# Find $\operatorname{ind}T$ where $T=\frac{d^2}{dx^2}$

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:

1. $$T: \mathscr{C}^2([a,b]) \longrightarrow \mathscr{C}([a,b])$$
2. $$T: \mathscr{C}^2(\mathbb{S}^1) \longrightarrow \mathscr{C}(\mathbb{S}^1)$$

My attempts:

1. $$\ker T =\{ u \in \mathscr{C}^2([a,b]) \quad | \quad Tu=0 \iff \dfrac{d^2u}{dx^2}=0 \} = \{u=mx+q| m,q \in \mathbb{R}\}$$, 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?

1. for the $$\ker$$ part it's the same and for the $$\operatorname{im}T$$ I don't even know how to start.
• In what sense is $T$ a continuous operator? – Omnomnomnom Nov 22 at 15:31
• By any reasonable definition, $T$ fails to be continuous. Note for instance that if we define $$f_k(x) = \frac{\cos(kx)}{k} \quad k = 1,2,3,\dots,$$ then $f_k \to 0$ as $k \to \infty$ while $Tf_k$ fails to converge. – Omnomnomnom Nov 22 at 15:37
• @Omnomnomnom the exercise says to take $T \in \mathscr{L}(\mathscr{C}^2([a,b]),\mathscr{C}([a,b]))$ – Phi_24 Nov 22 at 17:29
• How is the norm on $\mathcal C^2$ defined? – Omnomnomnom Nov 22 at 17:29

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.
• I think $f(x)=kx+c$ for integral values of $k$ is periodic on $\mathscr{C}^2(\mathbb{S}^1)$. – WE Tutorial School Nov 22 at 19:06
• @WETutorialSchool $\mathscr C^2(\Bbb S^1)$ consists of $\Bbb R$-valued functions on $\Bbb S^1$, not $\Bbb S^1$-valued functions – Omnomnomnom Nov 22 at 19:46
• @Omnomnomnom I fail to see that $\ker T$ is different. Can you explain it in details? – Phi_24 Nov 23 at 8:50