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.
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
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.