Gaussians and Young's inequality for convolutions

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.

• 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 $c\notin L^1(\mathbb R)$. – jonathan wolf May 22 at 8:43
• @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 $1 < p < \infty$. – Virtuoz May 22 at 10:01
• thank you for clarification! I marked your asnwer as the answer, and, indeed, upvote. Thanks a lot! – jonathan wolf May 22 at 10:54