How do we norm the space of Fréchet differentiable functions from a Banach space $E$ to $\mathbb R$?

Question

Quick question, I couldn't find an answer to: If $E$ is a $\mathbb R$-Banach space, and $C^1(E,\mathbb R)$ is the space of continuously Fréchet differentiable functions from $E$ to $\mathbb R$, how does one "usually" define a norm on $C^1(E,\mathbb R)$? I'd also be interested in Hölder spaces of functions from $E$ to $\mathbb R$, but all references I was able to find only consider $E=\mathbb R^d$.

Maybe there is some kind of duality to the space of signed measures which implicitly yields a norm.

EDIT: In particular, I'd like to show that if $d$ is a metric on $E$, equivalent to the canonical metric on $E$, and $\mu$ is a probability measure on $(E,\mathcal E)$, then $$\left\|f\right\|:=\sup_{x\ne y}\frac{|f(x)-f(y)|}{d(x,y)}+\left|\int f\:{\rm d}\mu\right|$$ is a norm on $C^1(E,\mathbb R)$.

This kind of norm is considered in this paper in equation (22) on p. 2063.

Answer 1

I'll leave aside any potential technicalities regarding how a probability measure on a Banach space is defined etc (because I really don't know enough measure theory to comment on that). Also, it is not clear to me that the "norm" so defined takes a finite value for all $f$. Aside from that, it seems like checking the other relevant properties of the norm is straight forward (but of course, take everything with a grain of salt, and perhaps someone else can provide their own answer/ suggest how to improve):

• It's clear that for all $f \in C^1(E, \Bbb{R})$, $\lVert f \rVert \geq 0$. Now, suppose that $\lVert f\rVert = 0$. Then, each summand defining $\lVert \cdot \rVert$ must be $0$. Fix an $x_0 \in E$, then, for all $y\neq x_0$, we have \begin{align} 0 \leq \dfrac{|f(y) - f(x_0)|}{d(y, x_0)} \leq \sup_{y \neq x} \dfrac{|f(y) - f(x)|}{d(y, x)} = 0 \end{align} Thus, $f(y) = f(x_0)$, which shows $f$ is constant on $E$. Next, $|\int_Ef \, d\mu| = 0$, together with the constancy of $f$, and the finiteness of the measure $\mu$ imply that the constant value for $f$ is actually $0$.
• It's fairly clear that for each $\lambda \in \Bbb{R}$, $\lVert \lambda f\rVert = |\lambda| \cdot \lVert f\rVert$.
• Finally, the triangle inequality follows because it holds for the absolute value on $\Bbb{R}$. More explicitly, \begin{align} \lVert f+ g\rVert &:= \sup_{x \neq y} \dfrac{|(f+g)(x) - (f+g)(y)|}{d(x,y)} + \left| \int_E (f+g) \, d\mu\right| \\ &\leq \sup_{x \neq y}\left( \dfrac{|f(x) - f(y)|}{d(x,y)} + \dfrac{|g(x) - g(y)|}{d(x,y)} \right) + \left| \int_E f\right| + \left| \int_E g\right| \\ &\leq \left[ \sup_{x \neq y}\left( \dfrac{|f(x) - f(y)|}{d(x,y)}\right) +\left| \int_E f\right| \right] + \left[ \sup_{x \neq y}\left( \dfrac{|g(x) - g(y)|}{d(x,y)}\right) +\left| \int_E g\right| \right] \\ &= \lVert f \rVert + \lVert g \rVert. \end{align}

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Edit:

Based on the comment provided, I corrected the argument for $\lVert f\rVert = 0 \implies f = 0$.