Calculating the product state capacity of the quantum depolarizing channel This is homework, so just let me know if I'm on the right track or where I went wrong, please.
So, we are asked to compute the product state capacity 
$$C_1(T) = \max_{\{p_j,\,\vert \psi_j\rangle\}}\left[ S\left(\sum_j p_j T(\vert \psi_j\rangle\langle \psi_j\vert)\right) - \sum_j p_jS(T(\vert \psi_j\rangle\langle \psi_j\vert))\right]$$
of the depolarizing channel $T(\rho)\equiv(1-p)\rho + p\frac{1}{d}\textbf{I}$ in $d$ dimensions. 
Here is how I proceded:


*

*See that for each pure state $\vert \psi\rangle$, we could write 
$$ T(\vert \psi\rangle\langle \psi\vert) \doteq \text{diag}\left(1-p + \frac{p}{d},\frac{p}{d},\ldots,\frac{p}{d}\right),$$

*which would then lead to 
$$
S(T(\vert \psi\rangle\langle \psi\vert))=-\left[\left(1-p+\frac{p}{d}\right)\log\left(1-p+\frac{p}{d}\right)+\frac{d-1}{d}p\log\left(\frac{p}{d}\right)
\right]$$

*Further, by the same token as in 1., we get
$$
\sum_x q_x T(\vert \psi\rangle\langle \psi\vert)\doteq
(1-p)\,\text{diag}\left(\sum_x q_x \lvert\langle\psi_x\vert 1\rangle \rvert^2,\ldots,\sum_x q_x \lvert\langle\psi_x\vert d\rangle \rvert^2
\right)+p\frac{1}{d}\textbf{I}.
$$

*Thus, we need to maximize the expression
\begin{align}
&-\sum_{i=1}^d\left[(1-p)\sum_x q_x \lvert\langle\psi_x\vert i\rangle \rvert^2 + \frac{p}{d}\right]\cdot\log\left((1-p)\sum_x q_x \lvert\langle\psi_x\vert i\rangle \rvert^2 + \frac{p}{d}\right)\\
&+\left(1-p+\frac{p}{d}\right)\log\left(1-p+\frac{p}{d}\right) + \frac{d-1}{d}p\log\left(\frac{p}{d}\right)
\end{align}
This looks kinda bogus and I don't even know what to do now.
 A: As far as I can tell, you didn't do anything wrong, but your straightforward attempt to optimisation is not very enlightning. Let's try to simplify this.
Hint 1:
As often in Physics, symmetries simplify the problem.
It does not help you ? Then let us move to hint 2
Hint 2:
At step 2., you can see that $S(T (|ψ\rangle))$ doesn't depend on $|ψ\rangle$. That allows you to simplify the quantity to optimise.
Still not enough ? Then move on to
Hint 3:
You now have to find $ρ=\sum_x p_x |ψ_x\rangle\langle ψ_x|$ maximising the output entropy $S (T (ρ))$.
A: Ok, so. Nothing wrong with what I did, and it thus follows (with the binary entropy $\mathcal{h}_2$)
\begin{alignat}{2}
C_1(T)&=\max_{\{q_x,\,\vert \psi_x\rangle\}} 
\left\{ S\left(\sum_xq_x T\left(\vert \psi_x\rangle\langle \psi_x\vert\right)\right)
+\left(1-\frac{d-1}{d}\cdot p\right)\log\left(1-\frac{d-1}{d}\cdot p\right)\right. \\[1em]
 &\qquad\qquad\qquad \left. + \frac{d-1}{d}\cdot p\log\frac{p}{d}\right\}\\[2em]
&=\left(1-\frac{d-1}{d}\cdot p\right)\log\left(1-\frac{d-1}{d}\cdot p\right) + \frac{d-1}{d}\cdot p\log\frac{p}{d}\\[1em]
&\quad+\max_{\{q_x,\,\vert \psi_x\rangle\}} 
\left\{S\left(\sum_xq_x T\left(\vert \psi_x\rangle\langle \psi_x\vert\right)\right)\right\}
\end{alignat}
And since $0\leq S \leq \log d$, it is easily seen that the maximum of the last term is obtained if the state $$\sum_x q_x T(\vert \psi_x\rangle\langle \psi_x\vert)$$ is the maximally entangled state $\frac{1}{d}\textbf{I}$, which can be achieved by pure states $\vert \psi_x\rangle$.
All in all, 
\begin{align}C_1(T)&=\log d + \left(1-\frac{d-1}{d}\cdot p\right)\log\left(1-\frac{d-1}{d}\cdot p\right) + \frac{d-1}{d}\cdot p\log\frac{p}{d}\\[1em]
&= \log d - h_2\left(\frac{d-1}{d}\cdot p\right) + \frac{d-1}{d}\cdot p\left(\log\frac{p}{d} -\log\left(\frac{d-1}{d}\cdot p\right)\right)\\[1em]
&= \log d - h_2\left(\frac{d-1}{d}\cdot p\right)-\frac{d-1}{d}\cdot p \log (d-1)
\end{align} 
And finally, a quick plot:

