Proof $f(x)\equiv 0$ Let $f\in C ((-\infty,+\infty)).
 $ If $ \forall  a,b\in(-\infty,+\infty),\int_{a}^{b}f^{2}(x)dx \leq  f(a)+f(b), $ then$f(x)\equiv 0$.


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*I can prove $f(x)\geq 0,$ so I want use the reduction to absurdity in the next proof that may refuse $f(x)>0.$ In my way of thinking ,I tried but it is seemingly difficult  to deny $f(x)>0.$ Could you please give me some hints?

 A: 1) We have $f(x)\geq 0$ for all $x$, as you have shown.
2) Let $a\in \mathbb{R}$ fixed, put for $x\geq a$ $\displaystyle F(x)=\int_a^x f(t)^2dt$. We have $\displaystyle F^{\prime}(x)=f(x)^2$, hence $\displaystyle f(x)=\sqrt{F^{\prime}(x)}$, and we get $\displaystyle F(x)-f(a)\leq \sqrt{F^{\prime}(x)}$ for $x\geq a$.
3) Suppose now that there exists  $b>a$ such that $F(x)-f(a)>0$. Then we have $F(x)-f(a)>0$ for all $x\geq b$, and so $\displaystyle \frac{F^{\prime}(x)}{(F(x)-f(a))^2}\geq 1$. We integrate from $b$ to $x>b$, we get 
$$-\frac{1}{F(x)-f(a)}+\frac{1}{F(b)-f(a)}\geq x-b$$
hence
$$\frac{1}{F(b)-f(a)}\geq x-b$$
If we let $x\to +\infty$, we get a contradiction. 
Hence for all $a\in \mathbb{R}$ and $x\geq a$, we have $F(x)\leq f(a)$.
4) Fix $b\in \mathbb{R}$, put $\displaystyle G(x)=\int_x^bf(t)^2dt$ for $x<b$. We have $G^ {\prime}(x)=-f(x)^2$ and $G(x)\leq f(x)$ by 3). Suppose that there exists $a<b$ such that $G(a)>0$. Then we have $G(x)>0$ for all $x\leq a$, and $G(x)\leq \sqrt{-G^{\prime}(x)}$. Hence $\displaystyle -\frac{G^{\prime}(x)}{G(x)^2}\geq 1$ for $a\geq x$. We integrate from $x<a$ to $a$, we get 
$$\frac{1}{G(a)}-\frac{1}{G(x)}\geq a-x$$
and $\displaystyle \frac{1}{G(a)}\geq a-x$. If $x\to -\infty$, we find a contradiction.
5) Hence $\displaystyle \int_a^b f(t)^2dt=0$ for all $b,a$ such that $b>a$, and this imply $f=0$.
