I am going through Terence Tao's "Nonlinear Dispersive Equations (Local & Global Analysis)" and trying to work through some of his exercises. However, I find myself being stumped by Exercise A.21. The question goes as follows:

Let $0<\alpha<1$ and $1\le p\le\infty$. If $f\in S_x(\mathbb{R}^d)$, we define the Hölder norm $\|f\|_{\Lambda_\alpha^p(\mathbb{R}^d)}$by the formula $$ \|f\|_{\Lambda_\alpha^p(\mathbb{R}^d)}:=\|f\|_{L^p_x(\mathbb{R}^d)} + \sup_{h\in\mathbb{R}^d; 0<|h|\le1}\frac{\|f^h-f\|_{L^p_x(\mathbb{R}^d)}}{|h|^\alpha} $$ where $f^h(x)=f(x+h)$ is the translate of $f$ by $h$. Show that $$ \|f\|_{\Lambda_\alpha^p(\mathbb{R}^d)} \sim_{p,\alpha,d} \|f\|_{L^p_x(\mathbb{R}^d)} + \sup_{N\ge 1}N^\alpha\|P_{N}f\|_{L^p_x(\mathbb{R}^d)} $$ (Hint: to control the former by the latter, obtain two bounds for the $L^p_x(\mathbb{R}^d)$ norm of $P_Nf^h-P_Nf$, using triangle inequality for the high frequency case $N\gtrsim |h|^{-1}$ and the fundamental theorem of calculus in the low frequency case $N\lesssim |h|^{-1}$.)

So when attempting to control the former with the latter in the high frequency case, I took his hint and applied triangle inequality, which landed me at $$ \|P_Nf^h-P_Nf\|_{L^p_x(\mathbb{R}^d)} \le 2\|P_Nf\|_{L^p_x(\mathbb{R}^d)} $$

Thereafter, $$ \frac{\|P_Nf^h-P_Nf\|_{L^p_x(\mathbb{R}^d)}}{|h|^\alpha} \lesssim N^\alpha\|P_Nf\|_{L^p_x(\mathbb{R}^d)} $$

From here, I found myself in a rather unpleasant dead-end since moving forward will require me to take the infinite sum over $N$'s, which in this present form will not converge.

May I know how should I side-step this problem? Or am I misunderstanding his hint? Thanks in advance!


1 Answer 1


You need to remember that $N$ is a dyadic number, so that $\sum_{N \geq c} N^{-\alpha}$ converges for positive $\alpha$.

So you have

$$ \sum_{N \geq |h|^{-1}} \frac{\|P_Nf^h-P_Nf\|_{L^p_x(\mathbb{R}^d)}}{|h|^\alpha} = \sum_{N \geq |h|^{-1}} N^\alpha\|P_Nf^h-P_Nf\|_{L^p_x(\mathbb{R}^d)}\cdot \frac{1}{N^\alpha|h|^\alpha} $$

Split the sum using Holder and put the first factor in $\ell^\infty$ and sum the second factor to get a constant (which is $\approx 1$, independently of $h$.)

  • 3
    $\begingroup$ No way! I can't believe an expert in this field actually bothered to answer my question! I have been following some of your works. Thank you for replying to my fairly trivial question, I'm so honoured! $\endgroup$
    – Tham
    Mar 17 at 17:21

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