Mean Value Theorem for Fréchet Derivatives

A linear function $T : R \to V$ is totally determined with its value at $1$ , since $T (t) = t T (1)$ , for $t \in R$ .

From this we get that the function $ι : B (R, V) \to V$ , with $ι (T) := T (1)$ . This is a vector isomorphism, and additionally it is an isometry since

∥ T ∥_{B (V, W)} = sup_{v \in S_{R}} ∥ T (v) ∥_{V} = ∥ T (1) ∥_{V}

Then we can identify that $B (R, V)$ with $V$ .

If $σ : (a, b) \to V$ is Fréchet-differentiable at the point $t_{0} \in (a, b)$ , then we can identify the derivative $σ^{'} (t_{0}) \in B (R, V)$ , with its value at $1$ , and then for simplicity we can write $σ^{'} (t_{0})$ to refers to $σ^{'} (t_{0}) (1)$ . Then we see that $σ^{'} (t_{0}) \in V$ , and

σ^{'} (t_{0}) = lim_{t \to 0} \frac{σ (t + t_{0}) - σ (t_{0})}{t} \in V

Which actually tells with the connection to the derivatives of functions from $R$ to $R^{n}$ , and why it behaves like that. This we can actually interpret as a velocity at $t_{0}$ of $σ$ .

Similarly we can look at the chain rule:
Let $σ : (a, b) \to Ω \subseteq V$ is differentiable at $t_{0} \in (a, b)$ and $φ : Ω \to W$ is differentiable at $u_{0} := σ (t_{0})$ , then we get

(φ \circ σ)^{'} (t_{0}) = φ^{'} (u_{0}) (σ^{'} (t_{0}))

The thing with the with the mean value in $R$ is that given $f : [a, b] \to R$ continuous and differentiable on $(a, b)$ , then $f (a) - f (b) = (b - a) f^{'} (c)$ for some $c \in (a, b)$ . The issue here is that we don't know which $c$ , so the most important is to have a bound for $| f^{'} (c) |$ , so we are going to do that in general:

Mean value theorem

Let $σ : [a, b] \to V$ is a continuous function. If $σ$ is differentiable for every $t \in (a, b)$ , and there's $M > 0$ , such that

∥ σ^{'} (t) ∥ \leq M \forall t \in (a, b)

Then

∥ σ (b) - σ (a) ∥ \leq M (b - a)

We get the corollaries:

Let $Ω$ be an open subset of $V$ , $φ : Ω \to W$ is of class $C^{1}$ on $Ω$ , and $u_{0}, u_{1} \in Ω$ such that $u_{t} = (1 - t) u_{0} + t u_{1} \in Ω$ for all $t \in [0, 1]$ , meaning $[u_{0}, u_{1}] \subseteq Ω$ , then

sup_{t \in [0, 1]} ∥ φ^{'} (u_{t}) ∥_{B (V, W)} = sup_{u \in [u_{0}, u_{1}]} ∥ φ^{'} (u) ∥_{B (V, W)} < \infty

and

∥ φ (u_{1}) - φ (u_{0}) ∥_{W} \leq (sup_{u \in [u_{0}, u_{1}]} ∥ φ^{'} (u) ∥_{B (V, W)}) ∥ u_{1} - u_{0} ∥

sup_{t \in [0, 1]} ∥ φ^{'} (u_{t}) - φ^{'} (u) ∥_{B (V, W)} < \infty

and it satisfies

∥ φ (u_{1}) - φ (u_{0}) - φ^{'} (u) (u_{1} - u_{0}) ∥ \leq (sup_{t \in [0, 1]} ∥ φ^{'} (u_{t}) - φ^{'} (u) ∥) ∥ u_{1} - u_{0} ∥

Let $Ω$ be an open and connected set of a Banach space $V$ , $φ : Ω \to V$ is differentiable on $Ω$ , and $φ^{'} (u) = 0$ for all $u \in Ω$ , the $φ$ is constant in $Ω$ .