Is it possible to prove that the sum of two independent r.v.'s $X$ and $Y$ with convex support cannot be uniformly distributed on an interval $[a,b]$, with $a < b$?
(Let us rule out the trivial case where $X$ is degenerate and $Y$ is uniform.) Note that $X$ and $Y$ need not be identically distributed.
I have a strong intuition supporting a negative answer because commutativity gets $X+Y$ to "pile up" away from the tails, but I cannot find a way to pin this down to an impossibility proof.
Related questions on Math.SE:
1) Can the sum of two independent identical random variables be uniform? and Can the sum of two i.i.d. random variables be uniformly distributed? answer negatively under the additional assumption that $X$ and $Y$ are identically distributed.
2) in Can sum of two random variables be uniformly distributed @GEdgar answers positively under the assumption that the two independent r.v.'s $X$ and $Y$ have non-convex support.
Your intuition is correct : the answer is negative. Here is a proof.
Suppose, by contradiction, that $X$ and $Y$ are independent, non-degenerate variables with convex support, say $[c,d]$ and $[e,f]$, and $X+Y$ is uniform on $[c+e,d+f]$. Replacing $(X,Y)$ with $(X-c,Y-e)$, we may assume without loss that $c=e=0$. Interchanging $X$ and $Y$ if necessary, we may further assume that $d\leq f$. If $d<f$ we can also replace $(X,Y)$ with $((2f)X+(3f+d)Y,(4f)X+(3f-d)Y)$ (remember that a linear bijective transformation preserves rv independence), so we may further assume that $d= f$. Finally, replacing $(X,Y)$ with $(\frac{X}{d},\frac{Y}{d})$ we may assume that $d=1$.
At this point, we then have $X,Y$ independent with support exactly equal to $[0,1]$, and $X+Y$ distributed uniformly on $[0,2]$. Now, define the following events :
$$ \begin{array}{cccl} X_1= \lbrace X\in[0,\frac{1}{3}] \rbrace, & X_2= \lbrace X\in(\frac{1}{3},\frac{2}{3}] \rbrace, & X_3= \lbrace X\in(\frac{2}{3},1] \rbrace, \\ Y_1= \lbrace Y\in[0,\frac{1}{3}] \rbrace, & Y_2= \lbrace Y\in(\frac{1}{3},\frac{2}{3}] \rbrace, & Y_3= \lbrace Y\in(\frac{2}{3},1] \rbrace, & \end{array} $$
$$ P_k=\cup_{i+j=k}X_i\times Y_j, Q_l=\cup_{k=2}^l P_k, L(x)=\lbrace X+Y \leq x \rbrace $$
We then have the inclusions :
$$ L\big(\frac{l-1}{3}\big) \subseteq Q_l \subseteq L\big(\frac{l}{3}\big) $$
Putting $q_l=P(Q_l)$, we deduce
$$ \frac{l-1}{6}\leq q_l \leq \frac{l}{6} \tag{1} $$
Next, put $x_k=P(X_k), y_k=P(Y_k), p_k=P(P_k)$. It follows from (1) that the following number is nonnegative :
$$ s=(1-x_1)\big(q_2-\frac{1}{6}\big)+x_1\big(\frac{2}{3}-q_4\big)+x_1x_2y_2 \tag{2} $$
Clearly $q_2=x_1y_1$. And using $x_3=1-(x_1+x_2)$ and $y_3=1-(y_1+y_2)$, we find that
$$ q_4 =x_1y_1+x_2y_1+x_1y_2+x_1y_3+x_2y_2+x_3y_1=x_1+y_1+x_2y_2-x_1y_1 \tag{3} $$
Reinjecting this back into (2), we obtain
$$ s=\bigg(x_1-\frac{1}{2}\bigg)\bigg(\frac{1}{3}-x_1\bigg) \tag{4} $$
It follows that $x_1\in[\frac{1}{3},\frac{1}{2}]$. Using natural symmetries of the problem, it follows that $x_3,y_1$ and $y_3$ are also in $[\frac{1}{3},\frac{1}{2}]$.
Next, consider
$$ \begin{array}{clc} t_1 &=& (x_1x_3-x_2^2)\big(\frac{5}{6}-q_5\big)+x_3x_2\big(\frac{2}{3}-q_4\big)+x_3^2\big(q_3-\frac{1}{3}\big) \\ t_2 &=& (x_1x_3-x_2^2)\big(q_2-\frac{1}{6}\big)+x_1x_2\big(q_3-\frac{1}{3}\big)+ x_1^2\big(\frac{2}{3}-q_4\big) \\ t &=& t_1+t_2 \end{array} $$
Note that since $x_1x_3 \geq \frac{1}{9} \geq x_2^2$, the numbers $t_1,t_2$ and $t$ are all nonnegative. Simplifying $t$ as we did for $s$, we obtain :
$$ t=-\frac{x_2}{4}(1+x_2)(\frac{1}{3}-x_2)-\frac{3(x_3-x_1)^2}{4}\bigg(\frac{4}{9}-x_2\bigg) \tag{5} $$
In (5) above we have a nonnegative LHS and a nonpositive RHS. Note also that $x_2>0$ because the support of $X$ is $[0,1]$. This forces $x_2=\frac{1}{3}$ and $(x_3-x_1)^2=0$, so that $x_1=x_2=x_3=\frac{1}{3}$, and similarly $y_1=y_2=y_3=\frac{1}{3}$. But then $q_2=\frac{1}{9}$ contradicts (1). This finishes the proof.
UPDATE: Since (5) was questioned in the comments, I append a computer verification of it.