Find the increasing function $f:\mathbb{R} \to \mathbb{R}$ such that $\int_{0}^{x}f(t)\,dt=x^2$ for all $x\in \mathbb{R}$ Find the increasing function $f:\mathbb{R} \to \mathbb{R}$ such that $$\int_{0}^{x}f(t)\,dt=x^2\text{ for all }x\in \mathbb{R}$$
My solution: Let $F:\mathbb{R} \to \mathbb{R}$,$F(x)=\int_{0}^{x}f(t)\,dt$.
Then it follows that $F$ is differentiable and that $F'(x)=2x$ for all $x\in \mathbb{R}$.
For $a\in \mathbb{R}$ we have that $$2a=F'(a)=\lim_{x\searrow a}\frac{F(x)-F(a)}{x-a}=\lim_{x\searrow a}\frac{\int_{a}^{x}f(t)\,dt}{x-a}\ge \lim_{x\searrow a} \frac{\int_{a}^{x}f(a)\,dt}{x-a}=\lim_{x\searrow a}\frac{(x-a)f(a)}{x-a}=f(a)$$
and $$2a=F'(a)=\lim_{x\nearrow a}\frac{F(x)-F(a)}{x-a}=\lim_{x\nearrow a}\frac{\int_{a}^{x}f(t)\,dt}{x-a}\le \lim_{x\nearrow a} \frac{\int_{a}^{x}f(a)\,dt}{x-a}=\lim_{x\nearrow a}\frac{(x-a)f(a)}{x-a}=f(a).$$
From these two relations it follows that $f(x)=2x$ for all $x\in \mathbb{R}$.
I think that my solution works, but what I would like to know is why we can't take $f$ to be decreasing. Assuming that it were decreasing, by reversing the inequalities from above we get that $f(x)=2x,\forall x \in \mathbb{R}$, which is a contradiction. What I want to find out is if there is any other way to reach a contradiction in this case. More precisely, I would like to know what may have motivated the authors of this problem to consider $f$ to be increasing instead of decreasing(I doubt that they just tried both cases and chose the one that actually worked).   
 A: By Lebesgue differentiation theorem, we have $F' = f$ almost everywhere. So we have $f(x) = 2x$ almost everywhere. In particular, for at least two different points $x_1 < x_2$, we have $f(x_1) = 2 x_1$ and $f(x_2) = 2 x_2$, so $f$ can't be non-increasing.
Now, if $f(x_0) \neq 2 x_0$ for some $x_0$, then either $f(x_0) < 2 x_0 - \varepsilon$ or $f(x_0) > 2 x_0 + \varepsilon$. In the former case, there is some $y \in [x_0 - \varepsilon / 2, x_0)$ s.t. $f(y) = 2 y$, and thus $f(y) > f(x_0)$ while $y < x_0$ - so $f$ can't be non-decreasing. Similarly, if $f(x_0) > 2 x_0$, for some $y$ we have $y > x_0$ but $f(x_0) < f(y)$.
So, if $f(x) \neq 2x$ for some $x$, then $f$ can't be non-decreasing. $f$ also can't be non-increasing, so if $f$ is monotone, it should be equal to $2x$.
More generally, if $F(0) = 0$, $F$ is everywhere differentiable and $F'$ is continuous and monotone, then there is only one monotone function - $F'$ s.t. $F(x) = \int_0^x f(t)\,dt$.
If $F$ is just absolute continuous, then there can be several solutions - for example, if $F(x) = |x|$, then for any $f$ s.t. $f(x) = \operatorname{sgn}(x)$ if $x \neq 0$ and $f(0) \in [-1, 1]$ the equality will hold, and $f$ is monotone.
A: $f(x)=2x$ whenever $f$ is continuous at $x$. Since $f$ is increasing it is continuous except at a countable number of points Hence it is continuous on a dense set. Given any $x$ choose an increasing sequence $(x_n)$ and a decreasing sequence $(y_n)$ both tending to $x$ such that $f$ is continuous at each $x_n$ and at each $y_n$. Then $2x_n=f(x_n)\leq f(x) \leq f(y_n)=2y_n$. Letting $n \to \infty$ you get $f(x)=2x$. 
A: In fact, the limit equalities that you wrote are valid if $f$ is supposed to be continuous.
The function $$f(x)=\begin{cases}
2x & x \notin I \\
0 & x \in I
\end{cases}$$
where $I=\{1/n \ ; \ n \in \mathbb N\}$satisfies the requested equality but is not increasing.
The function $f(x)=2x$ for all $x \in \mathbb R$ is the only increasing one satisfying the equality.
