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

Question

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?

Answer 1

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!

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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.