Proof of Hölder for Lorentz spaces (harmonic analysis)

So I was thinking about the proof of Hölder's inequality for Lorentz spaces

$$\left\lVert fg\right\rVert_{p,q} \lesssim \left\lVert f\right\rVert_{p_1,q_1} \left\lVert g\right\rVert_{p_2,q_2}$$

where the exponents are positive and finite ($q$ can be infinite, but let's ignore that) and $1/q = 1/{q_1} + 1/{q_2}, 1/p = 1/{p_1} + 1/{p_2}$.

We all know that a Lorentz function can be characterized in 2 ways:

• dyadic decomposition by height: $f = \sum_n f_n$, $2^n <|f_n| \leq 2^{n+1}$, $f||_{p,q} \sim || \; \left\lVert f_n\right\rVert_{L^p} ||_{l^q_n}$
• dyadic decomposition by width: $f = \sum_n f_n$, where $f^*(2^{n+1}) < |f_n| \leq f^*(2^n)$, and $\left\lVert f\right\rVert_{p,q} \sim \left\lVert \left\lVert f_n\right\rVert_{L^p} \right\rVert_{l^q_n}$

where $f^*$ is the decreasing rearrangement of $f$.

The standard proof of Holder for Lorentz spaces uses dyadic decomposition by width (as can be seen here, Theorem 6.9). So I guess my question really is: Can we use decomposition by height to tackle this one? I'd like to think that it's possible. I tried to adapt the width method but the supports wouldn't play nice.

If it is not possible, I'd like to know how one can figure out which characterization is better for a particular problem. Is there anything in the inequality that suggests the width approach would be better than the height approach?

• I'd like to know how one can figure out which characterization is better for a particular problem. Under normal circumstances both should work just fine. The preferences depend more on the set of tricks one has up his sleeve and that is highly individual. – fedja Nov 30 '17 at 19:18

1 Answer

You can certainly do it. Here is the trickery:

Let us assume that the norms of $f$ and $g$ are bounded by $1$ (scaling is trivial). Write the height decompositions $f=\sum_nf_n$, $g=\sum_m g_m$. The conditions on the norms rewrite as $\newcommand{\supp}{\operatorname{supp}}$ $$\sum_n 2^{nq_1}|\supp f_n|^{q_1/p_1}\lesssim 1,\qquad \sum_m 2^{mq_2}|\supp g_m|^{q_2/p_2}\lesssim 1$$ Now notice one funny thing: for $a_\ell\ge 0$, we always have the inequality $$\sum_{\ell}2^{-\varepsilon |\ell|}a_\ell^\beta\le \left(\sum_{\ell} a_\ell\right)^\beta\lesssim\sum_{\ell}2^{\varepsilon |\ell|}a_\ell^\beta$$ ($\beta,\varepsilon>0$). This is pretty well-known. A little bit less known (but equally obvious) is the inequality $$\sum_{\ell}2^{-\varepsilon |\ell-c\log a_\ell-d|}a_\ell^\beta\lesssim \left(\sum_{\ell} a_\ell\right)^\beta\lesssim\sum_{\ell}2^{\varepsilon |\ell-c\log a_\ell-d|}a_\ell^\beta$$ where $\log$ is understood as logarithm with respect to base $2$. These two allow us to do all sorts of "illegal" replacements of powers of sums by sums of powers any time we want as long as we have some geometric pre-factors.

Now the life becomes easy because the height decomposition of $fg$ is essentially $\sum_k \left[\sum_{n+m=k}f_ng_m\right]$ Let $a_{nm}$ be the size of the support of $f_ng_m$.

Thus, we want to estimate $$\sum_k 2^{kq}\left[\sum_{m+n=k}a_{nm}\right]^{p/q} \\ \lesssim \sum_k 2^{kq}\sum_{m+n=k}2^{\varepsilon|\alpha m-\beta n-c\log a_{nm}|}a_{nm}^{q/p}$$ Now observe that $$2^{(n+m)q}x^{q/p}=[2^{nq_1}x^{q_1/p_1}]^{\frac {q_2}{q_1+q_2}}[2^{mq_2}x^{q_2/p_2}]^{\frac {q_1}{q_1+q_2}}\le 2^{nq_1}x^{q_1/p_1}+2^{mq_2}x^{q_2/p_2}$$ and any deviation from the optimal regime $nq_1-mq_2-(\frac {q_2}{p_2}-\frac{q_1}{p_1})\log x=0$ results in an exponential penalty. So, we can write $$2^{(n+m)q}a_{nm}^{q/p}\le 2^{-\delta|nq_1-mq_2-(\frac {q_2}{p_2}-\frac{q_1}{p1})\log a_{nm}|}[2^{nq_1}a_{nm}^{q_1/p_1}+2^{mq_2}a_{nm}^{q_2/p_2}]$$ with some $\delta>0$ Now it is clear that we should match $\alpha,\beta, c$ with the coefficients we just figured out and to take $\varepsilon$ less than $\delta/2$. Then we get $$\sum_{m,n}2^{-\varepsilon|\alpha m-\beta n-c\log a_{nm}|}[2^{nq_1}a_{nm}^{q_1/p_1}+2^{mq_2}a_{nm}^{q_2/p_2}]$$ to bound, which is easy, because the loss that the Young inequality turned into the gain allows us to write $$\sum_{n}2^{nq_1}\sum_m2^{-\varepsilon|\alpha m-\beta n-c\log a_{nm}|}a_{nm}^{q_1/p_1} \\ \lesssim \sum_{n}2^{nq_1}\left[\sum_m a_{nm}\right]^{q_1/p_1}\le \sum_{n}2^{nq_1}|\supp f_n|^{q_1/p_1}\lesssim 1$$ and similarly for the other half.