In proving $\int_{-x}^xf^\alpha\le f(-x)+f(x)$ for nonnegative continuous $f$, $\alpha\gt1$ implies $f\equiv0$, why can we take $f$ nondecreasing? $\newcommand{\d}{\,\mathrm{d}}$I have a solution to this problem, but the TlDr is that my solution hinges on the assumption that $f$ is nondecreasing on $[0,\infty)$. It is obvious to me that proving the result for nondecreasing $f$ concludes the result for all $f$, but I have struggled to pin down an exact proof - this is what I'd like answered. I know there is an answer since another user on this site suggested it as a hint.
Some context: some users may recognise this (deleted) question here. The question received a lot of interest - $7$ upvotes and $5$ favourites at the time of deletion - but the OP (not myself!) deleted it for no clear reason, presumably (from the comments) because they'd figured out the answer or were embarrassed about not having done so. Considering that the post held interest for many other users I feel that the post should not be so wantonly deleted, so I am reposting it here with my own (partial) solution.
The question:

If $f:\Bbb R\to\Bbb R$ is a nonnegative continuous function and $\exists\alpha\gt1$ such that: $$\int_{-x}^x f(t)^\alpha\d t\le f(-x)+f(x),\,\forall x\in\Bbb R$$Prove that $f\equiv0$.

My solution, following a hint from Eyeballfrog left on the original post:

Let $f(x)=E(x)+O(x)$ be the standard even/odd decomposition. The condition is that, for all $x$: $$\begin{align}2\int_0^x E(t)^\alpha\d t&=\int_{-x}^xE(t)^\alpha\d t\\&=2^{-\alpha}\int_{-x}^x(f(t)^\alpha+f(-t)^\alpha)\d t\\&=2^{1-\alpha}\int_{-x}^xf(t)^\alpha\d t\\&\lt\int_{-x}^xf(t)^\alpha\d t\\&\le f(x)+f(-x)\\&=2E(x)\end{align}$$So if we demonstrate that the above inequality forces $E(x)\equiv0$ we force $f(x)\equiv0$ since $f$ is positive. It suffices to consider $f$ even then.

It is also sufficient to show the result for $f$ nondecreasing on $[0,\infty)$. I honestly have struggled to formalise this in a proof - I expect the answer is very simple, though - but it is strongly obvious. A strong proof of why, exactly, we can make this assumption is my question.
With that assumption in mind, we can continue:

I claim that if $f$ is not identically zero, it must be unbounded. First note that, for an $x\gt0$, $f(x)=0$ forces $f\equiv0$ on $[0,x]$, so for $f$ to not be identically zero there must eventually be an $x_0$ beyond which $f$ is never zero. Suppose $f\le M$ is a bound. For $x_1\gt x_0$ consider $f(x)\ge\int_{x_1}^x f(t)^\alpha\d t$ implies that $M\ge f(x_1)^\alpha(x-x_1)$ with the nondecreasing assumption. However this is obviously false for all $x$ as $f(x_1)\gt0$. Thus, $f$ is unbounded in this case.
Let $\lambda$ be an arbitrary positive real and $x$ such that $f(x)\gt1$ (possible by unboundedness). By the mean value theorem for integrals of continuous functions (and the nondecreasing assertion): $$\lambda f(x)^\alpha\le\lambda f(c)^\alpha=\int_x^{x+\lambda}f(t)^\alpha\d t\le f(x+\lambda)$$So let's integrate: $$\begin{align}f(x+\lambda)&\ge\int_x^{x+\lambda}f(t)^\alpha\d t\\&\ge\int_0^\lambda f(t+x)^\alpha\d t\\&\ge f(x)^{\alpha^2}\int_0^\lambda t^\alpha\d t\\&=\frac{\lambda^{\alpha+1}}{\alpha+1}f(x)^{\alpha^2}\end{align}$$This gives an induction: $$f(x+\lambda)\ge\frac{\lambda^{\alpha^2+\alpha+1}}{(\alpha+1)^\alpha(\alpha^2+\alpha+1)}f(x)^{\alpha^3}\\f(x+\lambda)\ge\frac{\lambda^{\alpha^3+\alpha^2+\alpha+1}}{(\alpha+1)^{\alpha^2}(\alpha^2+\alpha+1)^{\alpha}(\alpha^3+\alpha^2+\alpha+1)}f(x)^{\alpha^4}\\f(x+\lambda)\ge\cdots\ge\frac{\lambda^{1+\alpha+\cdots+\alpha^n}}{(\alpha+1)^{\alpha^{n-1}+2\alpha^{n-2}+3\alpha^{n-3}+\cdots+n}}f(x)^{\alpha^n},\quad\forall n\in\Bbb N$$But if $\lambda$ and $f(x)$ are sufficiently large (which can be arranged by unboundedness) then by $f(x)\gt1$, $\alpha\gt1$ we have that the inequality holds for all $n$ only if $f(x+\lambda)=\infty$, which is impossible. Therefore $f\equiv0$ must hold!

So we are done. However, I'd like some feedback on the correctness of the proof and more importantly I'd like guidance on why we can suppose $f$ is nondecreasing. It seems so obvious but I can't pin it down...
Many thanks in advance.
 A: Well, since I gave the hint, I suppose I should give my proof. It went like this.
First, we show that
$$
\int_{-x}^xf(t)^\alpha dt = \int_0^x[f(t)^\alpha + f(-t)^\alpha]dt \le \int_0^x[f(t)+ f(-t)]^\alpha dt.
$$
which means that if we can show that if there is an $\alpha$ such that for all $x$,
$$
\int_0^x g(t)^\alpha dt \le g(x)
$$
then $g\equiv 0$, that will be sufficient to prove the original statement. This is where the evenness hint comes in, although I now notice there's actually an irrelevant factor of $2$ between this statement and the original.
Once we have this, we consider the function $h(x) = \inf_{y > x} g(y)$. Clearly, for all $x$, $h(x) \le g(x)$. Secondly, since $\int_0^x g(t)^\alpha dt$ is a nondecreasing function of $x$ and $\int_0^x g(t)^\alpha dt \le g(x)$ for all $x$, we must also have $\int_0^x g(t)^\alpha dt \le h(x)$ for all $x$. Thus we have
$$
\int_0^x h(t)^\alpha dt \le \int_0^x g(t)^\alpha dt \le h(x).
$$
So if we can show that for a nondecreasing function $h$, the existence of some $\alpha$ such that for all $x$,
$$
\int_0^x h(t)^\alpha dt \le h(x)
$$
implies $h\equiv 0$, the above bounds on the integral force $g\equiv 0$.
