Let $c>0$ and $f:[0,\infty)\to[0,\infty)$ be Riemann integrable on every bounded interval in $[0,\infty)$.

Problem statement: Let $c>0$ and $f:[0,\infty)\to[0,\infty)$ be Riemann integrable on every bounded interval in $[0,\infty)$. If $$f(x)\leq\frac1c\int_0^1\left(\int_0^xf(ts)\,ds\right)dt$$ for all real $x\geq0$, prove that $f(x)=0$ for all $x\in[0,\infty)$.

MY solution

Here is my solution. I will use some techniques similar to Counting in Two Ways method.Define $\mathbb{I}=\int\limits_{0}^{1}{\left(\int\limits_{0}^{x}f(ts)ds\right)dt}$ where $t,s,x\in \mathbb{R^{+}}$ and rewrite the integral as: $\mathbb{I}=\int\limits_{0}^{1}\int\limits_{0}^{x}f(ts)dsdt$.Since $f(x)$ is a Riemann integrable function on $[0;+\infty)$, its Riemann integral summation should exist on every bounded interval, that is:$\int\limits_{a}^{b}f(t)dt=\lim\limits_{n\rightarrow +\infty}\sum\limits_{k=0}^{n}\dfrac{b-a}{n}f\left(a+\dfrac{k(b-a)}{n}\right)$ where $x\in (a;b)$.Take a simple interval and get: $\int\limits_{0}^{x}f(t)dt=\lim\limits_{n\rightarrow +\infty}\sum\limits_{k=0}^{n}\dfrac{x}{n}f\left(\dfrac{kx}{n}\right)$ where $x<+\infty$.Thus, $$\mathbb{I}=\int\limits_{0}^{1}\left(\int\limits_{0}^{x}f(ts)ds\right)dt=\int\limits_{0}^{x}\bigg(\lim\limits_{n\rightarrow +\infty}\sum\limits_{k=0}^{n}\dfrac{1}{n}f\left(\dfrac{sk}{n}\right)\bigg)=\lim\limits_{n\rightarrow +\infty}\bigg(\sum\limits_{k=0}^{n}\int\limits_{0}^{x}\dfrac{1}{n}f\left(\dfrac{sk}{n}\right)ds\bigg)=\lim\limits_{n\rightarrow +\infty}\bigg(\sum\limits_{k=0}^{n}\int\limits_{0}^{x}\dfrac{1}{k}f\left(\dfrac{sk}{n}\right)d\left(\dfrac{sk}{n}\right)\bigg)=\lim\limits_{n\rightarrow +\infty}\bigg(\sum\limits_{k=0}^{n}\int\limits_{0}^{\dfrac{xk}{n}}\dfrac{f(z)}{k}dz\bigg)<+\infty$$$\forall{x\in[0;+\infty)}$.Define $g(x)=\dfrac{1}{c}\int\limits_{0}^{1}\left(\int\limits_{0}^{x}f(st)ds\right)dt$.Because $f(x)\leq g(x)$$\forall{x\in [0;+\infty)}$, simply deducing $f(0)\leq g(0)=0$ and thus, $f(0)=0$ and $g(0)=0$.Since $f(x)$ is Riemann integrable on all bounded intervals in $[0;+\infty)$, applying Fubini's theorem to $g(x)$:$g(x)=\dfrac{1}{c}\int\limits_{0}^{1}\left(\int\limits_{0}^{x}f(st)ds\right)dt=\dfrac{1}{c}\int\limits_{0}^{x}\left(\int\limits_{0}^{1}f(st)dt\right)ds$.Taking the derivative of $g(x)$ in terms of $x$:$\dfrac{\partial}{\partial x}g(x)=\dfrac{\partial}{\partial x}\bigg(\dfrac{1}{c}\int\limits_{0}^{x}\left(\int\limits_{0}^{1}f(st)dt\right)ds\bigg)=\dfrac{1}{c}\int\limits_{0}^{1}f(xt)dt=\dfrac{1}{cx}\int\limits_{0}^{1}f(xt)d(xt)=\dfrac{1}{cx}\int\limits_{0}^{x}f(z)d(z)$.Therefore: $g(x)=\int\limits_{0}^{x}\dfrac{\int\limits_{0}^{t}f(z)d(z)}{ct}dx$ ($g(x)$ and $\dfrac{\partial}{\partial x}g(x)$ are valued at $x=0$).From the given condition: $g(x)\geq f(x)\implies \int\limits_{0}^{x}\dfrac{\int\limits_{0}^{t}f(z)d(z)}{ct}dx\geq f(x)$$\forall{x\in [0;+\infty)}$.Using Riemann's integral summation, $g(x)$ can be written as:$g(x)=\int\limits_{0}^{x}\dfrac{\int\limits_{0}^{t}f(z)d(z)}{ct}dx=\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{i=0}^{n}\left(\dfrac{1}{n}\dfrac{\int\limits_{0}^{\dfrac{ix}{n}}f(z)dz}{\dfrac{ix}{n}}\right)\right)=\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{i=0}^{n}\left(\dfrac{\int\limits_{0}^{\dfrac{ix}{n}}f(z)dz}{ix}\right)\right)=\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{i=1}^{n}\left(\dfrac{\int\limits_{0}^{\dfrac{ix}{n}}f(z)dz}{ix}\right)\right)$ (since $g(0)=0$).Rewrite $g(x)$ as:$g(x)=\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{i=1}^{n}\left(\dfrac{\int\limits_{0}^{\dfrac{ix}{n}}f(z)dz}{ix}\right)\right)=\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{i=1}^{n}(n+1-i)\left(\dfrac{\int\limits_{\dfrac{ix}{n}}^{\dfrac{(i+1)x}{n}}f(z)dz}{ix}\right)\right)$

Consider the following sequence: $\{a_{k}(x)\}=\bigg\{\left(\dfrac{n+1-k}{k}\right)\left(\int\limits_{\dfrac{kx}{n}}^{\dfrac{(k+1)x}{n}}f(z)dz\right)\bigg\}$.$0<\lim\limits_{k\rightarrow +\infty}\dfrac{a_{k+1}(x)}{a_{k}(x)}<1$$\forall{x\in [0;+\infty)}$$g(x)=\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{k=1}^{n}(n+1-k)\left(\dfrac{\int\limits_{\dfrac{kx}{n}}^{\dfrac{(k+1)x}{n}}f(z)dz}{kx}\right)\right)=\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{k=1}^{n}\dfrac{a_{k}(x)}{x}\right)$.Suppose that $f(x)\equiv 0$$\forall{x\in (0;+\infty)}$ isn't true. $g(x)\geq f(x)\implies xg(x)\geq xf(x)$ (1). Since $f(x)\neq 0$$\forall{x\in (0;+\infty)}$, taking limits on both sides of (1) and applying Squeeze Theorem:$\lim\limits_{x\rightarrow +\infty}xg(x)\geq \lim\limits_{x\rightarrow +\infty}xf(x)=+\infty$.Therefore, $\lim\limits_{n\rightarrow +\infty}\left(\sum\limits_{k=1}^{+\infty}a_{k}(x)\right)$ must be equal to $+\infty$ when $x\rightarrow +\infty$.Now, evaluating the limit of $\dfrac{a_{k+1}(x)}{a_{k}(x)}$ in the situation $k\rightarrow n$ and $n\rightarrow +\infty$ gives:$\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\dfrac{a_{k+1}(x)}{a_{k}(x)}=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\dfrac{\left(\dfrac{n-k}{k+1}\right)\left(\int\limits_{\dfrac{(k+1)x}{n}}^{\dfrac{(k+2)x}{n}}f(z)dz\right)}{\left(\dfrac{n+1-k}{k}\right)\left(\int\limits_{\dfrac{kx}{n}}^{\dfrac{(k+1)x}{n}}f(z)dz\right)}=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(\dfrac{\int\limits_{\dfrac{(k+1)x}{n}}^{\dfrac{(k+2)x}{n}}f(z)dz}{\int\limits_{\dfrac{kx}{n}}^{\dfrac{(k+1)x}{n}}f(z)dz}\right)\dfrac{(n-k)k}{(n+1-k)(k+1)}=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(\dfrac{\int\limits_{\dfrac{(k+1)x}{n}}^{\dfrac{(k+2)x}{n}}f(z)dz}{\int\limits_{\dfrac{kx}{n}}^{\dfrac{(k+1)x}{n}}f(z)dz}\right)\dfrac{(1-\dfrac{k}{n})\dfrac{k}{n}}{(1+\dfrac{1}{n}-\dfrac{k}{n})(\dfrac{k}{n}+1)}=0 < 1$.So, using d'Alembert's Convergence Criterion, $g(x)$ must converge for any finite $x\in [0;+\infty)$ and thus, $f(x)$ must be also convergent for finite $x\in [0;+\infty)$.This leads to $\int\limits_{0}^{x}f(z)dz$ being finite in $[0;+\infty)$.Consider the sequence $\{b_{k}\}=\bigg\{\int\limits_{0}^{\frac{xk}{n}}\dfrac{f(z)}{k}dz\bigg\}$.Because $f(x)=\lim\limits_{n\rightarrow +\infty}\bigg(\sum\limits_{k=0}^{n}b_{k}\bigg)$ converges for every finite $x\in (0;+\infty)$, applying d'Alembert's Convergence Criterion, $\lim\limits_{k\rightarrow +\infty}\left(\dfrac{b_{k+1}}{b_{k}}\right)$ must be smaller than 1.Which is equivalent to: $\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(\dfrac{\int\limits_{0}^{\frac{x(k+1)}{n}}\dfrac{f(z)}{k+1}dz}{\int\limits_{0}^{\frac{xk}{n}}\dfrac{f(z)}{k}dz}\right)<1$.Now, it suffices to deduce $\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(\dfrac{\int\limits_{0}^{\frac{x(k+1)}{n}}\dfrac{f(z)}{k+1}dz}{\int\limits_{0}^{\frac{xk}{n}}\dfrac{f(z)}{k}dz}\right)=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(\left(\dfrac{\int\limits_{0}^{\frac{x(k+1)}{n}}f(z)dz}{\int\limits_{0}^{\frac{xk}{n}}f(z)dz}\right)\bigg(\dfrac{k}{k+1}\bigg)\right)=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(\left(1+\dfrac{\int\limits_{\frac{xk}{n}}^{\frac{x(k+1)}{n}}f(z)dz}{\int\limits_{0}^{\frac{xk}{n}}f(z)dz}\right)\bigg(\dfrac{k}{k+1}\bigg)\right)=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(\left(1+\dfrac{\int\limits_{\frac{xk}{n}}^{\frac{x(k+1)}{n}}f(z)dz}{\int\limits_{0}^{\frac{xk}{n}}f(z)dz}\right)\bigg(\dfrac{\frac{k}{n}}{\frac{k}{n}+\frac{1}{n}}\bigg)\right)=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(1+\dfrac{\int\limits_{\frac{xk}{n}}^{\frac{x(k+1)}{n}}f(z)dz}{\int\limits_{0}^{\frac{xk}{n}}f(z)dz}\right)\times\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\bigg(\dfrac{\frac{k}{n}}{\frac{k}{n}+\frac{1}{n}}\bigg)=\lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(1+\dfrac{\int\limits_{\frac{xk}{n}}^{\frac{x(k+1)}{n}}f(z)dz}{\int\limits_{0}^{\frac{xk}{n}}f(z)dz}\right)$.Therefore: $1\leq \lim\limits_{n\rightarrow +\infty;\frac{k}{n}\rightarrow 1}\left(1+\dfrac{\int\limits_{\frac{xk}{n}}^{\frac{x(k+1)}{n}}f(z)dz}{\int\limits_{0}^{\frac{xk}{n}}f(z)dz}\right)<1$. This means contradiction, and $f(x)\equiv 0$$\forall{x\in (0;+\infty)}$ must hold.End of the solution. $\square$