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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

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.

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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 $N$.

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    Oh, it was simple after all. Thank you! (and by $\frac{1}{q_N}=\frac{1}{p_1}+\ldots+\frac{1}{p_{N-1}}$ I assume that you were trying to say $\frac{1}{q_N '}=\frac{1}{p_1 '}+\ldots+\frac{1}{p_{N-1} '}$?) – 2017-01-22
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    Yes, I meant that. – 2017-01-22
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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.