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