Upper bound for the norm of convolutions: $\Vert f_1\ast\cdots\ast f_N\Vert_r\leq\Vert f_1\Vert_{p_1}\cdots\Vert f_N\Vert_{p_N}$

Question

By definition, $(f\ast g)(x)=\int_{\mathbb{R}^n} f(y)g(x-y)dy$ for each $x\in\mathbb{R}^n$. Young's theorem states that if $p,q,r\in[1,\infty]$ satisfy $$\frac{1}{r}+\frac{1}{p'}+\frac{1}{q'}=1$$ ( ' denotes the Hölder conjugate), then for all $f\in L^p(\mathbb{R}^n)$ and $g\in L^q(\mathbb{R}^n)$ we have $$\Vert f\ast g\Vert_r\leq\Vert f\Vert_p \Vert g\Vert_q.$$

The proof of Young's theorem is done in three parts: Without loss of generality we may assume that all functions are non-negative and $f,g$ are normalized.

• If $r=\infty$, then this implies that $p,q$ are conjugates to each other, and Hölder's inequality gives the result.
• If $p'=\infty$ or $q'=\infty$, then the generalized Minkowski's inequality gives the result.
• Finally when none of $r,p',q'$ are infinite, we have $1<r,p,q<\infty$ and $1<r',p',q'<\infty$ so we may write $$(f\ast g)(x)=\int \left[f(y)^{p/r}g(x-y)^{q/r}\right]g(x-y)^{1-q/r}f(y)^{1-r/p}dy$$ and applying Hölder's inequality for three functions (with exponents $r,p',q'$ respectively) gives the result.

One of my exercises asks to generalize this result: Assume $1\leq p_k\leq\infty$ for $k=1,\ldots,N$ and $1\leq r\leq\infty$ and $$\frac{1}{r}+\frac{1}{p_1 '}+\cdots+\frac{1}{p_N '}=1.$$ Then, does it follow that $\Vert f_1\ast\cdots\ast f_N\Vert_r\leq\Vert f_1\Vert_{p_1}\cdots\Vert f_N\Vert_{p_N}$? I tried to mimic the proof of Young's theorem but could not see how to generalize the method. Funny enough, the biggest difficulty was that the function $f_1\ast\cdots\ast f_N$ involves so many integrals that I couldn't figure out a way to write it clearly...

So, does the inequality $\Vert f_1\ast\cdots\ast f_N\Vert_r\leq\Vert f_1\Vert_{p_1}\cdots\Vert f_N\Vert_{p_N}$ hold? If so, how do we prove such a result? Please enlighten me.

Answer 1

How about $$\left\|f_1*\ldots *f_N\right\|_r=\Big\|\big(f_1*\ldots*f_{N-1}\big)*f_N\Big\|_r\leq \left\|f_1*\ldots*f_{N-1}\right\|_{q_N}\,\left\|f_N\right\|_{p_N}\,,$$ where $\frac{1}{q_N}=\frac{1}{p_1}+\ldots+\frac{1}{p_{N-1}}$? Then, you can proceed by induction on $N$.

Answer 2

I think you may prove that the inequality holds for $N$ functions using the mathematical induction. It doesn't require writing $N$ integral anyway. The idea is to bring back the product of $N+1$ function to a product of two function where the second one is the result of the product of $N$, With respect to the holder constant that you will choose. So basically the demonstration is based on the result of the 2-fucntion inequality that you have already done.