Prove or disprove that $f\left ( x \right )\leq 1,$ where $f$ is a continuous function so that $$\int_{0}^{x}tf\left ( 2x- t \right ){\rm d}t= \frac{\tan^{-1}x^{2}}{2}$$ Source: AoPS/@Ji123_ https://artofproblemsolving.com/community/q1h144652p819322
I edited the original problem a little bit, my friend@twelve_sakuya also has a completed disproof in case $f\left ( x \right )< 1,$ we substitute $$t\mapsto 2x- t\Rightarrow\int_{0}^{x}tf\left ( 2x- t \right ){\rm d}t\equiv\int_{x}^{2x}\left ( 2x- t \right )f\left ( t \right ){\rm d}t$$ $$\frac{{\rm d}}{{\rm d}x}\int_{x}^{2x}\left ( 2x- t \right )f\left ( t \right ){\rm d}t= \frac{{\rm d}}{{\rm d}x}\left ( 2x\int_{x}^{2x}f\left ( t \right ){\rm d}t- \int_{x}^{2x}tf\left ( t \right ){\rm d}t \right )= 2\int_{x}^{2x}f\left ( t \right ){\rm d}t- xf\left ( x \right )$$ On the other hand $$\frac{{\rm d}}{{\rm d}x}\frac{\tan^{-1}x^{2}}{2}= \frac{x}{x^{4}+ 1}\quad\therefore 2\int_{x}^{2x}f\left ( t \right ){\rm d}t- xf\left ( x \right )= \frac{x}{x^{4}+ 1}$$ When $x> 0,$ we have $$f\left ( x \right )= \frac{2}{x}\int_{x}^{2x}f\left ( t \right ){\rm d}t- \frac{1}{x^{4}+ 1}= 2f\left ( c \right )- \frac{1}{x^{4}+ 1}\quad\left ( x< \exists c< 2x \right )$$ (According to Mean-value theorem), if $x\rightarrow +0\Rightarrow c\rightarrow +0\Rightarrow f\left ( 0 \right )= 2f\left ( 0 \right )- 1\quad\left ( x\rightarrow +0 \right ),$ so $f\left ( 0 \right )= 1,$ we have contractdiction.
Using Leibnitz integral rule:
$xf(x) = \dfrac{x}{x^4+1}$
Let $x \in \mathbb{R} - \{ 0\}.$ Now we can divide the above expression by $x$ as $x\ne0.$
$f(x) = \dfrac{1}{x^4+1}$ which is clearly less than one as denominator is always greater than numerator. Now lets include zero into the domain...
Substituting zero into the original equation we have:
$0f(0) = \dfrac{0}{0^4+1} =0 \implies f(0) \in \mathbb{R}$. But its given $f(x)$ is continuous. Hence LHL must equal RHL. $\lim_{x \to 0} \dfrac{1}{x^4+1} = 1$. Hence f(0) must equal 1.
Now we have: $f(x) \leq 1 \forall x \in \mathbb{R}$