Let $f\in \mathbb{L^1{\mathbb(R)}}$ and define for a fixed $h>0$, $f_h(x)=\frac{1}{2h}\int_{x-h}^{x+h} f(t)dt$. Prove that $\int_\mathbb{R}{|f_h(x)|}dx\leq \int_\mathbb{R}{|f(x)|}dx$.

Hint: Prove $f_h(x)=\int_\mathbb{R}f(x-t)\psi_h (t)dt$, where $\psi_h (t)=\frac{1}{2h}\mathbb{1}_{[-h,h]}$.

I think I will have to use Lebesgue differentiation theorem along with the dominated convergence theorem. Any help on how to proceed?

  • $\begingroup$ Nope. It was a typo. sorry about that. $\endgroup$ – SL_MathGuy Dec 9 '19 at 9:00
  • $\begingroup$ So in the end this is not at all an application of the Lebesgue différentiation theorem. Rather, it is a lemma, that is going to be used in the proof. $\endgroup$ – Giuseppe Negro Dec 11 '19 at 16:53

Using the hint you have that $$ \int|f_h(x)|\,\mathrm d x\leqslant \iint |f(x-t)||\psi _h(t)|\,\mathrm d t\,\mathrm d x $$ Now use Tonelli's theorem and the translation invariance of the Lebesgue measure to finish.


Assuming the hint, which is not hard to prove, one should use Young's inequality for convolutions, which states $$ 1 + \frac1r = \frac1p + \frac1q \implies \|f*g\|_{L^r} \le \|f\|_{L^p} \|g\|_{L^q}$$

since $ 1 + 1/1 = 1/1 + 1/1$ and $\|\phi_h\|_{L^1} = 1$, we get

$$\|f_h\|_{L^1} = \|f*\psi_h\|_{L^1} \le \|f\|_{L^1}\|\psi_h\|_{L^1} = \|f\|_{L^1},$$ exactly as needed.

  • $\begingroup$ True, but this is the easy case of Young's inequality, you can prove it very quickly with Fubini and Hölder as Masacroso did. $\endgroup$ – Giuseppe Negro Dec 9 '19 at 9:39
  • $\begingroup$ @GiuseppeNegro Yeah you're right. I also didn't see his answer because I pressed send many minutes after typing, but I'll just leave this here $\endgroup$ – Calvin Khor Dec 9 '19 at 9:47
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    $\begingroup$ That's fine, it is good to see many answers to the same question. $\endgroup$ – Giuseppe Negro Dec 9 '19 at 9:57
  • $\begingroup$ I've not learnt Young's inequality yet. Anyways, it's good to know alternative methods . $\endgroup$ – SL_MathGuy Dec 9 '19 at 10:06
  • $\begingroup$ @SL_MathGuy its pretty useful and not hard to remember, to the point where I reflexively used it without noticing its easy to do from first principles :) $\endgroup$ – Calvin Khor Dec 9 '19 at 10:08

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