Proofs of Young's inequality for convolution 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 artificial ("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 (pointwise) 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.
 A: Here's an alternative proof that is similar but also slightly different from the one proposed by Daniele Tampieri. We will attack this through a more general theorem. Throughout we will denote by $x$ an element of $\mathbb{R}^m$ and $y$ an element of $\mathbb{R}^n$. Throughout $r', s'$ will denote the Holder conjugates of $r, s$ respectively. 

Theorem
  Let $1 \leq r \leq s \leq \infty$, and suppose $k \in L^s_x L^r_y \cap L^s_y L^r_x$, with the intersection norm 
  $$ \| k\|_{\cap} = \max( \| k\|_{L^s_x L^r_y}, \|k\|_{L^s_y L^r_x}) $$
  Then for any $\theta\in [0,1]$, and 
  $$ \frac{1}p := \frac{1-\theta}{s'} + \frac{\theta}{r'},\qquad \frac{1}q := \frac{1-\theta}{r'} + \frac{\theta}{s'}$$
  we have
  $$ \int k(x,y) f(x) g(y) ~\mathrm{d}y ~\mathrm{d}x \leq \|k\|_\cap \|f\|_{L^p_x} \|g\|_{L^q_y}. $$

This theorem implies Young's inequality if we set $m = n$, $s = \infty$, and $k(x,y) = \tilde{k}(x-y)$. 
For this theorem to hold, it suffices we show that the function $f(x) g(y)$ belongs to the dual of $L^s_x L^r_y \cap L^s_y L^r_x$, or that it suffices to show
$$ f(x) g(y) \in (L^{s'}_x L^{r'}_y + L^{s'}_y L^{r'}_x) $$
Without loss of generality we can assume $\|f\|_p = \|g\|_q = 1$. 
Observe that the definition of $p$ and $q$ implies that
$$ 1 = (1-\theta) \frac{p}{s'} + \theta \frac{p}{r'} = (1-\theta) \frac{q}{r'} + \theta \frac{q}{s'} $$
therefore we can write
$$ f(x)g(y) = [ f(x)^{p/s'} g(y)^{q/r'}]^{1-\theta} \cdot [f(x)^{p/r'} g(y)^{q/s'}]^{\theta} $$
So Young's inequality (for products) implies the pointwise bound
$$ f(x) g(y) \leq (1-\theta)  f(x)^{p/s'} g(y)^{q/r'} + \theta f(x)^{p/r'} g(y)^{q/s'} $$
The first term on the right has $L^{s'}_x L^{r'}_y$ norm bounded by $(1-\theta)$ and the second term has $L^{s'}_y L^{r'}_x$ norm bounded by $\theta$. This proves the theorem. 
A: I found the proof given by Hormander in the Analysis of Linear Partial Differential Operators to be quite clear and instructive. It can be found in the first volume, pages 116-117, and for convenience I will reproduce it here.

$|u_1 * u_2 * \dots * u_k(0)| \leq \|u_1\|_{p_1}\dots\|u_k\|_{p_k}$ if $\sum 1/p_i = k - 1$ and $1 \leq p_i \leq \infty$.

When for some $p_i$ we have $p_i = \infty$, then every other $p_j = 1$, and thus by Fubini, we have the assertion.
The other cases can be reduced to this case by using a convexity argument: letting $v_i = |u_i|^{p_j}$, we have to prove $v_1^{t_1} *\dots *v_k^{t_k}(0) \leq 1$ assuming that $\int v_i = 1$, $0 \leq v_i$, $0 \leq t_i \leq 1$ and $\sum t_i = k - 1$.
However $v_1^{t_1} * \dots *v_k^{t_k}(0)$ as a function of $t$ is convex, since the integrand as a function of $t$ is convex.
Then, $(t_1, \dots t_k) = \sum (1 - t_i) f_i$ where $f_i$ is the vector with $i$-th component $0$ and everywhere else $1$, and this reduces the assertion to the first case.
Now letting $u_1 *\dots * u_k = u$, $|u * v(0)| \leq \|u_1\|_{p_1} \dots \|u_k\|_{p_k} \|v\|_{q'}$, where $1/p_1 + \dots + 1/p_k = k - 1 + 1/q$, and $q'$ is the holder conjugate of $q$.
Viewing $(u*v)(0)$ as a functional on $L^{q'}$, we have by duality that $\|u\|_q \leq \|u_1\|_{p_1} \dots \|u_k\|_{p_k}$.
The case $k = 2$ is the desired case.
A: 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 $1-\theta+\frac{\theta}{r} = \frac{1}{q}$.
