Why does $\int_{-x}^x f^\alpha\leq f(-x)+f(x)$ imply $f=0$? 
Let $0\leq f\in C(\mathbb{R})$, and for some $\alpha>1$, we have
\begin{equation*}\begin{aligned}
\int_{-x}^x f^\alpha\leq f(-x)+f(x), \forall\ x\in [0,+\infty).
\end{aligned}\end{equation*}
Prove that $f\equiv 0$.

Let $F(x)=\int_{-x}^x f^\alpha$, then $F'(x)=f^\alpha(x)+f^\alpha(-x)
\leq [f(x)+f(-x)]^\alpha$. How to use the assumption? What to do next? Any ideas?
 A: Consider the “even part” $g(x) = \frac 12(f(-x) + f(x))$ for $x \ge 0$. Then
$$
 g(x)^\alpha \le  \max(f(-x), f(x))^\alpha = \max(f(-x)^\alpha, f(x)^\alpha) \le f(-x)^\alpha+ f(x)^\alpha
$$
and therefore
$$
 \int_0^x g(t)^\alpha \, dt \le  \int_{-x}^x f(t)^\alpha \, dt \le f(-x)+f(x) = 2g(x) \, .
$$
The function
$$
 G(x) = \int_0^x g(t)^\alpha \, dt 
$$
is non-negative and differentiable on $[0, \infty)$ with
$$
 G'(x) = g(x)^\alpha  \ge 2^{-\alpha} G(x)^\alpha \, .
$$
Now assume that $G(x_1) > 0$ for some $x_1 > 0$. Then $G$ is positive and increasing on $[x_1, \infty)$, and for all $x_2 > x_1$ is
$$
\frac{1}{\alpha -1} \frac{1}{G(x_1)^{\alpha-1}} \ge
\frac{1}{\alpha -1} \left( \frac{1}{G(x_1)^{\alpha-1}}-\frac{1}{G(x_2)^{\alpha-1}}\right) =
 \int_{x_1}^{x_2} \frac{G'(t)}{G(t)^\alpha} \, dt\\
 \ge \int_{x_1}^{x_2} 2^{-\alpha} \, dt = 2^{-\alpha}(x_2 - x_1) \, .
$$
This is impossible because the right-hand side converges to $(+\infty)$ for $x_2 \to \infty$.
It follows that $G$ is identically zero, and that implies that $g$ is identically zero. Finally,
$$
\int_{-x}^x f(t)^\alpha \, dt \le 2 g(x) = 0
$$
for the non-negative function $f$ implies that $f$ is identically zero.
Remark: The proof shows that the assumption $ \int_{-x}^x f(t)^\alpha \, dt \le f(-x)+f(x) $ can be weakened to
$$
 \int_{-x}^x f(t)^\alpha \, dt \le C \bigl(f(-x)+f(x) \bigr)
$$
with an arbitrary constant $C > 0$.
