If $f \in C^1([0,a])$ and $f(0) = 0$, prove that $$\int_0^a |f(x)|^2|f'(x)|dx \le \frac{a^2}{3}\int_0^a |f'(x)|^3dx$$ when does the equality hold?

My attempt: Since $f(0) = 0$, we have $$f(x) = \int_0^x f'(t)dt$$ Plugging it into the LHS of the inequality the LHS becomes $$\int_0^a |f(x)|^2|f'(x)|dx = \int_0^a \left|\int_0^x f'(t)dt \right|^2 f'(x)dx$$ Due to the Cauchy-Bunyakovskiy inequality for $1$ and $f'(t)$, we have $$\begin{align} \int_0^a \left|\int_0^x f'(t)dt \right|^2 f'(x)dx &\le \int_0^a\left(\int_0^x 1^2 dt\right)\left(\int_0^x [f'(t)]^2dt \right)f'(x)dx \\ &= \int_0^a \int_0^x xf'(x)[f'(t)]^2dt dx \\ &= \int_0^x [f'(t)]^2dt\int_0^axf'(x)dx \\ &= \left(xf(x)|_0^a - \int_0^a f(x)dx \right)\int_0^x [f'(t)]^2dt \\ &= af(a)\int_0^x [f'(t)]^2dt - \int_0^a f(x)dx \int_0^x [f'(t)]^2dt \end{align}$$ I got stuck here. I cannot figure out the best way to observe the result.

Any help is greatly appreciated.


One can proceed similarly as in Let $f \in AC[0,1],f(0)=0 $. Show that $\int_0^1 \lvert f(x)f'(x)\rvert\,dx \leq \int_0^1\lvert f'(x)\rvert^2 \, dx$:

We have $$ |f(x)| = \left| \int_0^1 f'(t) \, dt \right| \le \int_0^x |f'(t)| \, dt =: F(x) \, , $$ with equality for all $x$ if and only if $f'$ has constant sign on $[0, a]$.

Then $F'(x) = |f'(x)|$ and $$ \int_0^a |f(x)|^2|f'(x)|dx \le \int_0^a F(x)^2 F'(x) \, dx = \frac 13 F(a)^3 \, . $$

Then $F(a)$ can be estimated Hölder's inequality (with $p=3/2$ and $q=3$), this gives $$ F(a) = \int_0^a 1 \cdot |f'(x)|\, dx \le a^{2/3} \cdot \left( \int_0^a |f'(x)|^3 \, dt \right)^{1/3} \, , $$ with equality if and only if $|f'|$ is constant on $[0, a]$.

Combining these estimates gives the desired result.

Remark: In the same way one can prove that $$ \int_0^a |f(x)|^p|f'(x)| \, dx \le \frac{a^p}{p+1}\int_0^a |f'(x)|^{p+1} \, dx $$ for any $p \ge 1$.


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