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According to Wikipedia Young's inequality for convolutions states that

For functions $f \in L^p$ and $g \in L^q$ one has

$|| f*g ||_r \leq ||f||_p ||g||_q$ $\hspace{6.75cm}$ (Eq. 1)

with $1/p + 1/q = 1 + 1/r$ and $1 \leq p,q,r$ . Here $*$ represents a Fourier convolution and $||\cdot||_p$ stands for the usual $L^p$-Norm.

Equivalently, if $1/p + 1/q + 1/r = 2$ and $1 \leq p,q,r$ then

$\int \int f(x) g(x-y) h(y) \mathrm d x \mathrm d y \leq ||f||_p ||g||_q ||h||_r$ $\hspace{2.5cm}$ (Eq. 2)

holds true.

My question: I am wondering what the connection between (Eq. 1) and (Eq. 2) actually is. Are they truly identical? Is it possible to derive (Eq. 2) from (Eq. 1) without too much effort?$^*$ What properties does $h$ have to fulfill? Can it be any suitably integrable function?

$^*$ I have already tried to read the original Papers of Young and Brascamp & Lieb but I was not able to fully understand the derivation.

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It is a result of Riesz representation theorem. It says that for $p\in [1,\infty]$, and $p^*=\frac{p}{p-1}$, it holds $$ \|f\|_{L^p}\le M \;\Longleftrightarrow \;\Big|\int fg \mathrm dx\Big|\le M \|g\|_{L^{p^*}},\quad\forall g\in L^{p^*}. \tag{*} $$ Now, note that $$ \int \int f(x)g(x-y)h(y)\mathrm dx\mathrm dy=\int f(x)\left[\int g(x-y)h(y)\mathrm dy\right]\mathrm dx\le \|f\|_{L^p}\|g\|_{L^q}\|h\|_{L^r} $$ implies $$ \Big|\int f(x)\left[\int g(x-y)h(y)\mathrm dy\right]\mathrm dx\Big|\le \|f\|_{L^p}\|g\|_{L^q}\|h\|_{L^r}\tag{**} $$ for all $f\in L^p$. (By considering the inequality for $-f$, we can strengthen the bound.) By $(*)$, it follows that $(**)$ is equivalent to $$ \Big\lVert\int g(\cdot-y)h(y)\mathrm dy\Big\rVert_{L^{p^*}}\le\|g\|_{L^q}\|h\|_{L^r}. $$ Finally note that $1+\frac{1}{p^*}=2-\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$.

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  • $\begingroup$ You should add this to the Wikipedia page! $\endgroup$ – Minze Jan 16 '19 at 14:13
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Hint: If $1/p+1/q=1+1/r$ and $1/p+1/q+1/r'=2$ then $$\frac 1r+\frac1{r' }=1.$$(Hence $||\phi||_r\le c$ is equivalent to $|\int \phi\psi|\le c||\psi||_{r'}.$)

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