Is the Gateaux differential an inner product if the Frechet derivative exists?

In $$\mathbb{R}^n$$ if the derivative of a function exists at a point $$x$$, then the directional derivative of that function at $$x$$ in the direction $$v$$ is an inner product between the derivative at $$x$$ and the direction $$v$$.

Does the same idea apply to Gateaux/Frechet derivatives? If the Frechet derivative exists is the Gateaux differential an inner product?

Let's concentrate on the Frechet derivative for now. From Wikipedia:

Let $$V$$ and $$W$$ be normed vector spaces, and $$U \subseteq V$$ be an open subset of $$V$$. A function $$f : U \to W$$ is called Fréchet differentiable at $$x \in U$$ if there exists a bounded linear operator $$A : V \to W$$ such that $$\lim_{\|h\| \to 0} \frac{ \| f(x + h) - f(x) - Ah \|_{W} }{ \|h\|_{V} } = 0.$$

What I want to highlight here is that $$A$$, which is the Frechet derivative, is defined to be a bounded linear operator.

In the particular case when $$V$$ is a Hilbert space, and $$W$$ is the scalar field (real or complex), then the derivative becomes a bounded linear functional. The Riesz representation theorem implies that such functionals take the form $$\langle \cdot, u \rangle$$, where $$u$$ is some fixed vector in $$V$$. So, in the case of maps from Hilbert spaces to their scalar fields, yes, the derivative is indeed expressed via an inner product. Although it's an abuse of notation, we sometimes call $$u$$ the Frechet derivative, even though it's really the functional $$\langle \cdot, u \rangle$$.

If $$W$$ is not the scalar field, then inner products are clearly not appropriate; the derivative is a linearisation and should map into $$W$$, not the scalar field, as an inner product does. This is true in finite-dimensions too.

Also, if $$V$$ is not an inner product space, then there's no inner product with which to express the derivative!

Now, the Gateaux derivative is a little bit more general. I've seen a couple of inequivalent definitions of the Gateaux derivative. The version I'm used to seeing looks quite similar to the above definition:

Let $$V$$ and $$W$$ be normed vector spaces, $$x \in V$$, and $$U \subseteq V$$ such that $$x \in \operatorname{core} U$$. A function $$f : U \to W$$ is called Gateaux differentiable at $$x \in U$$ if there exists a bounded linear operator $$A : V \to W$$ such that, for every $$h \in V$$, $$\lim_{t \to 0} \frac{ \| f(x + th) - f(x) - A(th) \|_{W} }{ t } = 0.$$

That is, it's much the same as the above definition, except that the difference quotient has to tend to $$0$$ invidividually along each line, rather than uniformly along all lines.

However, note that $$A$$ is still required to be bounded linear operator. By the same logic, you can express this as an inner product whenever $$V$$ is a Hilbert space and $$W$$ is the scalar field, otherwise no.

The other definition for Gateaux derivative drops the requirement that $$A$$ be linear or bounded, and defines it in terms of the directional derivative. The Gateaux derivative of $$f$$ at $$x$$ is defined to be the function $$h \mapsto \lim_{t \to 0} \frac{f(x + th) - f(x)}{t},$$ which may not be linear or continuous (even in finite dimensions). In this case, even when $$V$$ is an inner product space and $$W$$ is the scalar field, the Gateaux derivative may not be expressible as an inner product.