Show that $\int_0^\infty E[ 1\{f(X) \le f(X+t) \}] \, t dt =E[X^2]$ Consider a symmetric random variable $X$ with the pdf $f$. We want to study the following expression:
\begin{align}
\int_0^\infty E[ 1\{f(X) \le f(X+t) \}] t dt 
\end{align}
where $1\{\cdot \}$ is the indicator function. Can we show that
\begin{align}
\int_0^\infty E[ 1\{f(X) \le f(X+t) \}] \, t dt =E[X^2]?
\end{align}
What I have done:
Proof for the case when $f(x)$ is increase for $x<0$ and decreasing for $x>0$.
Under these conditions, we have that for $t\ge 0$
\begin{align}
1\{f(X) \le f(X+t) \}=1 \{ X \le 0, t \le 2|X|\}
\end{align}
Therefore,
\begin{align}
\int_0^\infty E[ 1 \{ X \le 0, t \le |X|\}] \, t dt &= E \left[1 \{ X \le 0\} \int_0^{2|X|} t dt  \right] \text{ by Tonelli_Fubini}\\
&=E \left[1 \{ X \le 0\} \frac{|X|^2}{2} \right]\\
&=E[X^2] \text{ by symmetry} 
\end{align}
The issue is that not all symmetric random variables have this increasing property for the pdf. Therefore, I am not sure if it always holds.

I'm assuming that $\{X: f(X) = t\}$ has measure $0$ for every $t$.
Let $\mu$ be the measure associated with $f$. Setting $Y = t + X$, we can rewrite the integral as
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 1_{X < Y} 1_{f(X) < f(Y)} (Y - X) d\mu(X) dY.$$
We split the integral into three parts: those with $X, Y \geq 0$, those with $X < 0 < Y$, and those with $X, Y \leq 0$. The first part is
$$\int_{0}^{\infty} \int_{0}^{\infty} 1_{X < Y} 1_{f(X) < f(Y)} (Y - X) d\mu(X) dY = \int_{0}^{\infty} \int_{0}^{\infty} 1_{X < Y} 1_{f(X) < f(Y)} (Y - X) f(X)dX dY.$$
The third part is, by symmetry
$$\int_{-\infty}^{0} \int_{-\infty}^{0} 1_{X < Y} 1_{f(X) < f(Y)} (Y - X) d\mu(X) dY = \int_{0}^{\infty} \int_{0}^{\infty} 1_{X > Y} 1_{f(X) < f(Y)} (X - Y) f(X)dX dY.$$
So the sum of the first and third part is
$$\int_{0}^{\infty} \int_{0}^{\infty} 1_{f(X) < f(Y)} |X - Y| f(X)dX dY.$$
On the other hand, the second part is
$$\int_{-\infty}^{0} \int_{0}^{\infty} 1_{f(X) < f(Y)} (Y - X) f(X) dY dX = \int_{0}^{\infty} \int_{0}^{\infty} 1_{f(X) < f(Y)} (Y + X) f(X)dX dY.$$
So we conclude by summing the three parts that
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 1_{X < Y} 1_{f(X) < f(Y)} (Y - X) d\mu(X) dY = \int_{0}^{\infty} \int_{0}^{\infty} 1_{f(X) < f(Y)}2\max(X,Y) f(X)dX dY.$$
We can manipulate the final expression as follows: swapping the role of $X$ and $Y$, we get
$$\int_{0}^{\infty} \int_{0}^{\infty} 1_{f(X) < f(Y)}2\max(X,Y) f(X)dX dY = \int_{0}^{\infty} \int_{0}^{\infty} 1_{f(Y) < f(X)}2\max(X,Y) f(Y)dX dY.$$
So, replacing a copy with LHS with a copy of RHS,
$$\int_{0}^{\infty} \int_{0}^{\infty} 1_{f(X) < f(Y)}2\max(X,Y) f(X)dX dY = \int_{0}^{\infty} \int_{0}^{\infty} \max(X,Y) \min(f(X), f(Y)) dX dY.$$
The right hand side is symmetric in $X$ and $Y$, so we can cut the region of integration by half,
$$\int_{0}^{\infty} \int_{0}^{\infty} \max(X,Y) \min(f(X), f(Y)) dX dY = 2\int_{0}^{\infty} \int_{0}^{X} X \min(f(X), f(Y)) dY dX.$$
Now, we have the inequality,
$$2\int_{0}^{\infty} \int_{0}^{X} X \min(f(X), f(Y)) dY dX \leq 2\int_{0}^{\infty} \int_{0}^{X} X f(X) dY dX = 2\int_{0}^{\infty} X^2 f(X) dX = E[X^2].$$
Thus we conclude that
$$\int_{-\infty}^{\infty} P[f(X) \leq f(X + t)] tdt \leq E[X^2]$$
with equality iff $f$ is monotonically decreasing almost everywhere on $[0,\infty]$.
