Norm of convolution of $f$ and $g$ where $f \in L^1(R)$ and $g \in L^p(R)$

Question

Here is the question:

For $f \in L^1(R)$ and $g \in L^p(R)$, define $f*g(x)=\int_{- \infty}^\infty f(x-y)g(y)dy$ .

prove that $f*g\in L^p(R)$ and $||f*g||_p\le||f||_1||g||_p$

This question is from Real and Complex analysis by Walter Rudin, on Chp 9 exercise Q4.

My thought:

imitate the proof in the book. I first prove that $f*g$ is measurable.

By consider the double integral on $|f(x-y)g(y)|$ , I can only reach $||f*g||_p\le||f||_p||g||_p$ instead of $||f*g||_p\le||f||_1||g||_p$

But p-norm and 1-norm of f are not related.

Thanks for any help.

Answer 1

Your problem is also a particular case of Young's inequality:

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Let $1\leq r,p,q\leq\infty$ satisfy $\tfrac1r=\tfrac1p+\tfrac1q-1$. If $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$ and $g\in\mathcal{L}_q(\mathbb{R}^n,\lambda_n)$, then $f*g\in\mathcal{L}_r(\lambda_n)$ and $$ \|f*g\|_r\leq\|f\|_p\|g\|_q $$

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Here is a short proof:

For any $s\geq1$, let $s'$ be its conjugate, that is $\frac1s+\frac1{s'}=1$. From $$ \frac1r+\frac{1}{q'}+\frac{1}{p'}=\frac1r+\Big(1-\frac{1}{q}\Big)+ \Big(1-\frac{1}{p}\Big)=1 $$ we get $$ \big(1-\frac{p}{r}\Big)q'=p\Big(\frac1p-\frac1r\Big)q'= p\Big(1-\frac1q\Big)q'=p\\ \big(1-\frac{q}{r}\Big)p'=q\Big(\frac1q-\frac1r\Big)p'= q\Big(1-\frac1p\Big)p'=q $$ If $1<r,p,r<\infty$, then by H"older's inequality \begin{aligned} &|(f*g)(x)|\leq \int\Big(|f(y)|^{p/r}|g(x-y)|^{q/r}\Big)|f(y)|^{1-p/r}|g(x-y)|^{1-q/r}\,dy\\ &\leq \Big(\int|f(y)|^p|g(x-y)|^q\,dy\Big)^{1/r} \Big(\int|f(y)|^{(1-p/r)q'}\,dy\Big)^{1/q'} \Big(\int|g(x-y)|^{(1-q/r)p'}\,dy\Big)^{1/p'}\\ &=\big(|f|^p*|g|^q(x)\big)^{1/r}\|f\|^{p/q'}_p\|g\|^{q/p'}_q \end{aligned} Hence $$ \int|f*g(x)|^r\,dx\leq\big(\int|f|^p*|g|^q(x)\,dx\big) \|f\|^{pr/q'}_p\|g\|^{qr/p'}_q\\ =\|f\|^p_p\|g\|^q_q\|f\|^{pr/q'}_p\|g\|^{qr/p'}_q =\|f\|^r_p\|g\|^r_q $$ If $r=\infty$ and $q=p'$, then a direct application of H"older's inequality and the symmetric and translation invariance properties of Lebesgue measure shows that $$ \|f*g(x)|\leq\|f\|_p\|g\|_q,\qquad x\in\mathbb{R}^n. $$ Hence $\|f*g\|_\infty\leq\|f\|_p\|g\|_q$

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The particular case $r=p$, $q=1$ can also be proved by an application of the generalized Minkowski inequality (Minkowski's integral inequality): For any $1\leq p<\infty$, and any measurable function $\phi:(X,\mathscr{B},\mu)\otimes(Y,\mathscr{F},\nu)\rightarrow\mathbb{R}$, and $\mu$, $\nu$ are $\sigma$--finite,

\begin{aligned} \Big(\int_X\Big|\int_Y \phi(x,y)\, \nu(dy)\Big|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\leq \int_Y \Big(\int_X |\phi(x,y)|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\,\nu(dy) \end{aligned}

with $\nu(dy)=f(y)\,dy$, $\mu(dx)=dx$ and $\phi(x,y)=f(x-y)$.

\begin{aligned}\Big(\int\Big|\int g(x-y)f(y)\,dy\Big|^p\,dx\Big)^{1/p}&\leq \int\Big(\int|g(x-y)|^p\,dx\Big)^{1/p}|f(y)|\,dy\\ &=\int\Big(\int|g(x)|^p\,dx\Big)^{1/p}|f(y)|\,dy=\|g\|_p\|f\|_1 \end{aligned}

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Answer 2

Let $q$ be the conjugate exponent of $p$. If $h\in L^q(\Bbb R)$ with $\|h\|_q \le 1$, then $$\int_{-\infty}^\infty |(f * g)(x)h(x)|\, dx \le \int_{-\infty}^\infty |g(y)| \left(\int_{-\infty}^\infty |f(x - y)h(x)|\, dx\right)\, dy$$ Use Hölder's inequality and the translational invariance of the Lebesgue measure to show that the innermost integral is dominated by $\|f\|_p \|h\|_q$, which is no greater than $\|f\|_p$ since $\|h\|_q \le 1$. So the integral above is dominated by $\int_{-\infty}^\infty |g(y)| \|f\|_p\, dy = \|f\|_p \|g\|_1$. Take the supremum over all such $h$ to deduce the result.