Proposition about unimodal symmetric probability distributions I am trying to prove the following proposition:
Proposition.
Given a unimodal probability distribution $f(u)$, symmetric around $u=0$, strictly increasing for $u&lt0$ and strictly decreasing for $u>0$, then for all $y \ge 0$,
$$ \int_{x-y}^x f(u)du - \int_x^{x+y} f(u)du = 0 \hspace{0.1in} \Leftrightarrow  \hspace{0.1in} x=0.$$
The proof of $\Leftarrow$ is simple (plug in $x=0$), but the proof of $\Rightarrow$ is escaping me. Even though it feels very intuitive, I can't seem to nail it down rigorously. Is it true? 
(In terms of strategy, I'm trying is to break it up in three cases: $x>0$, $x=0$, $x&lt0$, and try and find a contradiction for the two $x \ne 0$ cases, but I'm not getting it.)
(Assumptions about continuous differentiability of $f$ are fine if necessary.)
 A: This first proof uses strictly decreasing and shows you cannot have the equality for any $x$ and $y$ with $0 \lt y \le x$
If, for $x \gt 0$, $f(u)$ is strictly decreasing for positive $u$ then for $0 \lt z \le y \le 2x$ you have $f(x-z) \gt f(x) \gt f(x+z)$ so 
$$\int_{x-y}^x f(u)du \gt  \int_{x-y}^x f(x)du = y f(x) = \int_x^{x+y} f(x)du \gt \int_x^{x+y} f(u)du$$
so 
$$\int_{x-y}^x f(u)du - \int_x^{x+y} f(u)du \gt 0.$$
For $y\gt 2x \gt 0$  it gets a little longer as:
$$\int_{x-y}^x f(u)du -\int_x^{x+y} f(u)du $$ $$= \int_{-x}^x f(u)du + \int_{x-y}^{-x} f(u)du -\int_x^{x+y} f(u)du $$ $$= \int_{-x}^x f(u)du + \int_{x}^{y-x} f(u)du  - \int_x^{x+y} f(u)du $$ $$= \int_{-x}^x f(u)du - \int_{y-x}^{x+y} f(u)du$$  $$\gt 2x f(x) -2xf(x) = 0$$
If $x \lt 0$, do something similar, reversing the inequalities where necessary.

Alternatively, not using strictly decreasing, but using all $y \gt 0$,  
$$\lim_{y \to +\infty} \left( \int_{x-y}^x f(u)du - \int_x^{x+y} f(u)du \right) $$ $$ = \Pr(X \lt x)-\Pr(X \gt x) = \Pr(X \gt -x) - \Pr(X \gt x)  = \Pr(-x \lt X \le x) $$
which is positive if $x \gt 0$ and $X$ has a positive probability for any open interval including $0$, which it must do if $0$ is the mode.   
