How to prove a uniformly continuous function is bounded when its definite integral is bounded?

Question

Given a uniformly continuous functions $f:[0,+\infty)\rightarrow\Bbb R$, if $F(x)=\int_0^xf(y)dy$ is bounded on $[0,+\infty)$, how to prove that $f$ is also bounded on $[0,+\infty)$? I have tried to prove it by contradiction but it seems doesn't work here.

Answer 1

Chose $\epsilon$ and $\delta$ so that $\lvert x-y\rvert \lt \delta => \lvert f(x)-f(y)\rvert \lt \epsilon$ (this is possible since f is uniformly continous). Now assume f is unbounded, we then have that there are sequences ${x_n}, \ {y_n}$ such that for all n: $y_n-x_n = \delta/2$ and $f(x) \ge n $ on $(x_n,y_n)$ (f is uniformly continous and unbounded, so we can find points where f is greater than or equal to n + $\epsilon$, and it is $\ge n$ on the interval in question). Consider now the sequence $\frac{F(y_n)-F(x_n)}{y_n-x_n} = \frac{2}{\delta}(F(y_n)-F(x_n))$. We have on the one hand that the sequence is bounded because F is, but by definition we have: $\frac{F(y_n)-F(x_n)}{y_n-x_n} = \int_{x_n}^{y_n}\frac{f(x)}{y_n-x_n}dx \ge \int_{x_n}^{y_n}\frac{n}{y_n-x_n}dx = n$, which is a contradiction.