Uniqueness of the distribution maximizing the channel capacity
Suppose we have a binary symmetric channel: 

$x_1 \rightarrow y_1$ with probability $1-\epsilon$ and $y_2$ with probability $\epsilon$, 

$x_2 \rightarrow y_2$ with probability $1-\epsilon$ and $y_1$ with probability $\epsilon$. 

Now add a third input $x_3$ that goes to $y_1$ and $y_2$ with probability $\frac{1}{2}$ each.  

Then the optimal input probability distribution is $(\frac{1}{2}, \frac{1}{2}, 0)$. 

Proof: Let $X$ be the random variable describing the input, and $Y$ be the random variable describing the output. The capacity $C = H(Y) - H(Y|X)$, where $H(Y)$ is the entropy of $Y$, and $H(Y|X)$ is the conditional entropy of $Y$ given $X$. 

If you have an input distribution $(p_1, p_2, p_3)$ for $X$, you can replace it by the distribution $(p_1+p_3/2, p_2 + p_3/2, 0)$. This keeps $Y$ the same, so $H(Y)$ is unchanged, and it reduces $H(Y|X)$. Thus, capacity strictly increases.