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Let $p\ge1$, $f\in L^p(\mathbb R^d)$ and $g\in L^1(\mathbb R^d)$. I've read that the inequality $$\left\|g\ast f\right\|_{L^p}\le\left\|g\right\|_{L^1}\left\|f\right\|_{L^p}\tag1$$ would follow from applying the Minkowski inequality to $$F(x,y):=g(y)f(x-y)$$ and using that $$\left\|\lambda_yf\right\|_{L^p}=\left\|f\right\|_{L^p}\tag2,$$ where $(\lambda_yf)(x):=f(x-y)$.

However, I absolutely don't get why $(1)$ does hold. How can we show it

The "usual" version of Young's inequality, as stated on Wikipdia, is clear to me.

0xbadf00d
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1 Answers1

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$\|\int f(x-y)g(y)dy\|_p \leq \int \|f(x-y)g(y)\|_p dy$ where the norm inside the integral is w.r.t. the variable $x$. Since $g(y)$ is a constant we get $\|\int f(x-y)g(y)dy\|_p \leq \int |g(y)| \|f(x-y)\|_p dy=\|f\|_p \int |g(y)|dy=\|f\|_p \|g\|_1$.

Kavi Rama Murthy
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  • How did you obtain your first inequality? It seems like you've used that $\left|\int f(x-y)g(y):{\rm d}x\right|^p\le\int|f(x-y)g(y)|^p:{\rm d}x$, but why does this hold? Looks like Jensen's inequality, but we are dealing not with a finite measure here ... – 0xbadf00d Jan 09 '21 at 07:09
  • That is exactly Minkowski's Inequality for integrals. See https://math.stackexchange.com/questions/2355672/prove-minkowskis-inequality-for-integrals @0xbadf00d – Kavi Rama Murthy Jan 09 '21 at 07:15
  • Usual Minkowski's Inequality: $L^{p}$ norm of a sum is less the or equal to the sum of the $L^{p}$ norms. Minkowski's Inequality for integrals: $L^{p}$ norm of an integral is less the or equal to the integral of the $L^{p}$ norms. @0xbadf00d – Kavi Rama Murthy Jan 09 '21 at 07:18
  • Thank you for the link. I was only aware of the usual Minkowski inequality. – 0xbadf00d Jan 09 '21 at 08:07
  • The inequality you are trying to prove can easily be reduced to the one in which $f$ and $g$ are replaced by $|f|$ and $|g|$ in which case the proof works without that condition. @0xbadf00d – Kavi Rama Murthy Jan 09 '21 at 09:02