TITLE: Proofs of Young's inequality for convolution
QUESTION [20 upvotes]: For $1\leq p,q \leq \infty$ such that $\frac1p +\frac1q\geq 1$, Young's inequality states $\|f\star g\|_r\leq \|f\|_p\|g\|_q$ (we work on $\mathbf{R}^d$ here), where $1+\frac1r = \frac1p+\frac1q$. Equivalently
\begin{align*}
\|f\|_p=\|g\|_q=\|h\|_{r'}=1\Rightarrow \int_{\mathbf{R}^d}\int_{\mathbf{R}^d}f(x)g(y)h(x+y)\,\mathrm{d}x\,\mathrm{d}x \leq 1.
\end{align*}
The most elementary proof that I know is based on the (generalized) Hölder inequality on $\mathbf{R}^d\times\mathbf{R}^d$ (for three functions), applied on three "mixing" functions $\varphi(x,y)^a \psi(x,y)^b$ where $\{\varphi,\psi\}$ runs over the possible pairs of $\{(x,y)\mapsto f(x); (x,y)\mapsto g(y) ; (x,y)\mapsto h(x+y)\}$ and $a$ and $b$ are adequately chosen.
There is of course a way to guess the correct exponents, but I find this proof a bit tedious and, when it comes to teach it, a bit articial ("consider these three functions and ... the magic happens").
Instead, I am wondering if it is possible to prove it in a different way, remaining at the same level of knowledge. 
The relation between $p,q,r$ rewrites $\frac{1}{r'} = \frac{1}{p'}+\frac{1}{q'}$. This, together with Hölder inequality, proves that any element in $L^{r'}(\mathbf{R}^d)$ is the (ponctual) product of two elements respectively in $L^{p'}(\mathbf{R}^d)$ and $L^{q'}(\mathbf{R}^d)$.
Can we use this to prove (something like)
\begin{multline*}
\sup_{\|h\|_{r'}=1} \int_{\mathbf{R}^d}\int_{\mathbf{R}^d} f(x)g(y) h(x+y)\,\mathrm{d} x\,\mathrm{d} y \\
\leq \sup_{\|\varphi\|_{q'}=1,\|\psi\|_{p'}=1} \int_{\mathbf{R}^d}\int_{\mathbf{R}^d} f(x)g(y) \varphi(x)\psi(y)\,\mathrm{d} x\,\mathrm{d} y\quad ?
\end{multline*}
I did not succeed but still feel that the correspondance between the exponents in the convolution and poncutal products is not a coincidence.
Note that using (a bit of) interpolation theory (I did not check in details) :

Young's inequality can be obtained by Fourier transform (precisely using $\widehat{f\star g}=\widehat{f}\widehat{g}$), at least for exponents in $[1,2]$ and then all the other ones by a duality argument.
The case $\{p,q\}=\{1,\infty\}$ is straightforward and by a duality argument it is possible to recover then $\{p,q\}=\{1,r\}$, and then an interpolation argument should recover some intermediate exponents. 

However, I'd really much appreciate a proof without interpolation.

REPLY [2 votes]: Thanks Daniele and Willie for these nice answers. Willie : I tried this doubling variable thing but got stuck : I was writing $|h(x+y)|$ as the product of two elements respectively in $L^\infty_y(L^{p'}_x)$ and $L^\infty_x(L^{q'}_y)$, which was not the good point of view, nice insight that you got there ! Since the answer of Daniele came first and started the whole thing, I vote for him.
I found yet another way to present the proof, which is somehow linked to your answers. 
As I was writing in my original post, the "easy" case $L^\alpha \star L^{\alpha'}$ implies the $L^\alpha\star L^1$ one by duality because of the formula 
\begin{align*}
\int_{\mathbf{R^d}} (f\star g) h = \int_{\mathbf{R^d}} f(g\star \check{h}).
\end{align*}
Now as usual, w.l.o.g. we can assume $f,g\geq 0$ with $\|f\|_p=\|g\|_q=1$. 
The case $L^r \star L^1$ shows that $f^p \star g^{q/r}$ has  $L^r(\mathbf{R}^d)$ norm less than $1$ and the same thing holds for $f^{p/r}\star g^q$. Thanks to Hölder's inequality, this means that we only need to prove a.e. for some $\theta\in[0,1]$
\begin{align*}
f\star g \leq (f^p \star g^{q/r})^\theta (f^{p/r} \star g^{q})^{1-\theta},
\end{align*}
and another use of Hölder's inequality allows to reduce this to proving a.e.
\begin{align*}
f(x-y)g(y) \leq (f(x-y)^p g(y)^{q/r})^\theta (f(x-y)^{p/r}g(y)^{q})^{1-\theta},
\end{align*}
and in fact we even have an equality of this type. Indeed $p\in[1,r]$ so we have 
$\frac1p = \theta + \frac{1-\theta}{r}$ for some $\theta\in[0,1]$ and adding $\frac1q$ on both sides shows that $1-\theta+\frac{\theta}{r} = \frac{1}{q}$.