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.



Proofs of Young's inequality for convolution
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.


