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.