Young's convolution inequality states that, for $1/p+1/q=1/r+1$ ($1\leq p,\, q, r\leq \infty$), we have $$\lVert f * g \rVert_r \leq \lVert f\rVert_p \lVert g\rVert_q.$$ It is implicit here that the measure under which these norms are taken is the Lebesgue measure.

Let $\lVert\cdot\rVert_{L^p_w}$ denote the weighted norm defined by $$ \lVert f \rVert_{L^p_w}^p = \int f^p(x)w(x)dx,$$ where $dx$ denotes Lebesgue measure and $\int w(x)dx=1$.

Question. Is it true that $$ \lVert f * g \rVert_{L^r_w} \leq \lVert f\rVert_{L^p_w} \lVert g\rVert_{L^q_w}?$$

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    $\begingroup$ A counterexample for $w=\delta(x)$ is obvious. It also works for approximations of $\delta(x)$ by usual functions. $\endgroup$ Commented Apr 30, 2021 at 4:39
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    $\begingroup$ If $w$ is the standard gaussian measure, and the convolution is defined as $f*g(x) = \int f(x-y)g(y) w(y) dy$ then $\|f*g\|_{r}\leq \|f(\sqrt{2}\cdot )\|_{p} \|g\|_{q}$ holds if and only if $p,q,r \geq 1$ and $2\geq \frac{1}{p-1}\left( \frac{1}{q-1}+r-1\right)$. If you are interested about this result I may expand it into an answer. It follows by duality together with Gaussian Jensen’s inequality mathoverflow.net/a/363587 (assuming that I correctly computed the hessian matrix and the covariance matrix). $\endgroup$ Commented Apr 30, 2021 at 19:08

1 Answer 1


I explained the correct generalization in a blog post. You version cannot hold for two reasons:

  1. Your version doesn't have the right number of weights: on the left the weight only appears once and on the right it appears twice. To fix this you need to use a weighted convolution.
  2. Your version doesn't respect the lack of translation invariance. As Ilya Bogdanov's comment indicates, if your weight function is heavily localized, you can easily choose $f$ and $g$ to have small weighted norms (by having support where $w$ is small) while their convolution has large weighted norm (by being supported where $w$ is large).

The correct statement is

Notations: Let $(\Omega, \mathcal{E}, \mu)$ and $(\Omega', \mathcal{E}', \mu')$ be $\sigma$-finite measure spaces. Denote by $L_p$ and $L_p'$ the $p$-integrable functions on $\Omega$ and $\Omega'$ respectively. And denote by $L_pL_q'$ and $L_q'L_p$ the Bochner spaces $L_p(\Omega; L_q'(\Omega'))$ and $L_q'(\Omega'; L_p(\Omega))$.

Let $1 \leq s_1 \leq r_1 \leq \infty$ and $1 \leq s_2\leq r_2\leq \infty$. If $k_1, k_2\in \Omega\times\Omega'\to\mathbb{R}$ are kernels satisfying $$ k_1 \in L'_{r_1} L_{s_1}, \qquad k_2 \in L_{r_2} L'_{s_2} $$ Then the operator $$ T^\theta f(x) = \int_{\Omega'} k_1^\theta(x,y) k_2^{1-\theta}(x,y) f(y) ~d\mu' $$ defined for $\theta\in (0,1)$, is a bounded continuous mapping from $L'_p$ to $L_q$ provided $$ \frac{1}{q} = \frac{1-\theta}{r_2} + \frac{\theta}{s_1} $$ and $$ 1 - \frac{1}{p} = \frac{1-\theta}{s_2} + \frac{\theta}{r_1} $$

As you see, the integral operator is defined relative to the $\mu'$ measure, so in the end of the day both measure appears equally on the left and the right, that fixes problem 1. And to avoid the translation invariance problem, the integral kernel is measured in a Bochner space norm.

In your case, you want the two measure spaces to be equal, with $\Omega = \Omega' = \mathbb{R}^d$, and $\mu = \mu'$ being the Lebesgue measure with weight.

For the convolution case you are looking for $k_1 = k_2 = g(x-y)$. And so in order to copy the standard Young's inequality, what you need is

Let $g: \mathbb{R}^d \to \mathbb{R}$ be such that $$ \int |g(x-y)|^p w(y) ~dy $$ is uniformly bounded (in $x$) by some constant $M$. Then the weighted convolution $$ g*_w f(x) = \int g(x-y) f(y) w(y) ~dy $$ satisfies $$ \| g*_w f\|_{L^r_w} \leq M \|f\|_{L^q_w} $$ where $p, q, r$ satisfies the usual scaling relation.

In other words, we ask for the function $(x,y) \mapsto g(x-y)$ to be in $L^\infty_x L^p_y \cap L^\infty_y L^p_x$. In the usual case the translation invariance of the Lebesgue measure tells us that $g\in L^p$ is sufficient to imply the above. But for general weighted measures you need to insist this separately.

Incidentally: this is also related to how one should define convolutions and prove Young's inequalities for functions on Lie groups that are not unimodular.

  • $\begingroup$ Quick typo on your blog post: "We first note that commutation on general Lie groups are not commutative." should probably be "convolution … is". $\endgroup$
    – LSpice
    Commented May 1, 2021 at 1:39
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    $\begingroup$ @LSpice: much thanks. $\endgroup$ Commented May 1, 2021 at 2:28

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