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... is the result in $L_{p}$?

A remark in my notes says yes but I can't see how to verify it.

As was pointed out to me in a previous question I asked last night, I need to show that the following integral is finite:

$$\int_{-\pi}^{\pi}|\int_{-\pi}^{\pi}f(t-s)\phi_{n}(s)ds|^{p}dt < \infty$$.

One of the properties of a summability kernel is that there exists a $C > 0$ such that $\int_{-\pi}^{\pi}|\phi_{n}(t)|dt\leq C$ for every $n\geq 1$. I feel like this could help if I could get $\phi$ by itself

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Yes, apply Young's inequality with $r=p$ and $q = 1$, see also: en.wikipedia.org/wiki/Convolution#Integrable_functions –  t.b. Oct 5 '11 at 0:21
    
Thank you this is very helpful. –  Kyle Schlitt Oct 5 '11 at 0:28

1 Answer 1

up vote 10 down vote accepted

The way I usually prove Young's Inequality is using a couple of applications of Hölder's inequality to prove $$ \left|\int f(x)\;g(x)\;h(x)\;\mathrm{d}x\right|\le\|f\|_p\|g\|_q\|h\|_r\tag{1} $$ where $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2$. Then apply $(1)$ in a tricky way to show $$ \sup_{\|h\|_{L^r}=1}\left|\int\int f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\le\|f\|_p\|g\|_q\|h\|_r\tag{2} $$ Inequality $(2)$ says that $\|f\ast g\|_s\le\|f\|_p\|g\|_q$ where $\frac{1}{r}+\frac{1}{s}=1$, that is $\frac{1}{s}=\frac{1}{p}+\frac{1}{q}-1$.


Tricky Application of $\mathbf{(1)}$: Since $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2$, we have the following 7 relations: $$ \begin{align} \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= 1\\ \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{p}\text{ so that }p\left(1-\frac{1}{r}\right) + p\left(1-\frac{1}{q}\right) = 1\\ \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) &= \frac{1}{q}\text{ so that }q\left(1-\frac{1}{r}\right) + q\left(1-\frac{1}{p}\right) = 1\\ \left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{r}\text{ so that }r\left(1-\frac{1}{p}\right) + r\left(1-\frac{1}{q}\right) = 1 \end{align} $$ Therefore, $$ \begin{align} &\left|\int\int f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\\ &\le\int\int|f(x-y)|\;|g(y)|\;|h(x)|\;\mathrm{d}y\;\mathrm{d}x\\ &={\small\int\int|f(x-y)|^{p(1-1/r)}|g(y)|^{q(1-1/r)}\;|g(y)|^{q(1-1/p)}|h(x)|^{r(1-1/p)}\;|f(x-y)|^{p(1-1/q)}|h(x)|^{r(1-1/q)}\;\mathrm{d}y\;\mathrm{d}x}\\ &\le\left(\int\int|f(x-y)|^p|g(y)|^q\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/r}\\ &\times\left(\int\int|g(y)|^q|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/p}\\ &\times\left(\int\int|f(x-y)|^p|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/q}\\ &=\left(\int|f(x)|^p\;\mathrm{d}x\right)^{1/p}\left(\int|g(x)|^q\;\mathrm{d}x\right)^{1/q}\left(\int|h(x)|^r\;\mathrm{d}x\right)^{1/r} \end{align} $$

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♦ sorry could you explain the last passage you have done? Why is this true? I really don't get this. I've tried to interchange the variables thanks to fubini, but the last integral ((∫∫|f(x−y)|p|h(x)|rdydx)) depends only to x... –  Ale Jan 20 at 13:55
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@Ale: Use Fubini to see that $$\begin{align}\int\int|f(x-y)|^p|g(y)|^q\,\mathrm{d}y\,\mathrm{d}x&=\|f\|_p^p\i‌​nt|g(y)|^q\,\mathrm{d}y\\&=\|f\|_p^p\|g\|_q^q\end{align}$$ and similarly for the other integrals. –  robjohn Jan 20 at 14:44

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