Proofs of Young's inequality for convolution

Question

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.

Answer 1

Edit: I realized that the explanation of the former second step of proof below was a little bit obscure since, while entirely correct, did not clarify enough why the choice of integrability exponent is not done by guessing. Therefore I decided to substitute it by a similar but more direct procedure and put the former step 2 in the notes for a brief proof of their equivalence.

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I have not seen the proof of Young's inequality you allude to: however, the answer to your question i.e. whether it is possible to prove it in another way, remaining at the same level of knowledge, is yes. The proof given below is inspired by and follows the one in the nice monograph [1], pp. 26-27 and it is based on the standard and generalized (i.e. involving three or more functions) Hölder's inequalities and by a judicious choice of the integrability exponents associated to two or three factors expressing $| f(y)g(x-y) |$: the introduction of an auxiliary function $h$ is not required.

Young's Inequality. Let $p,q,r\in\Bbb R$ be such that $$ 1\le p\le q\le +\infty, \quad 1+\frac{1}{r}=\frac{1}{p}+\frac{1}{q},\label{1}\tag{1} $$ and let $f\in L^p(\mathbf{R}^d)$ and $g\in L^q(\mathbf{R}^d)$: if $$ f\ast g(x)=\int\limits_{\mathbf{R}^d} f(y)g(x-y)\mathrm{d}y, $$ then $$ \Vert f\ast g\Vert_{r}\le \Vert f\Vert_{p}\Vert g\Vert_q\label{2}\tag{2} $$ Comment. The strategy of the proof goes as follows: first, in every range of values of $p, q, r$ defined by conditions \eqref{1}, we will express $|f(y)g(x-y)|$ as the product of three factors $$ | f(y)g(x-y) |=\big(|f(y)|^p|g(x-y)|^q\big)^\frac{1}{s_1}|g(x-y)|^{\frac{q}{s_2}} |f(y)|^{\frac{p}{s_3}}.\label{step1}\tag{Step 1} $$ Since we want to estimate the the $L^r$ norm of the convolution, we assume $s_1=r$.
Now \ref{step1} implies that the coefficients $s_1, s_2, s_3$ must satisfy the following conditions $$ p\left(\frac{1}{s_1}+\frac{1}{s_3}\right)=1\quad q\left(\frac{1}{s_1}+\frac{1}{s_2}\right)=1,\label{c1}\tag{C1} $$ We thus have a non-homogeneous linear system in the $s_i^{-1}$ variables, $i=1,2,3$ which is uniquely solvable, provided $pq\neq0$, and the second step consist in solving it for the unknown exponents: explicitly $$ \begin{pmatrix} 1 & 0 & 0\\ q & q & 0\\ p & 0 & p \end{pmatrix} \begin{pmatrix} s_1^{-1}\\ s_2^{-1}\\ s_3^{-1}\\ \end{pmatrix}= \begin{pmatrix} {\frac{1}{r}}\\ {1}\\ {1}\\ \end{pmatrix}\iff \begin{pmatrix} s_1^{-1}\\ s_2^{-1}\\ s_3^{-1}\\ \end{pmatrix}= \begin{pmatrix} {\frac{1}{r}}\\ {\frac{1-\frac{q}{r}}{q}}\\ {\frac{1-\frac{p}{r}}{p}}\\ \end{pmatrix}\label{step2}\tag{Step 2} $$ The third and final step is to estimate the $L^r$ norm of the convolution $f\ast g$ by applying to equation \eqref{step1} one of the various forms of Hölder's inequality. This of course can be done since it is easily verified that $$ \frac{1}{s_1} + \frac{1}{s_2} + \frac{1}{s_3}=1.\label{c2}\tag{C2} $$ Proof. If $r=\infty$, then \eqref{2} is a direct consequence of the standard Holder's inequality, since $$ \frac{1}{p}+\frac{1}{q}=1. $$ Assuming $r<+\infty$, \eqref{2} must be verified for the three ranges defined by conditions \eqref{1}, i.e.

1. $1<p<r$ and $1<q< r$;
2. $p=1<q=r$;
3. $p=r$ and $q=1$.

• Case 1: this is the most general case. From \ref{step2} we have $$ \begin{cases} s_1=r\\ \\ s_2=\dfrac{q}{1-\frac{q}{r}}\\ s_3=\dfrac{p}{1-\frac{p}{r}} \end{cases}, $$ and thus equation \eqref{step1} becomes $$ | f(y)g(x-y) |=\big(|f(y)|^p|g(x-y)|^q\big)^\frac{1}{r}|g(x-y)|^{1-\frac{q}{r}} |f(y)|^{1-\frac{p}{r}}.\label{3}\tag{3} $$ Estimating the convolution $f\ast g$ by using \eqref{3} and the generalized Hölder inequality gives $$ |f\ast g(x)|\le \bigg(\int\limits_{\mathbf{R}^d} |f(y)|^p|g(x-y)|^q\mathrm{d}y\bigg)^{\!\frac{1}{r}}\Vert f\Vert_{p}^{1-\frac{p}{r}} \Vert g\Vert_q^{1-\frac{q}{r}}\label{4}\tag{4} $$ and applying the generalized Hölder inequality to \eqref{4} finally gives $$ \begin{split} \Vert f\ast g\Vert_r &\le \Vert f\Vert_{p}^{1-\frac{p}{r}} \Vert g\Vert_r^{1-\frac{q}{r}}\bigg(\int\limits_{\mathbf{R}^d}\mathrm{d}x \int\limits_{\mathbf{R}^d} |f(y)|^p|g(x-y)|^q\mathrm{d}y\bigg)^{\frac{1}{r}}\\  & = \Vert f\Vert_{p}^{1-\frac{p}{r}} \Vert g\Vert_q^{1-\frac{q}{r}} \bigg(\int\limits_{\mathbf{R}^d}|f(y)|^p \mathrm{d}y \int\limits_{\mathbf{R}^d} |g(x)|^q\mathrm{d}x\bigg)^{\!\frac{1}{r}}\\ & = \Vert f\Vert_{p}\Vert g\Vert_r \end{split} $$
• Case 2 and Case 3: in these cases, the right side of equation \eqref{3} reduces to the product of two terms and inequality \eqref{2} is obtained by means of the standard Hölder inequality. Explicitly, $$ | f(y)g(x-y)| = \begin{cases} \big(|f(y)||g(x-y)|^q\big)^\frac{1}{r}|f(y)|^{1-\frac{1}{r}}&\text{ in case 2}\\ \big(|f(y)|^p|g(x-y)|\big)^\frac{1}{r}|g(x-y)|^{1-\frac{1}{r}}&\text{ in case 3} \end{cases}.\qquad\blacksquare $$

Final notes

• This proof is entirely elementary and does not require the "guessing" of the coefficients $s_1, s_2, s_3$, which are instead well defined and calculable.
• Why the second step above is entirely equivalent to the formerly proposed one? Because the non-homogeneous linear system obtained by considering directly, without assuming a priori $s_1=r$, the conditions implied by equation \eqref{3} and by the necessity of using Hölder's inequality i.e. \eqref{c1} and \eqref{c2}, is perfectly equivalent to \eqref{step2}. To see this, is sufficient to write it down and solve it $$ \begin{pmatrix} p & 0 & p\\ q & q & 0\\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} s_1^{-1}\\ s_2^{-1}\\ s_3^{-1}\\ \end{pmatrix}= \begin{pmatrix} {1}\\ {1}\\ {1}\\ \end{pmatrix}\iff \begin{pmatrix} s_1^{-1}\\ s_2^{-1}\\ s_3^{-1}\\ \end{pmatrix}= \begin{pmatrix} {\frac{1}{p}+\frac{1}{p}-1}\\ {1-\frac{1}{p}}\\ {1-\frac{1}{q}}\\ \end{pmatrix}, $$ and then use \eqref{1} to express $s_1, s_2$ and $s_3$ respectively as functions of $r$, $r$ and $q$, $r$ and $p$. Finally, it is wort to note that, in the approach above, the truth of \eqref{c2} is a consequence of the implicit use of \eqref{1}.
• As a final remark, let me say that Besov, Il'in and Nikol’skiĭ prove \eqref{2} ([1], p. 27-28) first for $d=1$ (without showing explicitly \eqref{step1} and \eqref{step2}) and then for vector exponents and $d\ge 2$, i.e. $\mathbf{p}=(p_1,\ldots,p_d)$, $\mathbf{q}=(q_1,\ldots,q_d)$ and $\mathbf{r}=(r_1,\ldots,r_d)$ where each of their $i$-th component satisfies relation \eqref{1}: the result is used in the development of the theory of anysotropic function (Sobolev and Besov) spaces.

Bibliography
[1] Oleg V. Besov, Valentin P. Il’in, Sergei M. Nikol’skiĭ (1978), Integral representations of functions and imbedding theorems. Vol. I, Ed. by Mitchell H. Taibleson. Translation from the Russian. (English) Scripta Series in Mathematics. Washington, D.C.: V. H. Winston & Sons. New York-Toronto-London: John Wiley & Sons, ISBN: 0-470-26540-X, pp. VIII+345, MR0519341, Zbl 0392.46022.

Answer 2

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.

Answer 3

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 $1-\theta+\frac{\theta}{r} = \frac{1}{q}$.