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I'm taking a graduate engineering class about optimization methods where we have been talking about some things from analysis. This one about the Fréchet derivative has me confused:

Given two normed spaces $\mathcal{X}$ and $\mathcal{Y}$ then $f:\mathcal{X}\rightarrow \mathcal{Y}$ is differentiable if there is a bounded linear mapping $\mathcal{A}:\mathcal{X}\rightarrow \mathcal{Y}$ such that

\begin{equation} \lim_{z\rightarrow 0}\frac{||f(x+z)-f(x) -\mathcal (A)z ||}{||z||_y}=0 \end{equation} holds.

The bounded part of the linear mapping is what confuses me. We have been given this counter example of a bounded linear mapping.

Consider $\mathcal{X}$ as the space of $C^{\infty}$ functions on an open interval with the $L_{\infty}$ norm. Consider the linear mapping: $\frac{d}{dt}:C^{\infty}\rightarrow C^{\infty}$ and the function $h_k(t)=\sin(kt)$ Then \begin{equation} ||\frac{d}{dt} h_k(t)||_{\infty}=|k|\cdot ||\cos (kt)||_{\infty}=|k| \end{equation} But $||\sin(kt)||_{\infty}=1$ which implies that there is a not constant $c$ such that $||\frac{d}{dt} h_k(t)||_{\infty}\leq |c|\cdot||\sin(kt)||\text{ } \forall\text{ } k$. Thus, $\frac{d}{dt}$ is not a bounded linear operator.

There are a couple things I don't understand about this:

  • First, it seems like circular reasoning to take the derivative to find out if it's differentiable according the above definition.

  • Second my professor says the derivative is still $\cos(kt)$ which has left me incredibly confused. If $\sin(kt)$ still has a derivative, then what does the definition above actually mean?

I would really appreciate some insight into this and/or a pointer some good resources.

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It is a little confusing as presented. Hopefully this will not confuse more...

The operator is defined in terms of an ordinary derivative which is the source of confusion. The question is whether or not the operator has a Fréchet derivative.

First, note that if $f$ is a bounded linear operator, then the Fréchet derivative is just $f$ itself since $f(x+h) = f(x)+f(h)$, and $f$ is bounded.

Now consider the space given in the question, and define $f(x) = \dot{x}$ (that is, $f(x)(t)$ is just the ordinary 'time' derivative of $x$ evaluated at $t$). It is easy to see that $f$ is linear, so $f$ will have a Fréchet derivative iff $f$ is bounded.

Since $f$ is not bounded, it does not have a Fréchet derivative.

That is, the ordinary derivative operator does not have a Fréchet derivative (on the given space).

Note the difference between $f$ having a Fréchet derivative and just evaluating $f$ on a particular function (where, in this case, evaluation happens to mean 'take the ordinary derivative'). So we have $f(t \mapsto \sin (kt)) = t \mapsto k\cos (kt)$, but $f$ itself does not have a Fréchet derivative.

Aside: When dealing with functions defined on finite dimensional spaces, all linear operators are bounded, so this issue does not arise and the usual concern is just showing the existence of the linear operator.

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  • $\begingroup$ Thanks I think that makes some sense. I guess I didn't realize that the ordinary derivative operator is distinct from the Frechet derivitive. I will have to read about this more. $\endgroup$ – rabraker Feb 11 '15 at 17:55
  • $\begingroup$ Well, the ordinary derivative is a Fréchet derivative, but there is a difference between differentiating a function and differentiating the differentiation operator. Since the ordinary derivative arises on finite dimensional spaces, there is no need to specify the bounded part. $\endgroup$ – copper.hat Feb 11 '15 at 18:01
  • $\begingroup$ Here is another example that may help. Consider the space $L^\infty$ which includes non differentiable functions. The function $f(x)(t) = x(t)+1$ is an operator on $L^\infty$ and it is easy to see that $f$ has a Fréchet derivative (the identity operator). $\endgroup$ – copper.hat Feb 11 '15 at 18:06

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