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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?

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  • $\begingroup$ 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
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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.

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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.

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  • $\begingroup$ 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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