Let $T$ be a linear and continuous operator defined as $$T=\frac{d^2}{dx^2}$$ Determine $\dim \ker T$ and $\dim \operatorname{coker}T$ in this two cases:
- $T: \mathscr{C}^2([a,b]) \longrightarrow \mathscr{C}([a,b]) $
- $T: \mathscr{C}^2(\mathbb{S}^1) \longrightarrow \mathscr{C}(\mathbb{S}^1) $
My attempts:
- $\ker T =\{ u \in \mathscr{C}^2([a,b]) \quad | \quad Tu=0 \iff \dfrac{d^2u}{dx^2}=0 \} = \{u=mx+q| m,q \in \mathbb{R}\}$, so $\dim \ker T =2$ because the space has basis $(x,1)$
$\operatorname{im}T = \mathscr{C}([a,b]) \Rightarrow \dim \operatorname{coker}T =0$, so $\operatorname{ind}T = 2-0=2$. Is it right?
- for the $\ker$ part it's the same and for the $\operatorname{im}T$ I don't even know how to start.
Your steps and conclusion regarding 1 are correct.
Regarding question 2, you might have an easier time if you identify $\mathscr C^2(\Bbb S^1)$ with the functions in $\mathscr C^2([0,1])$ satisfying the constraint $f(0) = f(1)$. Another way to think about this is that we're considering the set of all $\mathscr C^2$ functions that are periodic with some fixed period.
With that said, the kernel will not be the same. Note in particular that the function $u = mx$ fails to be periodic. As for the image: in the same way that you deduced that $T$ was onto before, you should be able to deduce that $T$ is onto now.