No. You can get a higher $I(U;V)$ using asymmetric channels. Below I construct a counterexample, but first a more succinct restatement of the question.
Restatement
To summarize, there is an input $U$ distorted by three independent binary hops, each described by $2\times 2$ stochastic matrices $C_L,\ C,\ C_R.$ Labeling all your RV's, $$U\overset{C_L}{\to}X\overset{C}{\to}Y\overset{C_R}{\to}V.$$
We are interested in maximizing $I(U;V)$
subject to constraints:
- $C$ is determined by nature.
- $C_L$ is such that $I(U;X) \leq r_L$,
- $C_R$ is such that $I(Y;V) \leq r_R$,
- $U$'s distribution is such that the left channel's output (i.e. $X$) is $B(1/2)$.
Counterexample
You claim the $C_L,C_R$ which produce the maximum are binary symmetric.
- If $C_L$ is a BSC with $B(1/2)$ output then its input must also be $B(1/2).$ For a given rate $r_L \leq 1$, then there is at most one 'positive' (i.e. can't be improved by relabeling the outputs) BSC $C_L$ whose output is $B(1/2).$
- You have assumed $C$ is a BSC, so with a symmetric input its output is also symmetric.
- For a rate $r_R\leq 1$ there is only one positive choice for $C_R.$
So to say they are binary symmetric is to determine all of $U,C_L$ and $C_R$.
Now take $C$ a perfect channel,
$C= \left[\begin{smallmatrix}
1 & 0 \\
0 & 1
\end{smallmatrix}\right]$ and $r_L=r_R=0.4.$ The associated positive BSC for this rate has crossover probability $\approx 0.15,$ and the end-to-end mutual information can be computed:
$$I(U_{BSC}, V_{BSC})< 0.1895$$
However, trying randomly[1] you can find $U^\ast, C_L^\ast, C_R^\ast$ that satisfy all the mutual information constraints, but have greater $U$-to-$V$ mutual information. One I found happens to be quite close to a Z-channel:
\begin{equation}
I(U^\ast; V^\ast) > 0.19,
\end{equation}
\begin{equation}
C_L^\ast \approx \left[\begin{smallmatrix}0.2493 & 0.7507 \\ 0.9657 & 0.0343 \end{smallmatrix}\right], \qquad
C_R^\ast \approx \left[\begin{smallmatrix} 0.9821 & 0.0179 \\ 0.3374 & 0.6626 \end{smallmatrix}\right], \qquad
U^\ast \sim B(0.35)
\end{equation}
Discussion
This result is to be expected since there is a vague sense that uniform noise over a bounded space is the most degrading, even holding mutual information fixed. (by one heuristic "uniform noise means you can't precode to mitigate it")
A gentle introduction for a good, visualisable framework for studying binary symmetric channels is given in a short paper, Algebraic Information Theory for Binary Channels by Martin, Moskowitz and Allwein. Under this framework your maximization can be restated as a convex optimization problem for which I see no easy special cases.
An easier-to-investigate (and arguably more interesting) problem is one identical to yours that omits the fourth constraint that $X\sim B(1/2)$. But I could not find an easy path towards an answer for this either.
For both of these there might be some magical connection to KL divergence which I am not seeing.
Code
[1]: Below is a crude counterexample finder.
% Helper functions
% Binary entropy
fn_h = @(p) -p.*log2(p) - (1-p).*log2(1-p);
% MI across mtx_bc when v_distn is input
fn_I = @(mtx_bc,v_distn) fn_h(v_distn(1)) + fn_h(v_distn*mtx_bc(:,1)) ...
- nansum(nansum(-log2(diag(v_distn)*mtx_bc).*(diag(v_distn)*mtx_bc)));
% Channel matrix when P(out=0|in=0)=pa, P(out=0|in=1)=pb
fn_mtxBC = @(pa,pb) [pa, 1-pa; pb, 1-pb];
% Set params
d_r_L = 0.4;
d_r_R = 0.4;
d_xp = 0.146102; % solution to 1-H(p) = 0.4
mtxBSC = fn_mtxBC(d_xp, 1-d_xp);
% Search
while true
mtxL = fn_mtxBC(rand, rand);
mtxR = fn_mtxBC(rand, rand);
v_d = (mtxL'\[0.5, 0.5]')';
if (abs(sum(v_d)-1) > 0.001 || ...
min(v_d) < 0)
continue
end
if(fn_I(mtxL, v_d) > 0.4 || ...
fn_I(mtxR, v_d*mtxL) > 0.4)
continue;
end
fprintf('+\n');
if fn_I(mtxL*mtxR, v_d) > fn_I(mtxBSC*mtxBSC, [0.5, 0.5])
break;
end
end
