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I am currently studying Newton's method on Banach spaces, specifically as it pertains to the linearization of nonlinear PDEs. In a sample problem, the following operator between Banach spaces came up:

$$ \begin{align*} F: C^2_0[0,1] &\longrightarrow C^2[0,1] \\ u &\mapsto [ t \mapsto u''(t) - f(t,u(t)) ] \end{align*} $$

where $f: [0,1] \times \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function that is continuously differentiable in the second argument.

This said to have a Fréchet derivative $F'(u)$, for all $u \in C^2_0[0,1]$, which when taken as a linear operator, maps any other function $y \in C^2_0[0,1]$ to the function defined by

$$ [F'(u)](y)(t) = y''(t) - \frac{\partial f(t,u(t))}{\partial u} y(t)$$

I am trying to show that this is indeed true, but I have come up against a problem: the Fréchet derivative $F'(u)$ is supposed to be a bounded linear functional, yet it seems to me that it is possible to construct a sequence of functions in $C^2_0[0,1]$ which are bounded in the sup norm, but whose second derivative (and therefore the expression in the derivative above) grows beyond all limits in that same norm.

Specifically, the series of test functions

$$ \rho_{\epsilon}(x) := exp(-\frac{1}{1-(\frac{x-1/2}{\epsilon})^2})$$

for $\epsilon \rightarrow 0$ should do the trick, as $\| \rho_{\epsilon} \|_{C^2_0[0,1]} = 1/e$ for all $\epsilon$, yet $\rho''_{\epsilon}(1/2) = \frac{-2}{\epsilon^2e}$ grows arbitrarily large in the norm.

Can anyone see what I am missing? I don't think the book would make such a mistake.

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  • $\begingroup$ $C^2[0,1]$ is only a Banach space with a norm equivalent to $||f||:=||f||_\infty + ||f'||_\infty + ||f''||_\infty $. Is your sequence of test functions bounded in this norm ? $\endgroup$
    – Maksim
    Nov 26, 2021 at 12:42
  • $\begingroup$ Its not given what $f(.,.)$ is, in the case $f=0$ its not true that $F(u)=u''$ is a map $C^2_0\to C^2$. perhaps $C^2_0\to C^0$ is intended? If there is indeed a typo, you might want to share / reference the source so that others may check $\endgroup$ Nov 26, 2021 at 12:43
  • $\begingroup$ @Maksim Thank you, this already answers my question completely. The book said to take the norm $|| \cdot ||_{C^2_0[0,1]}$ and I simply didn't know the correct definition of that norm. Of course my sequence would then not be bounded anymore. Also, as CalvinKhor pointed out, I got the target space wrong, too. Actually that was not given in the book (Theoretical Numerical Analysis by Atkinson & Han), so I had to make my own assumption. Would $F: C^2_0[0,1] \rightarrow C[0,1]$ work? ($f$ is continous in $t$) $\endgroup$
    – Eriol
    Nov 27, 2021 at 11:17

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