# Young's convolution inequality: Equivalent representations

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.

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 $$1+\frac{1}{p^*}=2-\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$$.
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 $$|\int \phi\psi|\le c||\psi||_{r'}.$$)