Firstly, apologies if this is a duplicate question - I've looked, but can't find this question on SE (or elsewhere online); if it is, please let me know and I'll remove it.
I am trying to bound the Sobolev $s$-norm (for $f \in C^\infty_{per}([-\pi,\pi])$, the set of smooth 2$\pi$-periodic functions)
$$\| f \|_s^2 = \sum_{m \in \Bbb Z}(1+m^2)^s|\hat f(m)|^2$$
below by the sup-norm ($\sup|f|$), ie determine $\gamma_s$ such that $\sup|f| \le \gamma_s \|f\|_s$. Firstly, I am trying to do the case $s=1$, then determine whether it's true for general $s$.
I guess that it is not true for general $s$, as the next part of the question of the question asks to generalise the statement to an $n$-variable function, and then determine $\sigma(n)$ such that it is true if and only if $s > \sigma(n)$.
Also, similarly, trying to find $C_s > 0$ such that $\|u(t,\cdot)\|_s \le C_s \|u(0,\cdot)\|_0$, where $u(t,x)$ satisfies the heat equation $\partial_t u = \partial^2_x u$.
(I have a formula for $\|f\|_s$ in terms of $f$ and its derivatives.)
Any help would be most appreciated as I'm pretty stuck!
Thanks! :)
First, your definition of $||f||_s$ is wrong: the square root on the right-hand side is missing. Now, if we use the correct definition $$ ||f||_s=\sqrt{\sum_m(1+m^2)^s|\hat f(m)|^2}, $$ then everything is pretty straightforward: $$ \sup_x|f(x)|=\sup_x\left|\sum_m\hat f(m)e^{imx}\right|\le\sum_m|\hat f(m)|=\sum_m |\hat f(m)|(1+m^2)^{s/2}(1+m^2)^{-s/2} $$ (by Cauchy-Schwarz) $$ \le\sqrt{\sum_m |\hat f(m)|^2(1+m^2)^{s}}\sqrt{\sum_m (1+m^2)^{-s} }, $$ provided that $$ \gamma_s:=\sqrt{\sum_m (1+m^2)^{-s} }<\infty, $$ which is true if and only if $s>\sigma(1):=1/2$.
In the $n$-dimensional case we have $$ \gamma_s:=\sqrt{\sum_{m\in \mathbb{Z}^n} (1+m^2)^{-s} }\equiv \sqrt{\sum_{m_1,m_2,\dots,m_n\in \mathbb{Z}} (1+m_1^2+m_2^2+\dots+ m_n^2)^{-s} }<\infty $$ for $s>\sigma(n):=n/2$.
If $\hat u(t,m)$ are the Fourier coefficients of the solution $u(x,t)$ of the heat equation at time $t$, then $$ \hat u(t,m)=e^{-m^2t}\hat u(0,m). $$ This will help you to answer your last question yourself.