Now I'm reading the Young inequality. It says that if $f \in L^p(R)$, $g \in L^q(R)$, $1\leq p,q\leq \infty$, $1/p+1/q\geq 1$. Then how could we have the following inequalities:
$$\|f*g\|_q\leq \|g\|_q \|f\|_1$$
$$\|f*g\|_\infty\leq \|g\|_q \|f\|_{q^{'}},\quad (1/q+1/q^{'}=1)$$
$f*g$ means the convolution
The two inequalities seems not obvious, the second one is similar with the holder inequality, but actually not.
Could someone give me some hints or helpful links? Thanks a lot!
Here is the first inequality using Jensen instead of Young.
If $\|f\|_1=1$, we can use Jensen's inequality on the measure $\mathrm{d}\mu=f(x-y)\,\mathrm{d}y$ to get $$ \left|\int f(x-y)g(y)\,\mathrm{d}y\right|^p\le\int |f(x-y)||g(y)|^p\,\mathrm{d}y $$ Then integrate with respect to $x$ to get $$ \|f\ast g\|_p^p\le\|f\|_1\|g\|_p^p=\|f\|_1^p\|g\|_p^p $$ since $\|f\|_1=1$. This equation now scales nicely with $\|f\|_1$, so we can remove the restriction that $\|f\|_1=1$ to get $$ \|f\ast g\|_p\le\|f\|_1\|g\|_p $$
As mentioned by julien, the second inequality is simply Hölder's inequality.
The second inequality is easy. The first one is tricky, and was already proved by robjohn here as an answer to a different question. I just answer this for the sake of completeness because I couldn't find a proof of the second inequality here.
2) For the second inequality, I assume $1<q<\infty$ as the cases $q=1$ and $q=\infty$ are trivial. First note that $$ |(f\ast g)(x)|=\Big|\int f(x-y)g(y)dy\Big|\leq \int |f(x-y)||g(y)|dy $$ then by Hölder and variable change $$ \leq \left(\int|f(x-y)|^{q'} \right)^\frac{1}{q'} \left(\int|g(y)|^{q} \right)^\frac{1}{q}=\left(\int|f(y)|^{q'} \right)^\frac{1}{q'} \left(\int|g(y)|^{q} \right)^\frac{1}{q} $$ whence $$ \|f\ast g\|_\infty \leq \|f\|_{q'}\|g\|_q. $$
1) The first inequality is a particular case of Young's inequality $$ \|f\ast g\|_s\leq \|f\|_p\|g\|_q\qquad\mbox{when}\quad \frac{1}{p}+\frac{1}{q}=1+\frac{1}{s}, \quad 1\leq p,q,s\leq \infty $$ which is proved by robjohn in this thread. This does not cover the cases when $p,q,$ or $s$ are infinite. But $s=\infty$ is the case above, and your inequality has $p=1$, which yields $s=q$.
I'll give a proof just using Holder's Inequality so that it may easier for you to understand, but as the above comments said the proof for these inequalities is already somewhere else.
Proof for the first: By Holder's Inequality we have \begin{align*} \left|\int f(x-y)g(y)dy\right|\le&\int|f(x-y)|^{1/q}|g(y)||f(x-y)|^{1-1/q}dy \\ \le&\left(\int|f(x-y)||g(y)|^qdy\right)^{1/q}\left(\int|f(x-y)|dy\right)^{1-1/q} \\ =&\|f\|_1^{1-1/q}\left(\int|f(x-y)||g(y)|^qdy\right)^{1/q},\qquad\forall x. \end{align*} Then by Fubini Theorem we have \begin{align*} \|f*g\|_q^q\le&\|f\|_1^{q-1}\int\int|f(x-y)||g(y)|^qdydx \\ =&\|f\|_1^{q-1}\int\left(|g(y)|^q\int|f(x-y)|dx\right)dy \\ =&\|f\|_1^{q}\|g\|_q^q. \end{align*} Taking off $q$ we prove the first one.
The proof for the second one: By Holder's Inequality we have \begin{align*} \left|\int f(x-y)g(y)dy\right|\le\left(\int|f(x-y)|^{q'}dy\right)^{1/q'}\left(\int|g(y)|^qdy\right)^{1/q}=\|f\|_{q'}\|g\|_{q},\qquad\forall x. \end{align*} By taking the supremum we get the second one.
$$ \begin{align} \left(\int f g,\mathrm{d}x\right)^p &=|g|_q^{pq}\left(\int\frac{f}{g^{q-1}} \left(\frac{g}{|g|_q}\right)^q,\mathrm{d}x\right)^p\ &\le|g|_q^{pq}\int\left(\frac{f}{g^{q-1}}\right)^p \left(\frac{g}{|g|_q}\right)^q,\mathrm{d}x\ &=|g|_q^{pq-q}\int f^pg^{p+q-pq},\mathrm{d}x\ &=|g|_q^p|f|_p^p \end{align} $$
– robjohn May 15 '13 at 18:42
$|f\ast g|_r\le |f|_p\cdot|g|_q$, where $1+1/r=1/p+1/q$ and $p,q,r\in [1,\infty]$. Unfortunately, I can't remember its name (I was taught it was Holder's inequality, but wiki doesn't agree with this name).
– TZakrevskiy May 15 '13 at 14:42