Absolute value of definite integral: If $\big|\int^b_af(x) dx \big|\le(b-a)^2$ prove that $f(x)=0$ for all $x \in \mathbb{R}$.

Question

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous function. If for any bounded interval $[a,b]\subset\mathbb{R}$ we have $$\Bigg|\int^b_af(x) dx \Bigg|\le(b-a)^2$$ Prove that $f(x)=0$ for all $x \in \mathbb{R}$.

I've been looking at this statement for a few hours. It seems very simple but I can't really see a good place to start. Can someone point me in the right direction?

Answer 1

Hint: Consider the function \begin{align} F_a(x) = \int^x_a f(t)\ dt \end{align} then we see that $F_a(a) = 0$ and \begin{align} |F_a(x)-F_a(a)| \le (x-a)^2. \end{align} Now see what you can say about $f(a)$.

Remark: Under the suggestion of @zipirovich, I made some modification to the hint. The solution will be in the comments.

Answer 2

Proceeding by contradiction, suppose WLOG that $f(0) = \delta > 0$. By choosing $\varepsilon > 0$ small enough $$ \frac{1}{\varepsilon} \int_{-\varepsilon/2}^{\varepsilon/2} f(x) \,\mathrm{d}x $$ can be made arbitrarily close to $\delta$ (hint: mean value theorem for integrals) and greater than $\varepsilon$. This is a contradiction, of course.

Answer 3

For every $x\in\mathbb{R}$ we consider interval $[x,x+\epsilon]$ then by mean-value theorem for integrals there exists $x\leq\xi\leq x+\epsilon$ $$\Big|\int_{x}^{x+\epsilon}f(t)dt\Big|=\Big|\epsilon f(\xi)\Big|\leq\epsilon^2$$ now $\epsilon\to0$, then $f(x)=0$

Answer 4

We know that $\dfrac{1}{2\epsilon}\displaystyle\int_{B_{\epsilon}(x)}f(y)dy\rightarrow f(x)$ by Lebesgue Differentiation Theorem (I know this is a little bit over killed, since many people have provided answer here, so I come out with a little bit fancier one), as $\left|\displaystyle\int_{B_{\epsilon}(x)}f(y)dy\right|\leq 4\epsilon^{2}$, then $\dfrac{1}{2\epsilon}\displaystyle\int_{B_{\epsilon}(x)}f(y)dy\rightarrow 0$, so $f(x)=0$.