Consider a simplified version of Young's inequality: $$ ||f\ast g||_p\leq ||f||_1||g||_p, \quad 1< p\leq\infty $$ $$ f\ast g\equiv \int_{\mathbb R}dy f(y)g(x-y). $$ What strategy one should follow to determine for which functions $f\in L^1(\mathbb R)$ and $g\in L^p(\mathbb R)$ the equality holds?

I guess these should be Gaussian functions (as indicated in https://arxiv.org/abs/math/9704210), but I may be wrong.

Any hint appreciated!


This inequality can be deduced in the following way. Consider $$ (|f|\ast|g|)(x) = \int_{\mathbb{R}} |f(y)| |g(x-y)| dy. $$

Let's apply Hölder's inequality with respect to measure $|f(y)|dy$ to the functions $y \mapsto g(x - y)$ and $1$ with exponents $p$ and $p^\prime$ respectively. We get $$ (|f|\ast|g|)(x) \le \left( \int_{\mathbb{R}}|g(x-y)|^p |f(y)| dy \right)^{1/p} \left( \int_{\mathbb{R}}|f(y)| dy\right)^{1/p^\prime} $$

Then taking $L^p$ norms, we obtain $$ ||(|f|\ast|g|)||_{p} \le \left(||f||^{p-1}_1 \int_{\mathbb{R}} \int_{\mathbb{R}} |g(x-y)|^p |f(y)| dy dx \right)^{1/p}. $$

Using Fubini's theorem one can deduce that it's equal to $$ \left( ||g||_p^p ||f||_1 ||f||_1^{p-1}\right)^{1/p} = ||g||_p ||f||_1. $$

Now we can see that in order to obtain an equality one must have an equality in Hölder's inequality. In other words, it's necessary to have $$ |g|^{p} = c\cdot 1^{p^\prime} = 1 $$ for some constant $c$.

It cannot happen when $p < \infty$ since constant is not integrable on $\mathbb{R}$. One can see that for case $p = \infty$ desired equality holds for $g = c$.

So, equality never holds for $1 < p<\infty$ and holds with $g = c$ for $p = \infty$.

In the article which you pointed out, authors prove stronger ineqaulity $$ ||f\ast g||_p\leq C_p ||f||_1||g||_p $$ with $C_p < 1$ and apparently find functions $f,g$ for which equality takes place.

  • $\begingroup$ Thank you for your reply! There is one point that seems unclear to me: you stated that for $p<\infty$ the equality never holds. While it is obvious that for $p=1$ it does hold, since $||f\ast g||_1=\int dx \int dy |f(y)g(x-y)|=\int dy |f(y)|\int dx|g(x-y)|=||f||_1||g||_1$. Hence, for $p=1$ any pair $f, g\in L^1(\mathbb R)$ should satisfy the equality, while from your proof it follows that even in this case we don't have equality, since for a constant $c$ we have $c\notin L^1(\mathbb R)$. $\endgroup$ – jonathan wolf May 22 at 8:43
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    $\begingroup$ @jonathanwolf in the question you stated, that $p>1$. For $p = 1$ argument with Hölder's inequality does not work, since the conjugate exponent $p^\prime$ is equal to $\infty$. I added a clarification in answer: it never holds for $1 < p < \infty$. $\endgroup$ – Virtuoz May 22 at 10:01
  • $\begingroup$ thank you for clarification! I marked your asnwer as the answer, and, indeed, upvote. Thanks a lot! $\endgroup$ – jonathan wolf May 22 at 10:54

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