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Here is the question:

(a) Let $f\in C^1(\mathbb{T})$ and $\int_\mathbb{T} f(x)dx=0$. Show that for $1\leq p\leq\infty$: $\|f\|_p\leq\|f'\|_p$

(b) Let $f\in C^\infty(\mathbb{T})$, for $n\in\mathbb{N}$ define $$D^nf(x)=\sum_{j\in\mathbb{Z}}|2\pi j|^n\hat{f}(j)e^{2\pi ijx}$$ Let $1<p<\infty$, show that there exists a constant $C_{n,p}>0$ depends only on $n,p$ so that $$C_{n,p}^{-1}\|D^nf\|_p\leq\|f^{(n)}\|_p\leq C_{n,p}\|D^nf\|_p$$

Some comments: All functions are $1-$periodic. We define the Fourier coefficient for the $1-$periodic function as $$\hat{f}(j)=\int_{\mathbb{T}}f(x)e^{-2\pi ijx}dx=\int_{a}^{a+1}f(x)e^{-2\pi ijx}dx$$

for any $a\in\mathbb{R}$.

Part (a) is not hard to prove. I have trouble on part(b). I think it is supposed to use part (a) but I cannot figure it out. Any help will be appreciated.

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1 Answer 1

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You don't need the result for (a). Perhaps using conjugate function is easy to deal with.

First observe that if $r$ is even, then the result is trivial as in this case

$$f^{(r)}(x)=D^rf(x),\quad\forall r \text{ even}\quad (*)$$ Thus we only need to prove the case when $r$ is odd. To prove this, it suffices to verify the case $r=1$ because of $(*)$. For simplicity, let $Df(x)=D^1f(x)$. Denote $\tilde{f}(x)$ the conjugate function of $f$, which is defined as $$\tilde{f}(x)=\sum_{j\in\mathbb{Z}}-i (sgn(j))\hat{f}(j)e^{2\pi ijx}$$ Then we have the following facts

  • $f'(x)=-\widetilde{Df}(x)$
  • $\widetilde{\tilde{f}}(x)=-f(x)$
  • $f\mapsto\tilde{f}$ is a bounded linear map between $L^p(\mathbb{T})$ and $L^p(\mathbb{T})$ when $1<p<\infty$, i.e. $\exists C_p>0$ only depending on $p$ such that $\|\tilde{f}\|_p\leq C_p\|f\|_p$ for all $f\in L^p(\mathbb{T})$.

Therefore $$\|f'\|_p=\|\widetilde{Df}\|_p\leq C_p\|Df\|_p$$ and $$\|Df\|_p=\|\tilde{f'}\|_p\leq C_p\|f'\|_p$$ hence the result is proved.

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