Let $f:[a,b] \rightarrow \mathbb{R}$ with $f \in C^1$ and $f(a)=0$. We want to show that $$\int_{a}^{b}f^2(x)\mathrm{d}x \leq (b-a)^2\int_{a}^{b}[f'(x)]^2\mathrm{d}x$$

There is a hint namely to consider $$f(x)=\int_{a}^{x}f'(t)\mathrm{d}t$$ because $f(a)=0$ and then use the Cauchy-Schwarz inequality for integrals.

So I tried doing the following: $$f^2(x)=\left ( \int_{a}^{x}f'(t)\mathrm{d}t \right )^2\leq\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t\leq\left ( b-a \right )\int_{a}^{b}\left [ f'(t) \right ]^2\mathrm{d}t$$

and finally integrate the expression with respect to $x$ over $[a,b]$. My first question is whether my approach has any mistakes (I am mainly worried since in the textbook the last inequality has $x$ as instead of $b$ as the higher bound of integration and I am wondering whether it is a mistake, because when integrating again the result only follows if the integral in the last inequality is independent of $x$) and whether this this strategy works in general with higher powers of functions and their derivatives (for example using some variation of the Hölder inequality or something).

Then there is a challenge without hint or solution that asks us to prove that there is a smaller upper bound, namely $$\int_{a}^{b}f^2(x)\mathrm{d}x \leq \frac{(b-a)^2}{2}\int_{a}^{b}[f'(x)]^2\mathrm{d}x$$

Here I got stuck. I tried the same approach $$f^2(x)=\left ( \int_{a}^{x}f'(t)\mathrm{d}t \right )^2\leq\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t$$ and I tried finding a number $x$ such that $x-a=\frac{b-a}{2}$ which means $x=\frac{b+a}{2}$ so that I can say $$\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t\leq\left ( \frac{b-a}{2} \right )\int_{a}^{\frac{b+a}{2}}\left [ f'(t) \right ]^2\mathrm{d}t$$ and then integrate to get the desired result (as integration over $[a,b]$ would just multiply the right hand side with $b-a$ and we would be done). However this only works when $x\in [a,\frac{b+a}{2}]$ and I cannot seem to find a way to prove it for the rest of the interval. So how do I proceed here?

Finally, as I am new to integral inequalities, any suggestions on how one should study to learn how to solve basic problems or where to study from are welcome. Thanks in advance!


1 Answer 1


Your proof for the first inequality is correct.

For the second one: As you already showed, the inequality $$f^2(x)=\left ( \int_{a}^{x}f'(t)\mathrm{d}t \right )^2\leq\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t$$ holds. Now use this inquality inside the integral and integrate by parts \begin{align} \int_{a}^{b}f^2(x)\mathrm{d}x&\leq \int_{a}^{b}\left (\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t\right )\mathrm{d}x\\& =\frac{(b-a)^2}{2}\int_{a}^{b}\left [ f'(t) \right ]^2\mathrm{d}t-\int_{a}^{b}\frac{(x-a)^2}{2}\left [ f'(x) \right ]^2\mathrm{d}x\\& \leq \frac{(b-a)^2}{2}\int_{a}^{b}\left [ f'(x) \right ]^2\mathrm{d}x. \end{align} Here we have used that one term disappears while partial integration and the integrands of the term with the minus sign are all greater than zero for the last inequality.

  • $\begingroup$ ah thanks, it was in front of me this whole time and I couldn't see it $\endgroup$
    – Summand
    Commented Jan 6, 2021 at 18:16

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