# Frechet Derivative of Evaluation function

This is an exercise from Jack K.Hales book on ODEs

Let $X=C^1([0,1],\mathbb{R}^n) \times [0,1]$ and consider $$w:X\longrightarrow \mathbb{R}^n$$ where $w(f,t)=f(t)\in \mathbb{R}^n$. The point is to show that $w$ is Frechet differentiable and calculate it's derivative.

Some thoughts are that if we fix $f\in C^1([0,1],\mathbb{R}^n)$ then $$|f(t+h)-f(t)+D_{t}(f)h|\leq r(t,h)$$ with $\frac{ r(t,h)}{h}\longrightarrow 0$ for some function $r$ and $D_{t}(f)$ denoting the total derivative of the vector valued $f$ at $t$.

Now would it be correct to define the operator $$D:X\longrightarrow \mathbb{R}^n$$ with $D(f,t)=D_{t}(f)$ to be the Frechet derivative of $w$?

The way you started is unfortunately wrong. The Frechet derivative of $w$ is $Dw((f,t)) = (\delta_t , f'(t))$. Just for clarification how to interpret this equality: $$Dw((f,t))(g,s) = \big(\delta_t , f'(t)\big)\,(g,s) = g(t) + f'(t)s.$$

## Calculating the Frechet derivative

First of all we need a norm on $X$. I suggest $$\|(f,t)\|_X := \|f\|_\infty + \|f'\|_\infty + |t|$$ where $f'$ denotes the derivate of $f$ and $\|g\|_\infty = \sup_{s\in[0,1]} \|g(s)\|_p$ for any $p$-norm on $\mathbb{R}^n$.

Recall an equivalence of the definition of Frechet derivative:

Let $\phi: X \to Z$ and $D\phi : X \to L(X,Z)$ then $D\phi$ is the Frechet derivate of $\phi$ if $$\|\phi(x+y) - \phi(x)- D\phi(x)y\|_Z \in o(\|y\|_X)\tag{1}$$

In our case we have that each $x\in X$ has the representation $(f,t)$. An educated guess for the Frechet Derivate of $w$ is $$Dw(x)y = Dw((f,t)) (g,s) = \big(\delta_t, f'(t) \big)(g,s) = g(t) + f'(t)s$$ This comes from regarding $w(x+y)-w(x)$.

In order the show that our guess is in fact the Frechet derivative we will show the condition $(1)$. Let $x = (f,t) \in X$ and $y=(g,s) \in X$ \begin{align} \|w(x+y)-w(x) - Dw(x)y \|_p &= \|(f+g)(t+s)-f(t) - g(t) - f'(t)s\|_p \\ &= \|f(t+s) + g(t+s) - f(t) - g(t) - f'(t)s\|_p \\ &\leq \|f(t+s) - f(t) - f'(t)s\|_p + \|g(t+s) - g(t)\|_p \end{align} Since $\|y\|_X = \|g\|_\infty + \|g'\|_\infty + |s|$ we have from the well-known properties of differentiable functions that \begin{align} \|f(t+s) - f(t) - f'(t)s\|_p \in o(|s|) \subseteq o(\|(g,s)\|_X) \end{align} and \begin{align} \|g(t+s) - g(t)\|_p = \|g'(t)s + o(s)\|_p \in o(\|(g,s)\|_X) \end{align} which gives us that also the sum is in $o(\|(g,s)\|_X) = o(\|y\|_X)$. Hence we showed $(1)$ and consequently the desired statement.