Consider $V=C([a,b])$ and let $A:V\to V: f(t)\mapsto u(t)f(t)$ be the multiplication operator with fixed $u(t)\in V$.
Let $E$ denote the set of eigenvalues of operator $A$.
Let $S$ denote the spectrum of operator $A$. The spectrum is defined as the set of non-regular points; i.e. the points $\mu$ for which the resolvent $(A-\mu\mathbf{1})^{-1}$ is not well-defined or bounded on $V$.
It is clear that $S= \operatorname{im}(u)$, the image of $u$. By choosing $u(t)$ wisely, we can create 3 different situations: 1) $E=\emptyset$, 2) $E = S$ and 3) $\emptyset\ne E \subset S$. My question is regarding this third situation.
A possible example (let's work on $[0,1]$), would be to consider $u(t) = 0$ outside of $[1/3,2/3]$ and have it form a little baseless triangle within $[1/3,2/3]$ (to ensure $u$ is continuous); and then we can choose $f$ to be $0$ inside $[1/3,2/3]$, but non-zero outside of it (but still continuous). Then $u(t)f(t)=0 = 0f(t)$, so this way we have found $E=\{0\}\subset S=[0,1]$ (albeit without explicitly writing $f$).
Now, I want to find an example (just by drawing, not necessarily expressions) of $u(t)$ such that $|E|>1$ but $E\subset S$ strictly. I'm having trouble finding continuous functions for this to work. Any ideas?
Thanks.
You can show that, in general, $E$ will be the set of $x\in\mathbb{R}$ such that $\operatorname{int}(u^{-1}(x))\neq\varnothing$ (use an argument similar to the one you did with the little triangle). So just take a function whose graph looks like a stair to get more eigenvalues.
It will also follow that $E$ is always countable.