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.

  • 4
    $\begingroup$ William Beckner gave one proof using rearrangement techniques (with sharp constants). Later Brascamp-Lieb gave a different proof using heat flows. Also Franck Barthe has a new proof using optimal mass transportation arguments. Recently other proofs also appeared but essentially they are variations of these ones I think. My favorite one is Gaussian version of Young’s convolution inequality with constant 1 which recovers the usual one by taking a limit in a certain way. $\endgroup$ Commented Jul 19, 2019 at 13:42
  • $\begingroup$ By the way, note that interpolation covers all exponents: fixing $q$, you prove it first in the easy cases $p=1$ and $p=q'$. All other values of $p$ are intermediate between these two. However, you need the off-diagonal interpolation, which is far from being elementary. $\endgroup$
    – Mizar
    Commented Jul 20, 2019 at 14:28
  • $\begingroup$ "Ponctual" (of the product) means "pointwise", right? $\endgroup$
    – LSpice
    Commented Aug 10, 2022 at 20:20
  • $\begingroup$ @LSplice, yes pointwise ! I'll correct it. $\endgroup$ Commented Aug 13, 2022 at 6:27

4 Answers 4


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.

  • $\begingroup$ I must add that one can avoid convergence questions by taking $u_k \in C_0$ first, and then after achieving the inequality, take the $L^{p_k}$ limit for each $u_k$. $\endgroup$
    – user489890
    Commented Feb 11, 2023 at 18:20
  • $\begingroup$ Thanks for this tricky answer, it's really nice. $\endgroup$ Commented Feb 12, 2023 at 19:59

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.

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.

[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.

  • 4
    $\begingroup$ Awesome! This proof also works for general kernels instead of convolution. Replace $g$ by $k(x,y)$ and look for the operation $f(y) \mapsto Tf := \int f(y) k(x,y) dy$. Essentially the same proof shows that if $k \in L^\infty_x L^q_y \cap L^\infty_y L^q_x$, then the above mapping maps $L^p$ to $L^r$. $\endgroup$ Commented Jul 22, 2019 at 13:02
  • $\begingroup$ @WillieWong, nice observation. I did not noticed this application to general kernels $k$, because Besov et al. do not mention further applications, and for them this is only a tool for the later developments. Thanks for sharing your insight. $\endgroup$ Commented Jul 22, 2019 at 13:15
  • 2
    $\begingroup$ @DanieleTampieri Thank you for your great answer. The idea in step 1, which is decomposing $\vert\,f(x)g(x-y)\,\vert$ is innovative enough (I kept trying to find coefficients to fit in the generalized Minkowski Inequality but failed). May I ask what inspired you to write down the term $\vert\,f(x-y)\,\vert^p\vert\,g(y)\,\vert^q$? $\endgroup$ Commented Dec 10, 2020 at 16:23
  • $\begingroup$ @SanaeKochiya you are welcome. I found this idea in the monograph [1] and explained it in order to show that there is no magic in the choice of coefficients, provided you know the exponents of integrability $p$ and $q$ which characterize the single functions $f$ & $g$: and the term $|f(x-y)|^p |g(y)^q|$ is singled out since you can deal with it by using Ḧolder's inequality, since the integrals of each single term of the product is finite, obviously even before you extract their root and you can play with them by raising to the needed power. $\endgroup$ Commented Dec 10, 2020 at 16:45
  • $\begingroup$ @DanieleTampieri Good to know. Thank you for your input! $\endgroup$ Commented Dec 10, 2020 at 16:47

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.

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.

  • 1
    $\begingroup$ Note: the embedding $L^p \times L^q\ni (f,g) \mapsto f(x)g(y) \in L^{s'}_x L^{r'}_y + L^{s'}_y L^{r'}_x$ is I think the "correct version" of the inequality you asked about after the sentence "Can we use this to prove (something like)". $\endgroup$ Commented Jul 23, 2019 at 15:19
  • $\begingroup$ @ Willie Wong, can you please provide me some references on the above theorem? I am interested in a generalized Young inequality for functions with different dimensions $m \neq n$. $\endgroup$
    – Sia
    Commented Sep 9, 2019 at 4:00
  • $\begingroup$ @Sia: I don't know of any references except my answer above. Or perhaps a slightly more detailed write-up I posted on my blog. The basic idea is of course due to the three authors named in Daniele's answer; I just took the idea and ran with it from there. $\endgroup$ Commented Sep 9, 2019 at 14:42

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


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