I have an operator $A:H^{-1/2}([0,2\pi])\to H^{-1/2}([0,2\pi])$ where $H^{s}$ is the fractional Sobolev space of order $s$.
I want to bound the $||\cdot||_{H^{-1/2}([0,2\pi])\to H^{1/2}([0,2\pi])}$ operator norm with the $||\cdot||_{L^{2}([0,2\pi])\to L^{2}([0,2\pi])}$ operator norm. Here's what I've done so far: $$ \begin{align} ||A||_{H^{-1/2}([0,2\pi])\to H^{-1/2}([0,2\pi])}^2 & = \sup_{||\phi||_{H^{-1/2}([0,2\pi])}=1} ||A \phi||_{H^{-1/2}([0,2\pi])}^2 \\ & = \sup_{||\phi||_{H^{-1/2}([0,2\pi])}=1} \sum_{n \in \mathbb{Z}} (1+|n|^2)^{-1/2} |\widehat{A \phi}|^2 \\ & \le \sup_{||\phi||_{H^{-1/2}([0,2\pi])}=1} \sum_{n \in \mathbb{Z}} |\widehat{A \phi}|^2 \\ & = \sup_{||\phi||_{H^{-1/2}([0,2\pi])}=1} ||A \phi||_{L^{2}([0,2\pi])}^2. \end{align} $$
Now I need to somehow replace the $||\phi||_{H^{-1/2}([0,2\pi])}$ in the supremum with $||\phi||_{L^{2}([0,2\pi])}$ to have a bound of the operator norm $||A||_{H^{-1/2}([0,2\pi])\to H^{-1/2}([0,2\pi])}$ in terms of $||A||_{L^{2}([0,2\pi])\to L^{2}([0,2\pi])}$.
What properties of the operator $A$ would allow me to do this?
Or is this never possible?
All you need is that there are continuous inclusions $i_1:L^2\hookrightarrow H^{-1/2}$ and $i_2:H^{1/2}\hookrightarrow L^2$, hence the concatenation $A'=i_2\circ A \circ i_1:L^2\rightarrow L^2$ is continuous and hence bounded. This is exactly the bound you're looking for. (This uses the fact that a linear operator between normed spaces is bounded if and only if it is continuous.)