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Let $f:[a,b] \to \mathbb{R}$ be a differentiable function with $f(a)=0$ and so that $f'$ is continuous. Show that $$\int_a^b |{f(x)f'(x)}|\:\mathrm{d}x \leq \frac{b-a}{2} \int_a^b f'(x)^2\: \mathrm{d}x.$$

I received a hint for examining the function $G(x):=\int_a^x |f'(t)|\:\mathrm{d}t$ but haven't really gotten anywhere with it. Any tips would be greatly appreciated

Shuster
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1 Answers1

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Maybe too long for a comment, did you tried Holder's inequality ? $$\int_a^b|f(x)f'(x)|\mathrm{d}x \le \left( \int_{a}^{b}|f(x)|^2 \mathrm{d}x \right)^{\frac{1}{2}}\left( \int_{a}^{b}|f'(x)|^2 \mathrm{d}x \right)^{\frac{1}{2}}\tag{1}$$ Now write $$f(x)=\int_{a}^{x}f'(t)\mathrm{d}t \tag{1}$$ Now from (2) we may write $$|f(x)| \le \int_a^x|f'(t)|\mathrm{d}t \le (x-a)^{\frac{1}{2}}\left( \int_a^x |f'(t)|^2 \mathrm{d}t \right)^{1/2}$$ Then we have $$\int_a^b|f(x)|^2\mathrm{d}x \le \int_a^b(x-a)\int_a^x|f'(t)|^2\mathrm{d}t\mathrm{d}x \le \int_a^b |f'(t)|^2\mathrm{d}t\int_a^b(x-a)\mathrm{d}x$$ It's a few steps from here i guess

Siddhartha
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  • We aren't allowed to use Holder's inequality yet. Should have said that, sorry about it. – Shuster Apr 27 '20 at 17:01
  • Well this is mere Cauchy–Bunyakovsky–Schwarz inequality or Cauchy–Schwarz if you like, you'd like to try or should i add more ? – Siddhartha Apr 27 '20 at 17:03
  • The absolute values confuse me a bit. Could you add more please? – Shuster Apr 27 '20 at 17:08
  • Why does the 2nd inequality in the third line hold? – Shuster Apr 27 '20 at 17:48
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    @Shuster That's cauchy-schwarz again, write $$\begin{align}\int_a^x|f'(t)|\mathrm{d}t=\int_a^x|1 \cdot f'(t)|\mathrm{d}t & \le\left (\int_a^x|1|^2\mathrm{d}t \right)^{1/2} \left(\int_a^x |f'(t)|^2\mathrm{d}t\right)^{1/2} \ &\le(x-a)^{1/2}\left(\int_a^x |f'(t)|^2\mathrm{d}t\right)^{1/2}\end{align}$$ – Siddhartha Apr 28 '20 at 00:58
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    I am not sure if we are getting the $\frac{b-a}{2}$ constant from this. It seems like we get $\frac{b-a}{\sqrt{2}}$. Am I missing something here? – r9m Apr 29 '20 at 03:27
  • @r9m no i think you're right, i never calculated the whole thing, the bound i'm getting now is $\frac{b-a}{\sqrt{2}}$ not the desired one, i wonder what's wrong here – Siddhartha Apr 30 '20 at 03:40
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    I think in the 4th line we have $\int_a^b (x-a)\int_a^x |f'(t)|^2,dt ,dx = \frac{(b-a)^2}{2}\int_a^b |f'(t)|^2,dt \color{red}{-} (\frac{1}{2}\int_a^b (x-a)^2|f'(x)|^2,dx)$. Equality occurs in OP's inequality when $f(x) = \lambda(x-a)$. By ignoring the bracketed term additionally we are loosing sharpness here. – r9m Apr 30 '20 at 06:31
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    From eqn $(2)$, $|f(x)| \le \int_a^x |f'(t)|,dt = G(x)$ multiplying $G'(x) = |f'(x)|$ on both sides and integrating over $[a,b]$ seems to be fixing it. – r9m Apr 30 '20 at 06:37
  • Right, this does the trick, thank you. – Siddhartha Apr 30 '20 at 08:36