Property of the Gâteaux derivative

Question

Let $D \colon X \to Y$ be a map of Banach spaces and define

$$(\mathrm d_f D) g := \displaystyle\lim_{\varepsilon \to 0} \frac{D(f+\varepsilon g) - Df}\varepsilon$$ given that the limit exists.

Suppose that $(\mathrm d_f D) g$ does not depend on $f$, will the assignment $$g \mapsto (\mathrm d_f D) g$$ be additive?

A friend of mine has the hypothesis that the condition implies that $D$ is affine linear, i.e. linear up to a constant.

Answer 1

If $X = \mathbb{R}^n$ and $Y = \mathbb{R}$, then, if $D$ is differentiable any differentiable function (that would be in a Fréchet sense) we know: $$ D(f+\varepsilon g) = D(f) + (\varepsilon g)\cdot \nabla D(f) + o(\varepsilon g) \iff $$ $$ \frac{D(f+\varepsilon g)-D(f)}{\varepsilon} = g \cdot \nabla D(f) + \frac{o(\varepsilon g)}{\varepsilon} $$ It follows that $(d_fD)g = g \cdot \nabla D(f) $. So this is an additive relationship.

You can generalize this approach for other Banach spaces.

You see that $D$ need not be affine at all.

Answer 2

Let $F:X\to Y$ be Gateaux differentiable such that the Gateaux derivative is constant, i.e., $F'(x) = D$ for all $x$. Then using the mean-value theorem for Gateaux differentiable functions to the function $G(x):=F(x)-Dx$ yields $$ \|F(x) - F(0) - Dx\| = \|G(x)-G(0)\| = \sup_{t\in(0,1)} \|G'(tx)\| = 0. $$ So $$ F(x) = F(0) + Dx, $$ and $F$ is affine linear.

The above proof is part of the proof to show that continuous Gateaux differentiability implies Frechet differentiability.