap
U_{i}}\mu_{k}(\\{v\\})\,\bar{P}(v)\,.$ (6.27)

Putting together (6.25)-(6.27), and using (6.18), that is, (B5), we estimate

$\displaystyle r_{j}(\mu_{k})$
$\displaystyle\geq\sum^{N}_{i=1}\sum_{v\in\mathcal{A}_{k}\cap
U_{i}}\mu_{k}(\\{v\\})(1-\bar{P}(v))$
$\displaystyle\geq\kappa\sum^{N}_{i=1}\sum_{v\in\mathcal{A}_{k}\cap
U_{i}}\mu_{k}(\\{v\\})g(v,\bar{u}_{i})^{2}\geq\kappa\sum_{v\in\mathcal{A}_{k}\cap
U_{i_{0}}}\mu_{k}(\\{v\\})g(v,\bar{u}_{i_{0}})^{2}\,.$

Since $\mathcal{A}_{k}\cap U_{i_{0}}\neq\emptyset$ by Proposition 6.7, we have
$\mu_{k}(\bar{U}_{i_{0}})=\sum_{v\in\mathcal{A}_{k}\cap
U_{i_{0}}}\mu_{k}(\\{v\\})>0$. Due to the convexity of $(\cdot)^{2}$, we
conclude

$\displaystyle r_{j}(\mu_{k})$
$\displaystyle\geq\kappa\sum_{v\in\mathcal{A}_{k}\cap
U_{i_{0}}}\mu_{k}(\\{v\\})g(v,\bar{u}_{i_{0}})^{2}$ (6.28)
$\displaystyle\geq\frac{\kappa}{\mu_{k}(\bar{U}_{i_{0}})}\left(\sum_{v\in\mathcal{A}_{k}\cap
U_{i_{0}}}\mu_{k}(\\{v\\})g(v,\bar{u}_{i_{0}})\right)^{2}\,.$

Recall $\mu_{k}(\bar{U}_{i_{0}})\leq M_{0}$ by (6.15). Estimate (6.23) now
follows by (6.28) with $c:=\sqrt{M_{0}/\kappa}$.

We now show (6.24). Note that by Proposition 6.5 we have
$P_{k}(\widehat{v}_{k})=\max_{v\in\mathcal{B}}P_{k}(v)\geq 1$ for all
$k\in\mathbb{N}$ large enough. Therefore
$\widehat{v}_{k}\in\cup_{i=1}^{N}\bar{U}_{i}$ by (6.20). Recalling that
$\widehat{v}_{k}\in\operatorname{Ext}(B)$ and that
$U_{i}:=\bar{U}_{i}\cap\operatorname{Ext}(B)$ are pairwise disjoint, we deduce
that $\widehat{v}_{k}\in U_{\hat{\imath}_{k}}$ for some unique index
$\hat{\imath}_{k}$ in $\\{1,\dots,N\\}$. Utilizing (6.18), i.e. (B5), and the
fact that $\bar{P}(\bar{u}_{\hat{\imath}_{k}})=1$ by (6.16), we estimate

$\displaystyle\kappa\,g(\widehat{v}_{k},\bar{u}_{\hat{\imath}_{k}})^{2}$
$\displaystyle\leq
1-\bar{P}(\widehat{v}_{k})=\bar{P}(\bar{u}_{\hat{\imath}_{k}})-\bar{P}(\widehat{v}_{k})$
(6.29)
$\displaystyle\leq\bar{P}(\bar{u}_{\hat{\imath}_{k}})-\bar{P}(\widehat{v}_{k})+P_{k}(\widehat{v}_{k})-P_{k}(\bar{u}_{\hat{\imath}_{k}})\,,$

where in the last line we used that
$P_{k}(\widehat{v}_{k})=\max_{v\in\mathcal{B}}P_{k}(v)$. Using Assumption (B5)
as well as Proposition 5.5, the right-hand side in (6.29) is further bounded
by

$\displaystyle\bar{P}(\bar{u}_{\hat{\imath}_{k}})-\bar{P}(\widehat{v}_{k})$
$\displaystyle+P_{k}(v_{k})-P_{k}(\bar{u}_{\hat{\imath}_{k}})$ (6.30)
$\displaystyle=\langle\bar{p}-p_{k},\bar{u}_{\hat{\imath}_{k}}-\widehat{v}_{k}\rangle$
$\displaystyle=(\nabla F(\bar{y})-\nabla
F(y_{k}),K(\widehat{v}_{k}-\bar{u}_{\hat{\imath}_{k}}))_{Y}$
$\displaystyle\leq\|\nabla F(\bar{y})-\nabla
F(y_{k})\|_{Y}\,\|K(\widehat{v}_{k}-\bar{u}_{\hat{\imath}_{k}})\|_{Y}$
$\displaystyle\leq\tau\|\nabla F(\bar{y})-\nabla
F(y_{k})\|_{Y}\,g(\widehat{v}_{k},\bar{u}_{\hat{\imath}_{k}})\,,$

where $\tau>0$ is the constant from (3.12), which does not depend on $k$.
Finally, using (6.29) and (6.30) together with Proposition 6.6, we conclude
(6.24). ∎

####  6.2.8. Surrogate sequence

Let $M\in\mathbb{N}$ be sufficiently large so that all the above results hold.
For $k\geq M$ we denote by $\hat{\imath}_{k}\in\\{1,\dots,N\\}$ the index from
Proposition 6.8. Starting from the sequence $\\{\mu_{k}\\}_{k}$ generated by
Algorithm 2, we define the _surrogate sequence_ $\\{\widehat{\mu}_{k}\\}_{k}$
in $M^{+}(\mathcal{B})$ by setting

$\widehat{\mu}_{k}:=\mu_{k}(\bar{U}_{\hat{\imath}_{k}})\delta_{\widehat{v}_{k}}+\sum^{N}_{i=1,\leavevmode\nobreak\
i\neq\hat{\imath}_{k}}\mu_{k}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\bar{U}_{i}\,,$ (6.31)

where $\mu_{k}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\bar{U}_{i}$ is defined as in
Section 5.1. Notice that $\widehat{\mu}_{k}$ is just a local modification of
$\mu_{k}$ around $\widehat{v}_{k}$. In the following lemma we investigate the
properties of $\widehat{\mu}_{k}$. Most importantly, we establish the weak*
convergence of $\widehat{\mu}_{k}$ towards $\bar{\mu}$, and the crucial
estimate (6.11).

######  Lemma 6.9.

For all $k\geq M$ there holds

$\displaystyle\|\mu_{k}\|_{M(\mathcal{B})}=\|\widehat{\mu}_{k}\|_{M(\mathcal{B})}\,,$
(6.32)
$\displaystyle\llangle{P}_{k},\widehat{\mu}_{k}-\mu_{k}\rrangle=\mu_{k}(\bar{U}_{\hat{\imath}_{k}})(P_{k}(\widehat{v}_{k})-1)\,,$
(6.33) $\displaystyle\|\mathcal{K}(\widehat{\mu}_{k}-\mu_{k})\|_{Y}\leq
c\sqrt{r_{j}(\mu_{k})}\,,$ (6.34)

where $c>0$ does not depend on $k$. Moreover, as $k\to\infty$, we have

$\mathcal{K}\widehat{\mu}_{k}\to\mathcal{K}\bar{\mu}\,\,\text{ strongly in
}\,Y\,,\,\,\quad j(\widehat{\mu}_{k})\to
j(\bar{\mu})\,,\,\,\quad\widehat{\mu}_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}\,.$

In particular, $\widehat{\mu}_{k}\in E_{\mu_{0}}$ for all $k\in\mathbb{N}$
large enough, where $E_{\mu_{0}}$ is the set in (6.3).

###### Proof.

Recall that $\widehat{v}_{k}\in U_{\hat{\imath}_{k}}$ by Proposition 6.8.
Using (6.21), the definition of $\widehat{\mu}_{k}$, and the fact that the
sets $U_{i}$ are pairwise disjoint, it is immediate to check that (6.32)
holds. Noting that

$\widehat{\mu}_{k}-\mu_{k}=\mu_{k}(\bar{U}_{\hat{\imath}_{k}})\delta_{\widehat{v}_{k}}-\mu_{k}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\bar{U}_{\hat{\imath}_{k}}\,,$
(6.35)

we also obtain (6.33), since

$\displaystyle\llangle P_{k},\widehat{\mu}_{k}-\mu_{k}\rrangle$
$\displaystyle=\mu_{k}(\bar{U}_{\hat{\imath}_{k}})P_{k}(\widehat{v}_{k})-\int_{\bar{U}_{\hat{\imath}_{k}}}P_{k}(v)\,\mathrm{d}\mu_{k}(v)$
$\displaystyle=\mu_{k}(\bar{U}_{\hat{\imath}_{k}})(P_{k}(\widehat{v}_{k})-1)\,,$

where in the last equality we used that $P_{k}=1$ on
$\mathcal{A}_{k}=\operatorname{supp}\mu_{k}$ by Proposition 6.5. We now show
(6.34). By (6.21) and (6.35) we compute

$\displaystyle\widehat{\mu}_{k}-\mu_{k}=\sum_{v\in\mathcal{A}_{k}\cap{U}_{\hat{\imath}_{k}}}\mu_{k}(\\{v\\})(\delta_{\widehat{v}_{k}}-\delta_{v})\,.$

Note that
$\mathcal{I}(\delta_{\widehat{v}_{k}}-\delta_{v})=\widehat{v}_{k}-v$. Thus
linearity of $\mathcal{K}$, Proposition 5.3 ii), and triangle inequality imply

$\displaystyle\|\mathcal{K}(\widehat{\mu}_{k}-\mu_{k})\|_{Y}\leq\sum_{v\in\mathcal{A}_{k}\cap{U}_{\hat{\imath}_{k}}}\mu_{k}(\\{v\\})\|K(\widehat{v}_{k}-v)\|_{Y}\,.$

Recalling that $\widehat{v}_{k}\in{U}_{\hat{\imath}_{k}}$, by (B5) and (6.23),
(6.24) we estimate

$\displaystyle\sum_{v\in\mathcal{A}_{k}\cap{U}_{\hat{\imath}_{k}}}\mu_{k}(\\{v\\})$
$\displaystyle\|K(\widehat{v}_{k}-v)\|_{Y}$
$\displaystyle\leq\sum_{v\in\mathcal{A}_{k}\cap{U}_{\hat{\imath}_{k}}}\mu_{k}(\\{v\\})(\|K(\widehat{v}_{k}-\bar{u}_{\hat{\imath}_{k}})\|_{Y}+\|K(v-\bar{u}_{\hat{\imath}_{k}})\|_{Y})$
$\displaystyle\leq\tau\sum_{v\in\mathcal{A}_{k}\cap{U}_{\hat{\imath}_{k}}}\mu_{k}(\\{v\\})(g(\widehat{v}_{k},\bar{u}_{\hat{\imath}_{k}})+g(v,\bar{u}_{\hat{\imath}_{k}}))$
$\displaystyle\leq(\mu_{k}({U}_{\hat{\imath}_{k}})+1)\tau
c\,\sqrt{r_{j}(\mu_{k})}\leq(M_{0}+1)\tau c\,\sqrt{r_{j}(\mu_{k})}\,,$

where in the last inequality we used (6.15). This establishes (6.34).

As for the remaining part of the statement, first recall that
$r_{j}(\mu_{k})\to 0$ and
$\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$ along the
whole sequence by (6.6). Since $\mathcal{K}$ is weak*-to-strong continuous,
see Proposition 5.3 iii), we conclude that
$\mathcal{K}\mu_{k}\to\mathcal{K}\bar{\mu}$ strongly in $Y$. Thus,
$\mathcal{K}\widehat{\mu}_{k}\to\mathcal{K}\bar{\mu}$ from (6.34). We now show
that $j(\widehat{\mu}_{k})\to j(\bar{\mu})$. Recalling that
$\|\widehat{\mu}_{k}\|_{M(\mathcal{B})}=\|\mu_{k}\|_{M(\mathcal{B})}$ by
(6.32), we obtain

$\displaystyle 0\leq r_{j}(\widehat{\mu}_{k})$
$\displaystyle=F(\mathcal{K}\widehat{\mu}_{k})+\left\|\mu_{k}\right\|_{M(\mathcal{B})}-j(\bar{\mu})$
(6.36) $\displaystyle\leq
r_{j}(\mu_{k})+|F(\mathcal{K}\widehat{\mu}_{k})-F(\mathcal{K}\mu_{k})|\,.$

We have $r_{j}(\mu_{k})\to 0$ by Theorem 6.2, while
$|F(\mathcal{K}\widehat{\mu}_{k})-F(\mathcal{K}\mu_{k})|\to 0$ since
$\mathcal{K}\widehat{\mu}_{k},\mathcal{K}\mu_{k}\to\mathcal{K}\bar{\mu}$ and
$F$ is continuous. Therefore the right hand side of (6.36) converges to zero,
which implies $r_{j}(\widehat{\mu}_{k})\to 0$, that is,
$j(\widehat{\mu}_{k})\to j(\bar{\mu})$. We are left to show that
$\widehat{\mu}_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$.
Indeed, we have shown that $\\{\widehat{\mu}_{k}\\}_{k}$ is a minimizing
sequence for ($\mathcal{P}_{M^{+}}$). Since $j$ is weak* lower semicontinuous
and has weak* compact sublevels (Theorem 5.4), we infer the existence of a
subsequence $n_{k}$ and of $\hat{\mu}\in M^{+}(\mathcal{B})$ such that
$\mu_{n_{k}}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\hat{\mu}$, with
$\hat{\mu}$ minimizer of $j$. By uniqueness of the minimizer, see Proposition
5.7, we conclude that
$\mu_{n_{k}}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$. Moreover,
since the weak* limit does not depend on the subsequence $n_{k}$, we also
infer $\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$.
Finally, since $\mu_{0}$ is not a minimizer of $j$, we have
$\widehat{\mu}_{k}\in E_{\mu_{0}}$ for $k$ sufficiently large. ∎

####  6.2.9. Linear convergence of PDAP

Finally, we are in position to prove the linear convergence of Algorithm 2.

######  Theorem 6.10.

Assume (A1)-(A3) and (B1)-(B5). Let $\\{\mu_{k}\\}_{k}$ be generated by
Algorithm 2. Then, we have
$\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$ in
$M(\mathcal{B})$. Moreover, there is $\bar{k}\in\mathbb{N}$ and
$\zeta\in[3/4,1)$ such that

$r_{j}(\mu_{k+1})\leq\zeta\,r_{j}(\mu_{k})$ (6.37)

for all $k\geq\bar{k}$. In particular, there is $c>0$ with

$r_{j}(\mu_{k})\leq c\,\zeta^{k}$ (6.38)

for all $k\in\mathbb{N}$ sufficiently large.

###### Proof.

The fact that $\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$
in $M(\mathcal{B})$ along the whole sequence is already established in (6.6).
We have to show the improved convergence rate at (6.37). To this end, for a
fixed $s\in[0,1]$ define

$\mu^{s}_{k}:=\mu_{k}+s(\widehat{\mu}_{k}-\mu_{k})\,,$

with $\widehat{\mu}_{k}$ as in Definition (6.31). We will obtain (6.37) by
estimating the residual of $\mu^{s}_{k}$ and choosing an optimal value of $s$.
We start by noting that $\mathcal{A}_{k}\subset\cup_{i=1}^{N}U_{i}$ by (6.21),
with the sets $U_{i}$ pairwise disjoint. Therefore
$\mu_{k}=\sum_{i=1}^{N}\mu_{k}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\bar{U}_{i}$ and

$\mu^{s}_{k}=\left[\sum_{i=1,i\neq\hat{\imath}_{k}}^{N}\mu_{k}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\bar{U}_{i}\right]+s\mu_{k}(\bar{U}_{\hat{\imath}_{k}})\delta_{\widehat{v}_{k}}+(1-s)\mu_{k}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\bar{U}_{\hat{\imath}_{k}}\,.$

Since $\widehat{v}_{k}\in\bar{U}_{\hat{\imath}_{k}}$, from the above we deduce

$\displaystyle\left\|\mu_{k}^{s}\right\|_{M(\mathcal{B})}=\left\|\mu_{k}\right\|_{M(\mathcal{B})}\,.$
(6.39)

Recall that $\mu_{k+1}$ is optimal in (5.13) by definition of Algorithm 2. As
$\operatorname{supp}\mu^{s}_{k}\subset\mathcal{A}_{k}^{+}$, we infer

$j(\mu_{k+1})\leq j(\mu^{s}_{k})\quad\text{ for all }\,\,s\in[0,1]\,.$ (6.40)

Next, we estimate the residual of $\mu^{s}_{k}$. By (6.39) and regularity of
$F$ we infer

$\displaystyle j(\mu_{k}^{s})-j(\mu_{k})$
$\displaystyle=F(\mathcal{K}\mu_{k}+s(\mathcal{K}\hat{\mu}_{k}-\mathcal{K}\mu_{k}))-F(\mathcal{K}\mu_{k})$
(6.41) $\displaystyle\leq s(\nabla
F(\mathcal{K}\mu_{k}),\mathcal{K}\widehat{\mu}_{k}-\mathcal{K}\mu_{k})_{Y}+R(\mu_{k})$
$\displaystyle=-s\llangle
P_{k},\widehat{\mu}_{k}-\mu_{k}\rrangle+R_{s}(\mu_{k})\,,$

where $R_{s}(\mu_{k})$ is a remainder defined by

$R_{s}(\mu_{k}):=\int_{0}^{s}(\nabla F(\mathcal{K}\mu_{k}^{t})-\nabla
F(\mathcal{K}\mu_{k}),\\\ K\widehat{\mu}_{k}-K\mu_{k})_{Y}\leavevmode\nobreak\
\mathrm{d}t\,.$

In order to estimate $R_{s}(\mu_{k})$, first recall that
$\\{\mu_{k}\\}_{k}\subset E_{\mu_{0}}$ by (6.15). Moreover, by Lemma 6.9, we
have $\widehat{\mu}_{k}\in E_{\mu_{0}}$ for $k$ sufficiently large. Therefore,
as $E_{\mu_{0}}$ is convex, we also have $\mu^{s}_{k}\in E_{\mu_{0}}$ for all
$s\in[0,1]$ and $k$ sufficiently large. Note that the set
$\mathcal{K}E_{\mu_{0}}:=\\{\,\mathcal{K}\mu\;|\;\mu\in E_{\mu_{0}}\,\\}$ is
compact, given that $\mathcal{K}$ is weak*-to-strong continuous, see
Proposition 5.3 iii). Denote by $L_{\mu_{0}}$ the Lipschitz constant of
$\nabla F$ on the set $\mathcal{K}E_{\mu_{0}}$, which exists by Assumption
(A1). By Cauchy-Schwarz we then estimate

$\displaystyle|R_{s}(\mu_{k})|$
$\displaystyle\leq\left\|\mathcal{K}\widehat{\mu}_{k}-\mathcal{K}\mu_{k}\right\|_{Y}\int_{0}^{s}\|\nabla
F(\mathcal{K}\mu_{k}^{t})-\nabla
F(\mathcal{K}\mu_{k})\|_{Y}\leavevmode\nobreak\ \mathrm{d}t$ (6.42)
$\displaystyle\leq
L_{\mu_{0}}\left\|\mathcal{K}\widehat{\mu}_{k}-\mathcal{K}\mu_{k}\right\|_{Y}\,\int_{0}^{s}\|\mathcal{K}\mu_{k}^{t}-\mathcal{K}\mu_{k}\|_{Y}\leavevmode\nobreak\
\mathrm{d}t$
$\displaystyle=s^{2}\,\frac{L_{\mu_{0}}}{2}\left\|\mathcal{K}(\widehat{\mu}_{k}-\mu_{k})\right\|_{Y}^{2}\,.$

By minimality of $\bar{\mu}$ and (6.41)-(6.42), we infer

$\displaystyle r_{j}(\mu^{s}_{k})$ $\displaystyle=j(\mu^{s}_{k})-j(\bar{\mu})$
(6.43) $\displaystyle=j(\mu_{k})-j(\bar{\mu})-s\llangle
P_{k},\widehat{\mu}_{k}-\mu_{k}\rrangle+R_{s}(\mu_{k})$ $\displaystyle\leq
r_{j}(\mu_{k})-s\llangle
P_{k},\widehat{\mu}_{k}-\mu_{k}\rrangle+s^{2}\,\frac{L_{\mu_{0}}}{2}\|\mathcal{K}(\widehat{\mu}_{k}-\mu_{k})\|_{Y}^{2}.$

Recall that by construction
$P_{k}(\widehat{v}_{k})=\max_{v\in\mathcal{B}}P_{k}(v)$. Moreover
$\max_{v\in\mathcal{B}}P_{k}(v)\geq 1$ by Proposition 6.5. Thus
$P_{k}(\widehat{v}_{k})\geq 1$ and we can apply Proposition 6.1 to infer that
$M_{0}\delta_{\widehat{v}_{k}}$ minimizes in (6.4), that is,

$\llangle P_{k},\eta\rrangle-\lVert\eta\rVert_{M(\mathcal{B})}\leq
M_{0}(P_{k}(\widehat{v}_{k})-1)$ (6.44)

for all $\eta\in M^{+}(\mathcal{B})$ with
$\lVert\eta\rVert_{M(\mathcal{B})}\leq M_{0}$. We can now use convexity of $F$
to estimate

$\displaystyle r_{j}(\mu_{k})$
$\displaystyle=F(\mathcal{K}\mu_{k})-F(\mathcal{K}\bar{\mu})+\lVert\mu_{k}\rVert_{M(\mathcal{B})}-\lVert\bar{\mu}\rVert_{M(\mathcal{B})}$
$\displaystyle\leq(\nabla
F(\mathcal{K}\mu_{k}),\mathcal{K}(\mu_{k}-\bar{\mu}))_{Y}+\lVert\mu_{k}\rVert_{M(\mathcal{B})}-\lVert\bar{\mu}\rVert_{M(\mathcal{B})}$
$\displaystyle=\llangle
P_{k},\bar{\mu}-\mu_{k}\rrangle+\lVert\mu_{k}\rVert_{M(\mathcal{B})}-\lVert\bar{\mu}\rVert_{M(\mathcal{B})}$
$\displaystyle=\llangle
P_{k},\bar{\mu}\rrangle-\lVert\bar{\mu}\rVert_{M(\mathcal{B})}\,,$

where in the last equality we used Proposition 6.5. Since
$\|\bar{\mu}\|_{M(\mathcal{B})}\leq M_{0}$, see Section 6.2.3, we can apply
(6.44) with $\eta=\bar{\mu}$ to obtain

$\displaystyle r_{j}(\mu_{k})\leq M_{0}(P_{k}(\widehat{v}_{k})-1)\,.$

By (6.33) we then infer

$\llangle
P_{k},\widehat{\mu}_{k}-\mu_{k}\rrangle=\mu_{k}(\bar{U}_{\hat{\imath}_{k}})(P_{k}(\widehat{v}_{k})-1)\geq\frac{\mu_{k}(\bar{U}_{\hat{\imath}_{k}})}{M_{0}}\,r_{j}(\mu_{k})\,.$
(6.45)

Using (6.45) and (6.34) in (6.43) yields

$r_{j}(\mu_{k}^{s})\leq
r_{j}(\mu_{k})-s\,\frac{\mu_{k}(\bar{U}_{\hat{\imath}_{k}})}{M_{0}}\,r_{j}(\mu_{k})+s^{2}\,\frac{L_{\mu_{0}}c_{2}}{2}\,r_{j}(\mu_{k})\,,$
(6.46)

where $c_{2}>0$ is the square of the constant in (6.34). Define the constant

$\displaystyle
c_{1}:=\min_{i=1,\dots,N}\frac{\bar{\mu}(\bar{U}_{i})}{2M_{0}}=\min_{i=1,\dots,N}\frac{\bar{\lambda}_{i}}{2M_{0}}\,.$

Notice that $c_{1}>0$, since (B4) holds. Moreover, $c_{1}\leq 1/2$ given that
$\lVert\bar{\mu}\rVert_{M(\mathcal{B})}\leq M_{0}$. Invoking (6.22) we
conclude that

$\frac{\mu_{k}(\bar{U}_{\hat{\imath}_{k}})}{M_{0}}\geq\min_{i=1,\ldots,N}\frac{\mu_{k}(\bar{U}_{i})}{M_{0}}\to\min_{i=1,\ldots,N}\frac{\bar{\lambda}_{i}}{M_{0}}=2c_{1}\,,\,\,\,\text{
as }\,\,k\to\infty\,,$

which, together with (6.46) yields

$\displaystyle
r_{j}(\mu^{s}_{k})\leq\left(1-sc_{1}+s^{2}\,\frac{L_{\mu_{0}}c_{2}}{2}\right)r_{j}(\mu_{k})\,,$

for all $k$ sufficiently large and all $s\in[0,1]$. Subtracting $j(\bar{\mu})$
from both sides of (6.40) yields

$r_{j}(\mu_{k+1})\leq\varphi(s)r_{j}(\mu_{k})\,,\qquad\varphi(s):=1-sc_{1}+s^{2}\,\frac{L_{\mu_{0}}c_{2}}{2}\,,$
(6.47)

for all $k$ sufficiently large and all $s\in[0,1]$. It is immediate to check
that

$\min_{s\in[0,1]}\varphi(s)=\begin{cases}1-\displaystyle
c_{1}+\frac{L_{\mu_{0}}c_{2}}{2}&\text{ if }\,c_{1}>L_{\mu_{0}}c_{2}\,,\\\
1-\displaystyle\frac{c_{1}^{2}}{2L_{\mu_{0}}c_{2}}&\text{ if }\,c_{1}\leq
L_{\mu_{0}}c_{2}\,.\\\ \end{cases}$

Notice that $\min_{s\in[0,1]}\varphi(s)\leq\zeta$, where

$\displaystyle\zeta:=1-\frac{c_{1}}{2}\min\left\\{1,\frac{c_{1}}{L_{\mu_{0}}c_{2}}\right\\}\,,\,\,\,\,\,\,\frac{3}{4}\leq\zeta<1\,.$

Thus, from (6.47) we obtain an integer $\bar{k}\in\mathbb{N}$ such that
$r_{j}(\mu_{k+1})\leq\zeta\,r_{j}(\mu_{k})$ for all $k\geq\bar{k}$,
establishing (6.37). In particular, $r_{j}(\mu_{k})\leq
r_{j}(\mu_{\bar{k}})\zeta^{k-\bar{k}}$ for all $k\geq\bar{k}$, and the final
thesis (6.38) follows by setting $c:=r_{j}(\mu_{\bar{k}})\zeta^{-\bar{k}}>0$.
∎

####  6.2.10. Proof of Theorem 3.8

Assume that Algorithm 1 does not converge after finitely many steps and
generates a sequence $\\{u_{k}\\}_{k}$ in $\mathcal{M}$. By Proposition 5.7
both problems ($\mathcal{P}_{\mathcal{M}}$) and ($\mathcal{P}_{M^{+}}$) admit
a unique solution, given by
$\bar{u}=\sum_{i=1}^{N}\bar{\lambda}_{i}\bar{u}_{i}$ and
$\bar{\mu}=\sum_{i=1}^{N}\bar{\lambda}_{i}\delta_{\bar{u}_{i}}$, respectively.
Thanks to Theorem 5.10, there exists a sequence $\\{\mu_{k}\\}_{k}$ generated
by Algorithm 2 with $u_{k}=\mathcal{I}(\mu_{k})$. Invoking Theorem 5.10 as
well as Theorem 6.10 yields $r_{J}(u_{k})\leq r_{j}(\mu_{k})\leq
c\,\zeta^{k}$, for all $k$ sufficiently large, where $c>0$ and
$\zeta\in[3/4,1)$. This shows (3.14). In addition, Theorem 6.10 ensures that
$\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$ along the
whole sequence. Recalling that $\mathcal{I}$ is weak*-to-weak* continuous by
Proposition 5.2 iii), we infer
$u_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mathcal{I}(\bar{\mu})$.
Since $\bar{\mu}$ minimizes in ($\mathcal{P}_{M^{+}}$), by Theorem 5.4 ii), we
infer that $\mathcal{I}(\bar{\mu})$ minimizes in
($\mathcal{P}_{\mathcal{M}}$). As the minimizer is unique, we conclude
$\bar{u}=\mathcal{I}(\bar{\mu})$.

##  7\. Conclusions

We have introduced a fully-corrective generalized conditional gradient method
(FC-GCG) to solve a class of non-smooth minimization problems in Banach
spaces. For such algorithm we provided a global sublinear rate of convergence
under the mild Assumptions (A1)-(A3), as well as a local linear convergence
rate under the Assumptions (B1)-(B5). Several examples of applications were
considered, showing that it is possible to formulate a problem-dependent
natural set of assumptions which is easy to verify and implies Assumptions
(B1)-(B5), thus ensuring linear convergence of our method. We demonstrate
numerically the fast convergence for our first example, showing that, compared
to standard conditional gradient methods (GCG), our algorithm exhibits a
vastly improved rate of convergence. Additionally, we have discussed in
details the computational burden of each presented example, and we have
described viable strategies to solve the (partially) linearized problem (1.1).

As in the euclidean case of [52], we believe that all results in the present
paper are transferable to non-smooth regularizers of the form
$\Phi(\mathcal{G}(\cdot))$ where $\mathcal{G}$ is as in Assumptions (A1)-(A3),
and $\Phi$ is a suitable convex, monotone increasing function, e.g.,
$\Phi(\mathcal{G}(u))=(1/2)\mathcal{G}(u)^{2}$. This generalization was
omitted in the present paper for the sake of readability.

## Acknowledgements

KB and SF gratefully acknowledge support by the Christian Doppler Research
Association (CDG) and Austrian Science Fund (FWF) through the Partnership in
Research project PIR-27 “Mathematical methods for motion-aware medical
imaging” and project P 29192 “Regularization graphs for variational imaging”.
MC is supported by the Royal Society (Newton International Fellowship
NIF\R1\192048 Minimal partitions as a robustness boost for neural network
classifiers). The Institute of Mathematics and Scientific Computing, to which
KB and SF are affiliated, is a member of NAWI Graz
(https://www.nawigraz.at/en/). The authors KB and SF are further members
of/associated with BioTechMed Graz (https://biotechmedgraz.at/en/).

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##  Appendix A Complements to Section 3

In this section we state and prove two results. The first concerns well-
posedness for the minimization problem at (3.1), while the second shows
equivalence, in a suitable sense, of (3.2) and (3.4).

######  Lemma A.1.

Let $p\in\mathcal{C}$ be given. Then, there exists
$\bar{v}\in\operatorname{Ext}(B)$ with

$\displaystyle\langle p,\bar{v}\rangle=\max_{v\in\mathcal{B}}\,\langle
p,v\rangle=\max_{v\in B}\,\langle p,v\rangle\,.$

###### Proof.

In order to show the claimed result we first prove that the maximization
problem

$\max_{v\in B}\,\langle p,v\rangle$ (A.1)

admits a solution $\bar{v}\in\operatorname{Ext}(B)$. This statement is
classical, but for the reader’s convenience we produce a proof. Let $v^{*}$ be
a maximizer of (A.1), which exists since $v\mapsto\langle p,v\rangle$ is weak*
continuous and $B$ is weak* compact. Consider the set $H:=\\{\,v\in
B\;|\;\langle p,v\rangle=\langle p,v^{*}\rangle\,\\}$. Then
$\operatorname{Ext}(H)\neq\emptyset$ by Krein-Milman’s Theorem, since
$H\subset B$ is convex, non-empty, weak* closed and thus weak* compact. Thus,
let $\bar{v}\in\operatorname{Ext}(H)$ be given. We claim that
$\bar{v}\in\operatorname{Ext}(B)$. Indeed, assume that
$\bar{v}=sv_{1}+(1-s)v_{2}$ for $v_{1},v_{2}\in B$, $s\in(0,1)$. Notice that
$v_{1},v_{2}\in H$, otherwise we would have $\langle p,\bar{v}\rangle<\langle
p,v^{*}\rangle$, contradicting the maximality of $\bar{v}$. As
$\bar{v}\in\operatorname{Ext}(H)$ we then have $\bar{v}=v_{1}=v_{2}$, showing
that $\bar{v}\in\operatorname{Ext}(B)$. The claimed statement now follows from
$\operatorname{Ext}(B)\subset\mathcal{B}\subset B$ ending the proof. ∎

######  Proposition A.2.

Let
$\mathcal{A}^{u,+}_{k}=\\{u^{k}_{i}\\}^{N^{+}_{k}}_{i=1}\subset\operatorname{Ext}(B)$
be given. If $\widehat{\lambda}\in\mathbb{R}_{+}^{N^{+}_{k}}$ solves (3.4),
then $\widehat{u}=\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}u^{k}_{i}$ solves
(3.2). Conversely, if $\widehat{u}$ solves (3.2), then there exists a solution
$\widehat{\lambda}\in\mathbb{R}_{+}^{N^{+}_{k}}$ to (3.4) such that
$\widehat{u}=\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}u^{k}_{i}$.

###### Proof.

Let $\widehat{\lambda}\in\mathbb{R}_{+}^{N^{+}_{k}}$ be a solution to (3.4).
Set $\widehat{u}=\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}u^{k}_{i}$ and fix
an arbitrary $u\in\operatorname{cone}(\mathcal{A}^{u,+}_{k})$. Then there
exists $\lambda^{u}\in\mathbb{R}_{+}^{N^{+}_{k}}$ such that
$u=\sum^{N^{+}_{k}}_{i=1}\lambda^{u}_{i}u^{k}_{i}$ with
$\sum^{N^{+}_{k}}_{i=1}\lambda^{u}_{i}=\kappa_{\mathcal{A}^{+}_{k}}(u)$. We
estimate

$\displaystyle F(K\widehat{u})+\kappa_{\mathcal{A}^{+}_{k}}(\widehat{u})$
$\displaystyle\leq
F\left(\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}Ku^{k}_{i}\right)+\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}$
$\displaystyle\leq
F\left(\sum^{N^{+}_{k}}_{i=1}\lambda^{u}_{i}Ku^{k}_{i}\right)+\sum^{N^{+}_{k}}_{i=1}\lambda^{u}_{i}=F(Ku)+\kappa_{\mathcal{A}^{+}_{k}}(u)\,,$

showing that $\widehat{u}$ is a solution to (3.2). Conversely, let
$\widehat{u}$ be a minimizer to (3.2). Since
$\widehat{u}\in\operatorname{cone}(\mathcal{A}^{u,+}_{k})$, there exists
$\widehat{\lambda}\in\mathbb{R}_{+}^{N^{+}_{k}}$ with
$\widehat{u}=\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}u^{k}_{i}$,
$\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}=\kappa_{\mathcal{A}^{u,+}_{k}}(\widehat{u})$.
Now let $\lambda\in\mathbb{R}_{+}^{N^{+}_{k}}$ be given and set
$u^{\lambda}=\sum^{N^{+}_{k}}_{i=1}\lambda_{i}u^{k}_{i}$. From the optimality
of $\widehat{u}$ and the definition of the gauge function we get

$\displaystyle
F\left(\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}Ku^{k}_{i}\right)+\sum^{N^{+}_{k}}_{i=1}\widehat{\lambda}_{i}$
$\displaystyle\leq
F(Ku^{\lambda})+\kappa_{\mathcal{A}^{u,+}_{k}}(u^{\lambda})$
$\displaystyle\leq
F\left(\sum^{N^{+}_{k}}_{i=1}\lambda_{i}Ku^{k}_{i}\right)+\sum^{N^{+}_{k}}_{i=1}\lambda_{i}\,.$

Thus, $\widehat{\lambda}$ is a minimizer of (3.4). ∎

##  Appendix B Complements to Section 4

Here we collect the technical statements and the proofs of Section 4. For sake
of clarity we place the proofs of each example in different subsections.

###  B.1. Sparse source identification (Section 4.1)

The next lemma justifies the well-posedness of the definition of $K_{*}$.

######  Lemma B.1.

Let $\varphi\in L^{2}(\Omega)$ be given and let $z\in
L^{2}(0,T;H^{1}_{0}(\Omega))\cap H^{1}(0,T;H^{-1}(\Omega))$ the associated
unique solution of (4.3). Moreover, let $\Omega_{0}$ be a subdomain with
$\bar{\Omega}_{0}\subset\Omega$. Then, we have $z(0)\in C_{0}(\Omega)\cap
C^{2}(\Omega_{0})$ and

$\displaystyle\|z(0)\|_{\mathcal{C}}+\|z(0)\|_{C^{2}(\Omega_{0})}\leq
c\|\varphi\|_{L^{2}(\Omega)}$

for some $c>0$ independent of $\varphi\in L^{2}(\Omega)$.

###### Proof.

This follows from [50, Lemma 3.1] and the Sobolev embedding $H^{2}(\Omega)\cap
H^{1}_{0}(\Omega)\hookrightarrow C_{0}(\Omega)$. ∎

####  B.1.1. Proof of Lemma 4.5

We start by noting that $\nabla\bar{z}(0)(\bar{x}_{i})=0$ thanks to (C1) and
Proposition 4.2. Now fix $x\in\Omega$. By Taylor’s expansion we infer that

$\displaystyle\beta-|\bar{z}(0)(x)|$
$\displaystyle=|\bar{z}(0)(\bar{x}_{i})|-|\bar{z}(0)(x)|$
$\displaystyle=-(1/2)\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))((x-\bar{x}_{i}),\nabla^{2}\bar{z}(0)(x_{s})(x-\bar{x}_{i}))_{\mathbb{R}^{2}}\,,$

for some $x_{s}=(1-s)\bar{x}_{i}+sx$, $s\in(0,1)$. Due to the continuity of
$\nabla^{2}\bar{z}(0)$, see Lemma B.1, as well as Assumption (C2), there
exists $R>0$ such that $B_{R}(\bar{x}_{i})\subset\Omega$ and

$\displaystyle\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))((x-\bar{x}_{i}),\nabla^{2}\bar{z}(0)(x_{s})(x-\bar{x}_{i}))_{\mathbb{R}^{2}}\leq-(\gamma/2)|x-\bar{x}_{i}|^{2}\,,$

for all $x\in B_{R}(\bar{x}_{i})$. Combining both observations finally yields

$\displaystyle\beta-|\bar{z}(0)(x)|\geq(\gamma/4)|x-\bar{x}_{i}|^{2}\,,$

for all $x\in B_{R}(\bar{x}_{i})$, showing (4.10). Moreover for every $x\in
B_{R}(\bar{x}_{i})$ we readily verify

$\displaystyle\|K(\delta_{x}-\delta_{\bar{x}_{i}})\|_{Y}$
$\displaystyle=\sup_{\|y\|_{Y}\leq
1}(y,K(\delta_{x}-\delta_{\bar{x}_{i}}))_{Y}$
$\displaystyle=\sup_{\|y\|_{Y}\leq 1}[K_{*}y](x)-[K_{*}y](\bar{x}_{i})$
$\displaystyle\leq\sup_{\|y\|_{Y}\leq
1}\|{K_{*}y}\|_{\operatorname{Lip}(B_{R}(\bar{x}_{i}))}|x-\bar{x}_{i}|$
$\displaystyle\leq\|K_{*}\|_{Y,\operatorname{Lip}(B_{R}(\bar{x}_{i}))}|x-\bar{x}_{i}|\,.$

To finish, $R>0$ can be chosen small enough to ensure

$\operatorname{sign}(\bar{z}(0)(x))=\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))\,,$

for all $x\in B_{R}(\bar{x}_{i})$, due to $\bar{z}(0)(\bar{x}_{i})\neq 0$ and
$\bar{z}(0)\in C_{0}(\Omega)$. ∎

####  B.1.2. Proof of Lemma 4.6

First assume that $g(u_{k},\bar{u}_{i})\to 0$. Then, we have
$x_{k}\rightarrow\bar{x}_{i}$ in $\Omega$ as well as
$\sigma_{k}\rightarrow\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))$.
Consequently, for all $p\in C_{0}(\Omega)$ we have

$\displaystyle\lim_{k\to\infty}\ \langle p,u_{k}\rangle$
$\displaystyle=\lim_{k\to\infty}\ \sigma_{k}\beta^{-1}p(x_{k})$
$\displaystyle=\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))\beta^{-1}p(\bar{x}_{i})=\langle
p,\bar{u}_{i}\rangle\,.$

This implies $u_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{u}_{i}$.
For the other direction assume that
$u_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{u}_{i}$. Note that
from every subsequence of $(\sigma_{k},x_{k})$ we can extract a further
convergent subsequence still denoted by $(\sigma_{k},x_{k})$, relabeling the
indices. Let $(\bar{\sigma},\bar{x})\in\\{-1,+1\\}\times\bar{\Omega}$ denote
its limit. As $\bar{x}\in\partial\Omega$ would imply
$u_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}0\neq\bar{u}_{i}$, we
conclude $\bar{x}\in\Omega$. Then, using again the weak* convergence of
$u_{k}$, we get

$\displaystyle\bar{\sigma}\beta^{-1}p(\bar{x})=\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))\beta^{-1}p(\bar{x}_{i})\,,$

for all $p\in C_{0}(\Omega)$, from which we immediately conclude
$\bar{\sigma}=\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))$ and
$\bar{x}=\bar{x}_{i}$. Since the initial subsequence was chosen arbitrary we
get
$(\sigma_{k},x_{k})\rightarrow(\operatorname{sign}(\bar{z}(0)(\bar{x}_{i})),\bar{x}_{i})$
for the whole sequence and thus, $g(u_{k},\bar{u}_{i})\to 0$. Finally, since
$\sigma_{k}\in\\{-1,+1\\}$ for all $k\in\mathbb{N}$ and
$\sigma_{k}\rightarrow\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))$, we
necessarily have $\sigma_{k}=\operatorname{sign}(\bar{z}(0)(\bar{x}_{i}))$ for
all $k\in\mathbb{N}$ sufficiently large. ∎

###  B.2. Rank-one matrix reconstruction by trace regularization (Section 4.2)

####  B.2.1. Proof of Lemma 4.8

Recall that every extremal point $\mathcal{U}$ of $B$ necessarily satisfies
$\operatorname{Tr}(\mathcal{U})=\beta^{-1}$ or
$\operatorname{Tr}(\mathcal{U})=0$. We first focus on those extremal points
with $\operatorname{Tr}(\mathcal{U})=\beta^{-1}$. According to Lidskii’s
theorem, for every $\mathcal{U}\in B$ there holds
$\operatorname{Tr}(\mathcal{U})=\sum_{i\in\mathbb{N}}\sigma^{\mathcal{U}}_{i}$.
From these observations, we already see that every extremal point of $B$ has
at most rank one. Indeed, if $\mathcal{U}\in B$,
$\operatorname{Tr}(\mathcal{U})=\beta^{-1}$, is at least of rank two, i.e.
$\sigma^{\mathcal{U}}_{1}\geq\sigma^{\mathcal{U}}_{2}>0$, then we can
decompose $\mathcal{U}$ as

$\displaystyle\mathcal{U}$
$\displaystyle=(\beta\sigma^{\mathcal{U}}_{1})\beta^{-1}h^{\mathcal{U}}_{1}\otimes
h^{\mathcal{U}}_{1}+\left(\beta\sum_{j\in\mathbb{N},j\neq
1}\sigma^{\mathcal{U}}_{j}\right)\sum_{i\in\mathbb{N},i\neq
1}\left(\beta\sum_{j\in\mathbb{N},j\neq
1}\sigma^{\mathcal{U}}_{j}\right)^{-1}\sigma^{\mathcal{U}}_{i}h^{\mathcal{U}}_{i}\otimes
h^{\mathcal{U}}_{i}$
$\displaystyle=(\beta\sigma^{\mathcal{U}}_{1})\beta^{-1}h^{\mathcal{U}}_{1}\otimes
h^{\mathcal{U}}_{1}+(1-\beta\sigma^{\mathcal{U}}_{1})\sum_{i\in\mathbb{N},i\neq
1}\left(\beta\sum_{j\in\mathbb{N},j\neq
1}\sigma^{\mathcal{U}}_{j}\right)^{-1}\sigma^{\mathcal{U}}_{i}h^{\mathcal{U}}_{i}\otimes
h^{\mathcal{U}}_{i}$
$\displaystyle:=(\beta\sigma^{\mathcal{U}}_{1})\mathcal{U}_{1}+(1-\beta\sigma^{\mathcal{U}}_{1})\mathcal{U}_{2}$

where $(\beta\sigma^{\mathcal{U}}_{1})\in(0,1)$ and the second equality
follows from

$\displaystyle\beta^{-1}=\operatorname{Tr}(\mathcal{U})=\sum_{i\in\mathbb{N}}\sigma^{\mathcal{U}}_{i}.$

Noting that $\mathcal{U}_{1},\mathcal{U}_{2}\in B$, as well as
$\mathcal{U}_{1}\neq\mathcal{U}_{2}$, we conclude that $\mathcal{U}$ is not
extremal. Hence, if $\mathcal{U}\in B$, $\mathcal{G}(\mathcal{U})=1$, is
extremal then we have $\mathcal{U}=\beta^{-1}h\otimes h$, $\|h\|_{H}=1$. It
remains to show that every operator of this form is actually an extremal
point. For this purpose assume that there is $\mathcal{U}=\beta^{-1}h\otimes
h$, $\|h\|_{H}=1$, $s\in(0,1)$ as well as $\mathcal{U}_{1},\mathcal{U}_{2}\in
B$, with $\mathcal{U}=(1-s)\mathcal{U}_{1}+s\mathcal{U}_{2}$. Then we also
have

$\displaystyle
1=(1-s)\sum_{i\in\mathbb{N}}\beta\sigma^{\mathcal{U}_{1}}_{i}(h^{\mathcal{U}_{1}}_{i},h)^{2}_{H}+s\sum_{i\in\mathbb{N}}\beta\sigma^{\mathcal{U}_{2}}_{i}(h^{\mathcal{U}_{2}}_{i},h)^{2}_{H}\,,$
(B.1)

as well as

$\displaystyle\max\left\\{\sum_{i\in\mathbb{N}}\beta\sigma^{\mathcal{U}_{1}}_{i},\sum_{i\in\mathbb{N}}\beta\sigma^{\mathcal{U}_{2}}_{i}\right\\}\leq
1.$ (B.2)

Moreover, we have $(h^{\mathcal{U}_{j}}_{i},h)^{2}_{H}\leq 1$ for all
$i\in\mathbb{N}$, $j=1,2$. Together with (B.1) and (B.2), this yields

$\displaystyle
1=\sum_{i\in\mathbb{N}}\beta\sigma^{\mathcal{U}_{1}}_{i}(h^{\mathcal{U}_{1}}_{i},h)^{2}_{H}=\sum_{i\in\mathbb{N}}\beta\sigma^{\mathcal{U}_{2}}_{i}(h^{\mathcal{U}_{2}}_{i},h)^{2}_{H}.$
(B.3)

We will now prove that these observations imply
$\mathcal{U}_{1}=\beta^{-1}h\otimes h$, the proof for $\mathcal{U}_{2}$
follows by the same steps. Note that $(h^{\mathcal{U}_{1}}_{i},h)^{2}_{H}=1$
holds if and only if we have $h^{\mathcal{U}_{1}}_{i}=h$ or
$h^{\mathcal{U}_{1}}_{i}=-h$. Recalling that
$\\{h^{\mathcal{U}_{1}}_{i}\\}_{i\in\mathbb{N}}$ is an orthonormal set, (B.3)
implies the existence of a unique index $\bar{\imath}$ with

$\displaystyle(h^{\mathcal{U}_{1}}_{\bar{\imath}},h)^{2}_{H}=1\quad\text{as
well as}\leavevmode\nobreak\
(h^{\mathcal{U}_{1}}_{i},h)^{2}_{H}=0,\leavevmode\nobreak\ i\neq\bar{\imath}.$

Together with (B.2) and (B.3), we finally get

$\displaystyle
1=\sum_{i\in\mathbb{N}}\beta\sigma^{\mathcal{U}_{1}}_{i}(h^{\mathcal{U}_{1}}_{i},h)^{2}_{H}=\beta\sigma^{\mathcal{U}_{1}}_{\bar{\imath}}$

as well as $\sigma^{\mathcal{U}_{1}}_{i}=0$, $i\neq\bar{\imath}$.
Consequently, there holds

$\displaystyle\mathcal{U}_{1}=\sigma^{\mathcal{U}_{1}}_{\bar{\imath}}h^{\mathcal{U}_{1}}_{\bar{\imath}}\otimes
h^{\mathcal{U}_{1}}_{\bar{\imath}}=\beta^{-1}h\otimes h.$

Repeating these arguments for $\mathcal{U}_{2}$ implies that $\mathcal{U}$ is
an extremal point.

Finally, $\operatorname{Tr}(\mathcal{U})=0$ holds if and only if
$\mathcal{U}=0$. Assume that $\mathcal{U}=0$ is not extremal, i.e., there are
operators $\mathcal{U}_{1},\mathcal{U}_{2}\in B$,
$\mathcal{U}_{1}\neq\mathcal{U}_{2}$, as well as $s\in(0,1)$ with
$0=(1-s)\mathcal{U}_{1}+s\mathcal{U}_{2}$. Then, without loss of generality,
we have $\sigma^{\mathcal{U}_{1}}_{1}>0$ and thus

$\displaystyle
0=(h^{\mathcal{U}_{1}}_{1},((1-s)\mathcal{U}_{1}+s\mathcal{U}_{2})h^{\mathcal{U}_{1}}_{1})_{H}\geq\sigma^{\mathcal{U}_{1}}_{1}>0$

yielding a contradiction. Hence $0$ is an extremal point and the proof is
finished.

It remains to characterize $\mathcal{B}$. Since $H$ is infinite dimensional,
for every $h\in H$, $\|h\|_{H}\leq 1$, there exists a weakly convergent
sequence $\\{h_{k}\\}_{k}$ with limit $h$ such that $\|h_{k}\|_{H}=1$ for all
$k$. Setting $\mathcal{U}_{k}=\beta^{-1}h_{k}\otimes h_{k}$ and
$\mathcal{U}=\beta^{-1}h\otimes h$, we then also conclude

$\displaystyle\langle
P,\mathcal{U}_{k}\rangle=\beta^{-1}(h_{k},Ph_{k})\rightarrow\beta^{-1}(h,Ph)=\langle
P,\mathcal{U}\rangle$

for all $P\in K(H)$, using compactness of $P$. Hence,
$\mathcal{U}_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}U$ and thus
$\mathcal{U}\in\mathcal{B}$. For the converse inclusion, let
$\mathcal{U}_{k}=\beta^{-1}h_{k}\otimes h_{k}\in B$ with $\|h_{k}\|_{H}=1$
denote a weak* convergent sequence with limit $\mathcal{U}$. Then there exists
a subsequence of $\\{h_{k}\\}_{k}$, denoted by the same index, which converges
weakly to some $h\in H$, $\|h\|_{H}\leq 1$. Repeating previous arguments and
noting the uniqueness of weak* limits, we then arrive at
$\mathcal{\mathcal{U}}=\beta^{-1}h\otimes h$. ∎

####  B.2.2. Proof of Proposition 4.9

The optimality condition in (4.16) follows immediately from Proposition 2.3,
the characterization of the extremal points as well as
$\sigma^{\bar{P}}_{1}=\sup_{\|h\|_{H}=1}(h,\bar{P}h)_{H}$. Now, let
$\bar{\mathcal{U}}$ be a minimizer of (4.13) and assume that
$\sigma^{\bar{P}}_{1}=\beta$. Then we have

$\displaystyle\operatorname{Tr}(\bar{P}\bar{\mathcal{U}})=\sum_{i\in\mathbb{N}}\sigma^{\bar{\mathcal{U}}}_{i}(h^{\bar{\mathcal{U}}}_{i},\bar{P}h^{\bar{\mathcal{U}}}_{i})_{H}=\beta\operatorname{Tr}(\bar{\mathcal{U}})=\beta\sum_{i\in\mathbb{N}}\sigma^{\bar{\mathcal{U}}}_{i},$

as well as
$(h^{\bar{\mathcal{U}}}_{i},\bar{P}h^{\bar{\mathcal{U}}}_{i})_{H}\leq\beta$,
$i\in\mathbb{N}$. Consequently, we conclude that
$\sigma^{\bar{\mathcal{U}}}_{i}>0$ implies
$\bar{P}h^{\bar{\mathcal{U}}}_{i}=\beta h^{\bar{\mathcal{U}}}_{i}$, i.e.,
$h^{\bar{\mathcal{U}}}_{i}$ is an eigenfunction for the leading eigenvalue of
$\bar{P}$. Thus, since $\\{h^{\bar{\mathcal{U}}}_{i}\\}_{i}$ form an ONB, and
possibly after a change of basis, $\bar{\mathcal{U}}$ is of the form (4.17). ∎

####  B.2.3. Proof of Theorem 4.10

The statement in (4.18), the structure of $\bar{\mathcal{U}}$ and its
uniqueness as well as the Lipschitz result on $K$ follow immediately from
Assumption (D1), the assumptions on $K$ and the strict convexity of $F$, see
Assumption 2.2. Let $\mathcal{U}\in B$ with
$\operatorname{Tr}(\mathcal{U})=\beta^{-1}$ and $\mathcal{U}\geq 0$ be
arbitrary but fixed. For the proof of the quadratic growth behavior we follow
similar steps as in the finite dimensional setting, see [42, Lemma 4]. Setting
$\delta=\sigma^{\bar{P}}_{1}-\sigma^{\bar{P}}_{2}>0$, we estimate

$\displaystyle 1-\langle\bar{P},\mathcal{U}\rangle$
$\displaystyle=1-\beta(h^{\bar{P}}_{1},\mathcal{U}h^{\bar{P}}_{1})_{H}-\sum_{i\in\mathbb{N},i>1}\sigma^{\bar{P}}_{i}(h^{\bar{P}}_{i},\mathcal{U}h^{\bar{P}}_{i})_{H}$
$\displaystyle\geq
1-\beta\sum_{i\in\mathbb{N}}(h^{\bar{P}}_{i},\mathcal{U}h^{\bar{P}}_{i})_{H}+\delta\sum_{i\in\mathbb{N},i>1}(h^{\bar{P}}_{i},\mathcal{U}h^{\bar{P}}_{i})_{H}$
$\displaystyle=\delta\left(\beta^{-1}-(h^{\bar{P}}_{1},\mathcal{U}h^{\bar{P}}_{1})_{H}\right)=\delta\beta\left(\beta^{-2}-(\mathcal{U},\bar{\mathcal{U}}_{1})_{\text{HS}}\right)$
$\displaystyle=\delta\beta\left(\frac{1}{2}\operatorname{Tr}(\mathcal{U})^{2}+\frac{1}{2}\operatorname{Tr}(\bar{\mathcal{U}}_{1})^{2}-(\mathcal{U},\bar{\mathcal{U}}_{1})_{\text{HS}}\right)$
$\displaystyle\geq\delta\beta\left(\frac{1}{2}\|\mathcal{U}\|_{\text{HS}}^{2}+\frac{1}{2}\|\bar{\mathcal{U}}_{1}\|_{\text{HS}}^{2}-(\mathcal{U},\bar{\mathcal{U}}_{1})_{\text{HS}}\right)$
$\displaystyle=(\delta\beta/2)\|\mathcal{U}-\bar{\mathcal{U}}_{1}\|^{2}_{\text{HS}},$

where we use $\operatorname{Tr}(\mathcal{U})\geq\|\mathcal{U}\|_{\text{HS}}$
for all $\mathcal{U}\in B$ in the last inequality. This finishes the proof. ∎

###  B.3. Minimum effort problems (Section 4.3)

####  B.3.1. Proof of Lemma 4.12

Without loss of generality assume $\alpha=1$. The characterization (4.22) is
classical [30, page 144]. Moreover the inclusion
$\overline{\operatorname{Ext}(B)}^{*}\subset B$ follows from weak* lower semi-
continuity of the norm. Let us now prove that
$B\subset\overline{\operatorname{Ext}(B)}^{*}$. To this end, set
$R:=(0,1)^{d}$ and $R_{\lambda}:=(0,\sqrt[d]{\lambda})^{d}$ for any
$\lambda\in[0,1]$. Define the map $h^{\lambda}$ as $h^{\lambda}:=1$ in
$R\setminus R_{\lambda}$, $h^{\lambda}:=-1$ in $R_{\lambda}$, and extend it
$R$-periodically to the whole $\mathbb{R}^{d}$. For $k\in\mathbb{N}$,
$x\in\mathbb{R}^{d}$, set $h^{\lambda}_{k}(x):=h^{\lambda}(kx)$. Then

$h^{\lambda}_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\int_{R}h^{\lambda}\,dx=1-2\lambda\,\,\,\text{
weakly* in }\,\,L^{\infty}(A)$ (B.4)

as $k\to\infty$, for all open, bounded and non-empty sets
$A\subset\mathbb{R}^{d}$, see [6, Lemma A.1]. Let
$\\{Q_{n,j}\\}_{j\in\mathbb{N}}$ be a family of pairwise disjoint and open
$d$-cubes of side length $1/n$ covering
$(\mathbb{R}\setminus(\mathbb{Z}/n))^{d}$. Define
$\tilde{Q}_{n,j}:=Q_{n,j}\cap\Omega$ and let $J_{n}$ be the collection of
indices $j\in\mathbb{N}$ such that $\tilde{Q}_{n,j}\neq\emptyset$. Let
$\mathcal{P}_{n}$ be the set of maps $u\in B$ such that $u=c_{n,j}$ in
$\tilde{Q}_{n,j}$ for some $c_{n,j}\in[-1,1]$ and for all $j\in J_{n}$. Fix
$u\in\mathcal{P}_{n}$ and $n\in\mathbb{N}$. For every $j\in J_{n}$,
$k\in\mathbb{N}$ define $u_{k}(x):=h_{k}^{\lambda_{n,j}}(x)$ for all
$x\in\tilde{Q}_{n,j}$, where $\lambda_{n,j}:=(1-c_{n,j})/2$. Therefore
$\\{u_{k}\\}_{k}\subset\operatorname{Ext}(B)$. By (B.4) we have that
$u_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}c_{n,j}$ weakly* in
$L^{\infty}(\tilde{Q}_{n,j})$ as $k\to\infty$, for all $j\in J_{n}$. Thus
$u_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}u$ in $L^{\infty}(\Omega)$.
In particular we have shown that
$\mathcal{P}_{n}\subset\overline{\operatorname{Ext}(B)}^{*}$. Therefore
$\mathcal{P}\subset\overline{\operatorname{Ext}(B)}^{*}$, where
$\mathcal{P}:=\cup_{n\in\mathbb{N}}\,\mathcal{P}_{n}$. As clearly
$\overline{\mathcal{P}}^{*}=B$, the proof is concluded. ∎

####  B.3.2. Proof of Lemma 4.13

The optimality condition in (4.21) follows immediately from Proposition 2.3 by
pointing out

$\displaystyle\int_{\Omega}|\bar{p}(x)|\leavevmode\nobreak\ \mathrm{d}x$
$\displaystyle=\alpha\,\max_{v\in
B}\int_{\Omega}\bar{p}(x)v(x)\leavevmode\nobreak\ \mathrm{d}x$ (B.5)
$\displaystyle=\alpha\,\max_{v\in\operatorname{Ext}(B)}\
\langle\bar{p},v\rangle\leq\alpha.$ (B.6)

This also implies

$\displaystyle\alpha\|\bar{u}\|_{\infty}=\int_{\Omega}\bar{p}(x)\bar{u}(x)\leavevmode\nobreak\
\mathrm{d}x\leq\|\bar{u}\|_{\infty}\int_{\Omega}|\bar{p}(x)|\leavevmode\nobreak\
\mathrm{d}x\leq\alpha\|\bar{u}\|_{\infty},$

hence, equality holds everywhere. If $\bar{u}\neq 0$, this can only be the
case if

$\displaystyle\int_{\Omega}|\bar{p}(x)|\leavevmode\nobreak\
\mathrm{d}x=\alpha\quad\text{and}\quad\frac{\bar{u}(x)}{\|\bar{u}\|_{\infty}}\in\begin{cases}\\{1\\}&\text{if
}\,\bar{p}(x)>0\\\ \\{-1\\}&\text{if }\,\bar{p}(x)<0\\\
[-1,1]&\text{else},\end{cases},$

for a.e. $x\in\Omega$, yielding (4.24).∎

##  Appendix C Complements to Section 5 and Section 6

###  C.1. Proofs of Section 5

In this section we exhibit proofs for Propositions 5.2, 5.3. In order to
define the map $\mathcal{I}$ at Proposition 5.2 we employ the classical
Choquet’s Theorem [51, Page 14], which we recall for reader’s convenience.

######  Theorem C.1 (Choquet).

Let $X$ be a locally convex space and $K\subset X$ be a metrizable compact
convex subset. For each $v_{0}\in K$ there exists a probability measure $\mu$
over $X$ concentrated on $\operatorname{Ext}(K)$ which represents $v_{0}$,
that is,

$T(v_{0})=\int_{X}T(v)\,\mathrm{d}\mu(v)\,,$

for all $T$ in the topological dual of $X$.

We are now ready to prove Proposition 5.2.

####  C.1.1. Proof of Proposition 5.2

Let $\mu\in M^{+}(\mathcal{B})$ and consider the embedding
$f\colon\mathcal{B}\to\mathcal{M}$, $f(v):=v$. We have that $f$ is weak*
$\mu$-measurable, since $v\mapsto\langle p,v\rangle$ is weak* continuous and
hence, $\mu$-measurable for all $p\in\mathcal{C}$. Moreover, $v\mapsto\langle
p,v\rangle$ is $\mu$-integrable for each $p\in\mathcal{C}$, since

$\displaystyle\int_{\mathcal{B}}|\langle p,v\rangle|\leavevmode\nobreak\
\mathrm{d}\mu(v)$
$\displaystyle\leq\left\|p\right\|_{\mathcal{C}}\int_{\mathcal{B}}\left\|v\right\|_{\mathcal{M}}\leavevmode\nobreak\
\mathrm{d}\mu(v)$
$\displaystyle\leq\left\|p\right\|_{\mathcal{C}}\left\|\mu\right\|_{M(\mathcal{B})}\max_{v\in\mathcal{B}}\left\|v\right\|_{\mathcal{M}}\,,$

and the last term is finite, since $\mathcal{B}$ is norm bounded. Therefore,
$f$ is Gelfand integrable over $\mathcal{B}$ (see [1, Thm 11.52]), that is,
there exists a unique $u\in\mathcal{M}$ such that (5.2) holds. We thus set
$\mathcal{I}(\mu):=u$. Recalling that the weak* topology separates points, we
immediately conclude that $\mathcal{I}$ is well defined and

$\mathcal{I}(\lambda_{1}\mu_{1}+\lambda_{2}\mu_{2})=\lambda_{1}\mathcal{I}(\mu_{1})+\lambda_{2}\mathcal{I}(\mu_{2})\,,$

for all $\lambda_{1},\lambda_{2}\in\mathbb{R}_{+}$, $\mu_{1},\mu_{2}\in
M^{+}(\mathcal{B})$. We now show i). Notice that $\mathcal{I}(0)=0$ and the
estimate holds trivially. If $\mu\neq 0$, we can apply Theorem 11.54 in [1] to
obtain

$\frac{1}{\lVert\mu\rVert_{M(\mathcal{B})}}\,\mathcal{I}(\mu)\in\overline{\operatorname{conv}(\mathcal{B})}^{*}\,.$

By recalling that $B$ is weak* compact and convex, we conclude that
$\overline{\operatorname{conv}(\mathcal{B})}^{*}\subseteq B$. Thus, by
1-homogeneity of $\mathcal{G}$ we conclude
$\mathcal{G}(\mathcal{I}(\mu))\leq\lVert\mu\rVert_{M(\mathcal{B})}$ and, in
particular, $\mathcal{I}(\mu)\in\operatorname{dom}\mathcal{G}$.

For the proof of ii), let $u\in\operatorname{dom}\mathcal{G}$ be fixed. If
$\mathcal{G}(u)=0$ then $u=0$, and the measure $\mu=0$ satisfies the
statement. Assume now $\mathcal{G}(u)>0$. Apply Theorem C.1 with
$X=(\mathcal{M},weak^{*})$, $K=B$, $v_{0}=u/\mathcal{G}(u)$ to obtain a
probability measure $\tilde{\mu}$ over $X$ concentrated on
$\operatorname{Ext}(B)$ and such that

$\langle p,u\rangle=\int_{\mathcal{M}}\langle
p,v\rangle\mathcal{G}(u)\leavevmode\nobreak\ \mathrm{d}\tilde{\mu}(v)\,,$
(C.1)

for all $p\in\mathcal{C}$, since $(\mathcal{M},weak^{*})^{*}=\mathcal{C}$. Set
$\mu:=\mathcal{G}(u)\tilde{\mu}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\mathcal{B}$. Since $\mathcal{B}$
is weak* closed, we have that $\mu\in M^{+}(\mathcal{B})$. Moreover,
$\left\|\mu\right\|_{M(\mathcal{B})}=\mathcal{G}(u)$, $\mu$ is concentrated on
$\operatorname{Ext}(B)$ and it represents $u$. As the weak* topology separates
points, from (C.1) we conclude that $\mathcal{I}(\mu)=u$. This shows in
particular that $\mathcal{I}$ is surjective, achieving the proof of ii).

Finally, $\mathcal{I}$ is weak*-to-weak* continuous: indeed if
$\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu$ weakly* in
$M(\mathcal{B})$, then by definition of weak* convergence we have $\llangle
P,\mu_{k}\rrangle\to\llangle P,\mu\rrangle$ for all $P\in C(\mathcal{B})$.
Note that the map $P(v)=\langle p,v\rangle$, $v\in\mathcal{B}$,
$p\in\mathcal{C}$ belongs to $C(\mathcal{B})$. Thus, for all
$p\in\mathcal{C}$,

$\int_{\mathcal{B}}\langle
p,v\rangle\,\mathrm{d}\mu_{k}(v)\to\int_{\mathcal{B}}\langle
p,v\rangle\,\mathrm{d}\mu(v)\,.$

Thanks to (5.3) the above reads
$\mathcal{I}(\mu_{k})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mathcal{I}(\mu)$
weakly* in $\mathcal{M}$, concluding the proof. ∎

Finally, we prove Proposition 5.3.

####  C.1.2. Proof of Proposition 5.3

For $\mu\in M(\mathcal{B})$ define the functional $T_{\mu}\colon
Y\to\mathbb{R}$ by setting

$T_{\mu}(y):=\int_{\mathcal{B}}(Kx,y)_{Y}\,\mathrm{d}\mu(x)\,,$

for all $y\in Y$. Notice that $T_{\mu}$ is well defined, since the map
$v\mapsto(Kv,y)_{Y}=\langle K_{*}y,v\rangle$ is weak* continuous over
$\mathcal{M}$, and hence $\mu$-measurable. It is clear that $T_{\mu}$ is
linear. Moreover, $T_{\mu}$ is continuous. Indeed, it holds

$\displaystyle\left\|T_{\mu}(y)\right\|$
$\displaystyle\leq\int_{\mathcal{B}}|(Kv,y)_{Y}|\,\mathrm{d}|\mu|(v)$
$\displaystyle\leq\left\|K\right\|\left\|y\right\|_{Y}\,\sup_{v\in\mathcal{B}}\,\left\|v\right\|_{\mathcal{M}}\int_{\mathcal{B}}1\,\mathrm{d}|\mu|(v)$
$\displaystyle=C\left\|K\right\|\left\|\mu\right\|_{M(\mathcal{B})}\left\|y\right\|_{Y},$

where we recalled that the constant $C$ is defined at i) and is finite, since
$\mathcal{B}$ is norm bounded. Therefore, by Riesz’s Theorem, there exists a
unique element in $Y$ representing $T_{\mu}$, thus defining a map
$\mathcal{K}\colon M(\mathcal{B})\to Y$. In particular, $\mathcal{K}$ is
linear and satisfies (5.4). Moreover, by (5.4) we have

$\left\|\mathcal{K}\mu\right\|_{Y}=\sup_{\left\|y\right\|_{Y}\leq
1}\left\|T_{\mu}(y)\right\|\leq
C\left\|K\right\|\left\|\mu\right\|_{M(\mathcal{B})},$

by the above, so $\mathcal{K}$ is bounded and i) follows.

As for ii), assume that $\mu\in M^{+}(\mathcal{B})$ represents
$u\in\mathcal{M}$. By (5.2) and (5.4) we have

$\displaystyle(Ku,y)_{Y}$ $\displaystyle=\langle
K_{*}y,u\rangle=\int_{\mathcal{B}}\langle K_{*}y,v\rangle\,\mathrm{d}\mu(v)$
$\displaystyle=\int_{\mathcal{B}}(Kv,y)_{Y}\,\mathrm{d}\mu(v)=(\mathcal{K}\mu,y)_{Y}\,,$

which holds for all $y\in Y$, thus showing that $\mathcal{K}\mu=Ku$.

In order to show iii), assume that
$\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu$ in $M(\mathcal{B})$.
First notice that by (5.4),

$\displaystyle\lim_{k\rightarrow\infty}(\mathcal{K}\mu_{k},y)_{Y}$
$\displaystyle=\lim_{k\rightarrow\infty}\int_{\mathcal{B}}(Kv,y)_{Y}\,\mathrm{d}\mu_{k}(v)$
$\displaystyle=\int_{\mathcal{B}}(Kv,y)\,\mathrm{d}\mu(v)=(\mathcal{K}\mu,y)_{Y}\,,$

for every $y\in Y$, where the second equality follows because the map
$v\mapsto(Kv,y)_{Y}$ is weak* continuous, as $K$ is weak*-to-weak continuous
in $\mathcal{M}$, and hence, belongs to $C(\mathcal{B})$. Thus,
$\mathcal{K}\mu_{k}\rightharpoonup\mathcal{K}\mu$ in $Y$. Moreover, by
applying (5.4) twice we see that

$\displaystyle\|\mathcal{K}\mu_{k}\|^{2}_{Y}$
$\displaystyle=\int_{\mathcal{B}}(Kv,\mathcal{K}\mu_{k})_{Y}\leavevmode\nobreak\
\mathrm{d}\mu_{k}(v)$ (C.2)
$\displaystyle=\int_{\mathcal{B}}\int_{\mathcal{B}}(Kv,Kw)_{Y}\leavevmode\nobreak\
\mathrm{d}\mu_{k}(w)\mathrm{d}\mu_{k}(v)\,.$ (C.3)

Note that the function $(v,w)\mapsto(Kv,Kw)_{Y}$ is an element of
$C(\mathcal{B}\times\mathcal{B})$. Indeed, given a sequence
$(v_{k},w_{k})\in\mathcal{B}\times\mathcal{B}$ such that
$(v_{k},w_{k})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}(v,w)$ for
$(v,w)\in\mathcal{B}\times\mathcal{B}$ we estimate

$\displaystyle|(Kv_{k},Kw_{k})_{Y}$ $\displaystyle-(Kv,Kw)_{Y}|$
$\displaystyle\leq
C\|K\|_{\mathcal{L}(\mathcal{M},Y)}(\|Kw_{k}-Kw\|_{Y}+\|Kv_{k}-Kv\|_{Y})\,,$

where $C$ is defined at i). As, $v_{k},w_{k},v,w\in\mathcal{B}$, using the
weak*-to-strong continuity of $K$ on ${\rm dom}(\mathcal{G})$, cf. Assumptions
(A1)-(A3), we conclude that $(Kv_{k},Kw_{k})_{Y}\rightarrow(Kv,Kw)_{Y}$
proving that $(v,w)\mapsto(Kv,Kw)_{Y}$ is an element of
$C(\mathcal{B}\times\mathcal{B})$. Moreover, we have
$\mu_{k}\otimes\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu\otimes\mu$
in $M(\mathcal{B}\times\mathcal{B})$, which, combined with (C.2) and the fact
that $(v,w)\mapsto(Kv,Kw)_{Y}\in C(\mathcal{B}\times\mathcal{B})$ implies the
convergence $\|\mathcal{K}\mu_{k}\|_{Y}\to\|\mathcal{K}\mu\|_{Y}$. Combined
with the weak convergence $\mathcal{K}\mu_{k}\rightharpoonup\mathcal{K}\mu$ we
finally conclude $\mathcal{K}\mu_{k}\rightarrow\mathcal{K}\mu$ in $Y$, so that
$\mathcal{K}$ is weak*-to-strong continuous.

Last, note that the operator $\mathcal{K}_{*}$ is well-defined, linear and
continuous. For any $y\in Y$ and $\mu\in M(\mathcal{B})$ we obtain

$\displaystyle\llangle\mathcal{K}_{*}y,\mu\rrangle$
$\displaystyle=\int_{\mathcal{B}}\langle K_{*}y,v\rangle\leavevmode\nobreak\
\mathrm{d}\mu(v)$
$\displaystyle=\int_{\mathcal{B}}(Kv,y)_{Y}\leavevmode\nobreak\
\mathrm{d}\mu(v)=(\mathcal{K}\mu,y)_{Y}\,,$

where again, we used (5.4). This concludes the proof of iv).

Assume now that $Y$ is separable and fix $\mu\in M(\mathcal{B})$. Notice that
$f\colon\mathcal{B}\to Y$ defined by $f(v):=Kv$ is weakly $\mu$-measurable,
since the map $v\mapsto(Kv,y)_{Y}=\langle K_{*}y,v\rangle$ is weak* continuous
and hence, $\mu$-measurable, for each $y\in Y$ fixed. As $Y$ is separable, we
also have that $f$ is essentially separably valued. Therefore, Pettis’ Theorem
([34, Sec II.1, Thm 2]) implies that $f$ is strongly measurable with respect
to $\mu$. Moreover

$\int_{\mathcal{B}}\left\|Kv\right\|_{Y}\leavevmode\nobreak\
\mathrm{d}\mu(v)\leq
C\left\|K\right\|_{\mathcal{L}(\mathcal{M},Y)}\left\|\mu\right\|_{M(\mathcal{B})}<\infty\,,$

where $C$ is defined at i), showing that $f$ is Bochner integrable with
respect to $\mu$. Hence, by (5.4) and Hille’s Theorem ([34, Sec II.1, Thm 6]),
we infer

$(\mathcal{K}\mu,y)_{Y}=\int_{\mathcal{B}}(Kv,y)_{Y}\leavevmode\nobreak\
\mathrm{d}\mu(v)=\left(\int_{\mathcal{B}}Kv\leavevmode\nobreak\
\mathrm{d}\mu(v),y\right)_{Y}$

for all $y\in Y$, concluding (5.5) and the proof. ∎

###  C.2. Proofs of Section 6

####  C.2.1. Proof of Proposition 6.5

Let $k\geq 1$. By construction, $\lambda^{k}\in\mathbb{R}^{N_{k}}$ solves

$\displaystyle\min_{\lambda\in\mathbb{R}_{+}^{N_{k}}}\left[F\left(\sum^{N_{k}}_{i=1}\lambda_{i}Ku^{k}_{i}\right)+\sum^{N_{k}}_{i=1}\lambda_{i}\right].$

Deriving the first order necessary optimality conditions for this problem, we
obtain that

$\displaystyle P_{k}(u^{k}_{i})\leq 1\quad\text{ for all
}\,\,u^{k}_{i}\in\mathcal{A}_{k}\,,\quad\leavevmode\nobreak\
\sum^{N_{k}}_{i=1}\lambda^{k}_{i}P_{k}(u^{k}_{i})=\sum^{N_{k}}_{i=1}\lambda^{k}_{i}\,.$

Since $\lambda^{k}_{i}>0$ by construction, we deduce that $P_{k}(u^{k}_{i})=1$
for $i=1,\dots,N_{k}$. In particular,

$\displaystyle\llangle
P_{k},\mu_{k}\rrangle=\sum^{N_{k}}_{i=1}\lambda^{k}_{i}P_{k}(u^{k}_{i})=\sum^{N_{k}}_{i=1}\lambda^{k}_{i}=\|\mu_{k}\|_{M(\mathcal{B})}\,.$

Finally,
$\max_{v\in\mathcal{B}}P_{k}(v)\geq\max_{v\in\mathcal{A}_{k}}P_{k}(v)=1$,
concluding.

####  C.2.2. Proof of Proposition 6.6

Note that $\\{\mu_{k}\\}_{k}\subset E_{\mu_{0}}$ by (6.15). Arguing as in the
proof of Theorem 6.2, we get

$\displaystyle\|\nabla F(y_{k})-\nabla F(\bar{y})\|_{Y}$
$\displaystyle+\|P_{k}-\bar{P}\|_{C(\mathcal{B})}$ $\displaystyle\leq
L_{\mu_{0}}(1+C\|K\|_{\mathcal{L}(\mathcal{M},Y)})\|y_{k}-\bar{y}\|_{Y}\,,$

where we recall that $L_{\mu_{0}}>0$ is the Lipschitz constant of $\nabla F$
in $\mathcal{K}E_{\mu_{0}}$, and $C>0$ is the constant in Proposition 5.3 i).
Hence, it suffices to prove (6.19) for $\|y_{k}-\bar{y}\|_{Y}$. Let
$\mathcal{N}(\bar{y})\subset Y$ and $\theta>0$ denote the neighbourhood and
constant from Assumption (B1), respectively. Recall that
$\mu_{k}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\bar{\mu}$ along the
whole sequence by (6.6). By weak*-to-strong continuity of $\mathcal{K}$, see
Proposition 5.3 iii), we then conclude $y_{k}\rightarrow\bar{y}$ in $Y$. Thus,
there exist $M\in\mathbb{N}$ such that $y_{k}\in\mathcal{N}(\bar{y})$ for all
$k\geq M$. Using the strong convexity of $F$ in $\mathcal{N}(\bar{y})$, we
estimate

$\displaystyle j(\mu_{k})$
$\displaystyle=F(y_{k})+\|\mu_{k}\|_{M(\mathcal{B})}$ $\displaystyle\geq
F(\bar{y})+\theta\|y_{k}-\bar{y}\|_{Y}^{2}/2+(\nabla
F(\bar{y}),\mathcal{K}\mu_{k}-\mathcal{K}\bar{\mu})_{Y}+\|\mu_{k}\|_{M(\mathcal{B})}$
$\displaystyle=j(\bar{\mu})+\theta\|y_{k}-\bar{\mu}\|_{Y}^{2}/2+\llangle\bar{P},\bar{\mu}-\mu_{k}\rrangle+\|\mu_{k}\|_{M(\mathcal{B})}-\|\bar{\mu}\|_{M(\mathcal{B})}$
$\displaystyle\geq j(\bar{\mu})+\theta\|y_{k}-\bar{y}\|_{Y}^{2}/2\,,$

where we used (5.11) in the final inequality. Thus (6.19) follows by
rearranging the terms in the above estimate, recalling that
$r_{j}(\mu_{k})=j(\mu_{k})-j(\bar{\mu})$. The final part of the statement
holds since $r_{j}(\mu_{k})\to 0$ by Theorem 6.2.