Norm of convolution of $f$ and $g$ where $f \in L^1(R)$ and $g \in L^p(R)$ 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.
 A: Your problem is also a particular case of Young's inequality:

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
$$
\|f*g\|_r\leq\|f\|_p\|g\|_q
$$

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 $\|f*g\|_\infty\leq\|f\|_p\|g\|_q$

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}

A: 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.
