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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.
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  • $\begingroup$ In what sense is $T$ a continuous operator? $\endgroup$ – Omnomnomnom Nov 22 at 15:31
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Nov 22 at 15:37
  • $\begingroup$ @Omnomnomnom the exercise says to take $T \in \mathscr{L}(\mathscr{C}^2([a,b]),\mathscr{C}([a,b]))$ $\endgroup$ – Phi_24 Nov 22 at 17:29
  • $\begingroup$ How is the norm on $\mathcal C^2$ defined? $\endgroup$ – Omnomnomnom Nov 22 at 17:29
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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.

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  • $\begingroup$ I think $f(x)=kx+c$ for integral values of $k$ is periodic on $\mathscr{C}^2(\mathbb{S}^1)$. $\endgroup$ – WE Tutorial School Nov 22 at 19:06
  • $\begingroup$ @WETutorialSchool $\mathscr C^2(\Bbb S^1)$ consists of $\Bbb R$-valued functions on $\Bbb S^1$, not $\Bbb S^1$-valued functions $\endgroup$ – Omnomnomnom Nov 22 at 19:46
  • $\begingroup$ @Omnomnomnom I fail to see that $\ker T$ is different. Can you explain it in details? $\endgroup$ – Phi_24 Nov 23 at 8:50
  • $\begingroup$ If the derivative of a periodic function is constant, then the function itself must also have been constant $\endgroup$ – Omnomnomnom Nov 23 at 19:57

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