How do higher-order Fréchet derivatives look like in case of $f:\mathbb R \to \mathbb R, x \mapsto x^3$?

Question

Recently, I have come across the Taylor's formula for functions between Banach spaces. To better understand the machinery, I try to see how higher-order derivatives look like in a simple example, i.e., $f:\mathbb R \to \mathbb R, x \mapsto x^3$.

• Let $E,F$ be Banach spaces and $f:E \to F$ be $m$-times continuously differentiable. Here we use Fréchet derivative, i.e., $\partial^m f : E \to \mathcal L^m(E, F)$ with $\mathcal L^m(E, F)$ the space of continuous multilinear maps from $E^m$ to $F$. For clarity, I will enclose the arguments of multilinear maps in square brackets.

• Let $E = F := \mathbb R$. We take $f: x \mapsto x^3$ as an example. It's easy to see that $\partial f : x \mapsto (h_3 \mapsto 3x^2h_3)$. I pretend that $\partial f$ has an untrue form $x \mapsto 3x^2h_3$, which is much simpler than its true form $\partial f:E \to \mathcal L(E, F)$. I differentiate this untrue form and get an untrue $\partial^2 f: x \mapsto (h_2 \mapsto 6xh_3h_2)$. I guess the correct form is $\partial^2 f : x \mapsto \big ( (h_2, h_3) \mapsto 6xh_3h_2 \big)$.

To make sure that above process is valid. I aim at proving

$$ \partial^m f(x)[h_1, \ldots, h_m] = \partial g(x)[h_1] \text{ with }g:y \mapsto \partial^{m-1}f(y)[h_2,\ldots, h_m] $$ for all $x,h_1,\ldots, h_m \in E$.

Assume we know $\partial^{m-1} f$, this result reduces the computation of $\partial^m f:E \to \mathcal L^m(E, F)$ to that of $g:E \to F$.

Could you have a check on my attempt?

I posted my proof separately so that I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.

Answer 1

First, we need a lemma.

Let $E,F$ be Banach spaces and and $f:E \to F$ such that $\partial^2 f(x)$ exists and is not necessarily continuous. Then the continuous bilinear map $\partial^2 f(x)$ is symmetric. [A proof can be found here]

By our Lemma, $$\partial^m f(x)[h_1, \ldots, h_m] = \partial^m f(x)[h_2, \ldots, h_m, h_1] = \partial^m f(x)[h_2, \ldots, h_m] [h_1].$$

To make the notation compact, let's write $[\mathbf h]$ for $[h_2, \ldots, h_m]$. So it suffices to show that $\partial^m f(x)[\mathbf h] = \partial g(x)$. By definition of Fréchet derivative, we need to prove that $$ \lim_{y \to y_0} \frac{g(y)-g(y_0) - \partial^m f(y_0)[\mathbf h][y-y_0]}{|y-y_0|} = 0. $$

This is equivalent to $$ \lim_{y \to y_0} \frac{\partial^{m-1}f(y)[\mathbf h] - \partial^{m-1}f(y_0)[\mathbf h] - \partial^m f(y_0)[\mathbf h][y-y_0]}{|y-y_0|} = 0. $$

By our Lemma again, we get $\partial^m f(y_0)[\mathbf h][y-y_0] = \partial^m f(y_0)[y-y_0][\mathbf h]$. Hence we need to prove $$ \lim_{y \to y_0} \frac{\partial^{m-1}f(y)[\mathbf h] - \partial^{m-1}f(y_0)[\mathbf h] - \partial^m f(y_0)[y-y_0][\mathbf h]}{|y-y_0|} = 0. $$

Clearly, it suffices to prove $$ \lim_{y \to y_0} \frac{\partial^{m-1}f(y) - \partial^{m-1}f(y_0) - \partial^m f(y_0)[y-y_0]}{|y-y_0|} = 0. $$

This is trivially true because $\partial^m f = \partial (\partial^{m-1} f)$ by definition. This completes the proof.

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Let's apply this result to compute $\partial^3 f$ in our example. We have $\partial^3 f(x)[h_1, h_2, h_3] = \partial g(x)[h_1]$ with $g:y \mapsto \partial^{2}f(y)[h_2, h_3]$. In particular, $g: y \mapsto 6yh_3h_2$. Hence $\partial^3 f(x)[h_1, h_2, h_3] = 6h_3h_2h_1$.