1
$\begingroup$

Let $E,F$ be Banach spaces, $X$ an open set in $E$, $x_0\in X$ and $f:X\to F$. Then the $n$th Fréchet derivative $\partial^nf(x_0)$ of $f$ at $x_0$ is an $n$-linear map in $\mathcal{L}^n(E,F)$. So what is the meaning of $\partial^nf(x)(h_1,\ldots,h_n)$? How does $\partial^nf(x)(h_1,\ldots,h_n)$ depend on the $h_i$'s?

For example, when $E=\mathbb{R}^n$ and $(h_1,\ldots,h_n)=(e_{i_1},\dots,e_{i_k})$ are standard basis vectors, then $\partial^nf(x)(e_{i_1},\dots,e_{i_k})$ is equal to some higher order partial derivative of $f$: $$\partial^nf(x)(e_{i_1},\dots,e_{i_k}) = \frac{\partial^nf(x)}{\partial x_{i_1}\cdots\partial x_{i_n}}.$$ However, I could not see why this is true after contemplating the definition for about an hour. Can anyone help explain this formula? Why is it true?

Please let me know if my question is unclear or too soft. Thanks in advance!

$\endgroup$
1
  • $\begingroup$ Define the "diference operator" $\Delta_h$ to be $\Delta_hf(x) = f(x + h) - f(x).$ Then, $\Delta_{h_1, \ldots, h_n} f(x) = f^{(n)}(x) \cdot (h_1, \ldots, h_n) + o(h_1, \ldots, h_n),$ where $o$ is a function that vanishes faster than $\|h_1\| \cdots \|h_n\|.$ $\endgroup$
    – William M.
    Commented Jun 24, 2018 at 23:06

2 Answers 2

2
$\begingroup$

$\partial^n f(x_0)$ is the value of $\partial^n f$ at $x_0$. And $\partial^n f(x_0)$ is a multilinear map from $E \times \dots \times E$ into $F$. $\partial^nf(x)(h_1,\ldots,h_n)$ is the value of this multilinear map at the $n$-uple $(h_1, \dots, h_n)$, i.e. a vector in $F$. See Fréchet - Higher derivatives.

Now regarding $$\partial^nf(x)(e_{i_1},\dots,e_{i_n}) = \frac{\partial^nf(x)}{\partial x_{i_1}\cdots\partial x_{i_n}}.$$

First a comment. The fact that you're using $n$ as the dimension of $E$ and the rank of Fréchet derivative is really not helping... and confusing.

Anyhow, this is coming from the way to denote a $n$-multinear map $A$ on $E=\mathbb R^k$ in a matrix way. If $h_i = \sum\limits_{j_i=1}^k h_i^{j_i} e_{j_i}$ for $1 \le i \le n$, then $$A(h_1, \dots, h_n) = A(\sum\limits_{j_1=1}^k h_1^{j_1} e_{j_1},\dots,\sum\limits_{j_n=1}^k h_n^{j_n} e_{j_n}) = \sum\limits_{j_1=1}^k \dots \sum\limits_{j_n=1}^k h_1^{j_1} \times \dots \times h_n^{j_n}A(e_{j_1}, \dots e_{j_n})$$

$\frac{\partial^nf(x)}{\partial x_{i_1}\cdots\partial x_{i_n}}$ is just a way in that case to denote $A(e_{i_1}, \dots e_{i_n})$ for the multilinear map $\partial^n f(x)$.

$\endgroup$
7
  • $\begingroup$ Thanks for the explanation! $\frac{\partial^nf(x)}{\partial x_{i_1}\cdots\partial x_{i_n}}$ is a partial derivative, while $\partial^nf(x)(e_{i_1},\dots,e_{i_n})$ is a value of the Fréchet derivaitve. What I'm asking is why these are the same. In other words, I'm confused about the actual meaning of the values of the Fréchet derivaitve. For example, by the same token, can I conclude that $\partial^nf(x)(h_1,\ldots,h_n)$ is a "directional derivative of order $n$"? $\endgroup$
    – Yuxiao Xie
    Commented Jun 20, 2018 at 14:23
  • $\begingroup$ What I explained in my answer is that those two objects are the same. The partial derivative is the value of the Fréchet derivative on vectors of the canonical basis. $\endgroup$ Commented Jun 20, 2018 at 14:29
  • $\begingroup$ This answer is pretty bad. The partial derivatives are related with higher derivatives in the particular case of a product of Banach spaces. $\endgroup$
    – William M.
    Commented Jun 24, 2018 at 23:07
  • $\begingroup$ @WillM. I don't understand your comment. In the second paragraph of the question, a product of spaces - namely $\mathbb R^n$ is explicitly mentioned. $\endgroup$ Commented Jun 25, 2018 at 6:39
  • $\begingroup$ Sorry for the late response. My question is actually how to prove that $\partial^nf(x)(e_{i_1},\dots,e_{i_n}) = \frac{\partial^nf(x)}{\partial x_{i_1}\cdots\partial x_{i_n}}$? I perfectly understand that $\partial^nf$ is an $n$-linear map. The problem is that I'm not sure what happens if I plug in the $e_i$'s... So how is the above formula derived? Could you please include a proof? $\endgroup$
    – Yuxiao Xie
    Commented Jul 2, 2018 at 0:08
0
$\begingroup$

The answer of mathcounterexamples.net is spot on. However, I think it may help your understanding if we look at something a little more specific. I will address two things:

  1. What is the intuitive meaning of the $n$th-order Frechet derivative at $x_0$?
  2. How can we convince ourselves that the action of $\partial^kf(x_0)$ on a $k$-tuple of basis vectors is equal to one of the usual $k$th-order partial derivatives of $f$?

To address item 1: In short, the $n$th order Frechet derivative of $f$ at $x_0$ is the $n$th-order approximating map of $f$ at at $x_0$. I will illustrate this with $n = 1$ and $n = 2$.

Let $E = X = \mathbb R^2$, let $F = \mathbb R$ and consider the second-order Taylor expansion of a $C^\infty$ function $f:\mathbb R^2\to \mathbb R$ at $x_0 = 0$: $$ f(x) \sim f(0) + \partial^1f(0)[x] + \frac 12\partial^2f(0)[x, x]. $$ Here $\partial^1f(0)[x]$ is the action of $f$'s first Frechet derivative at the origin on $x\in \mathbb R^2$ and $\partial^2f(0)[x,x]$ is the action of $f$'s second Frechet derivative at the origin on $(x,x)\in \mathbb R^2\times \mathbb R^2$.

Once we fix a basis for $\mathbb R^2$, the linear map $\partial^1f(0):\mathbb R^2\to \mathbb R$ and the bilinear map $\partial^2f(0):\mathbb R^2\times \mathbb R^2\to\mathbb R$ can be represented concretely as (multiplication by) matrices of appropriate sizes. For example, if we fix the standard basis on $\mathbb R^2$ then $\partial^1f(0)$ is represented by the matrix $[\frac{\partial f}{\partial x_1}(0), \frac{\partial f}{\partial x_2}(0)]$ (or its transpose depending on your convention). That is, the action of $\partial^1f(0)$ on $x\in \mathbb R^2$ is \begin{eqnarray*} \partial^1f(0)[x] & = & [\frac{\partial f}{\partial x_1}(0), \frac{\partial f}{\partial x_2}(0)] \begin{bmatrix} x_1\\ x_2 \end{bmatrix}\\ & = & \frac{\partial f}{\partial x_1}(0) x_1 + \frac{\partial f}{\partial x_2}(0) x_2. \end{eqnarray*} Similarly, $\partial^2f(0)$ can be represented by the matrix \begin{equation*} \begin{bmatrix} \frac{\partial^2f}{\partial x_1^2}(0)& \frac{\partial^2 f}{\partial x_1\partial x_2}(0)\\ \frac{\partial^2f}{\partial x_2\partial x_1}(0) & \frac{\partial^2 f}{\partial x_2^2}(0) \end{bmatrix}. \end{equation*} In particular, the acton of $\partial^2f(0)$ on $(x,y)\in \mathbb R^2\times \mathbb R^2$ is \begin{equation*} \partial^2f(0)[x,y] = [x_1, x_2] \begin{bmatrix} \frac{\partial^2f}{\partial x_1^2}(0)& \frac{\partial^2 f}{\partial x_1\partial x_2}(0)\\ \frac{\partial^2f}{\partial x_2\partial x_1}(0) & \frac{\partial^2 f}{\partial x_2^2}(0) \end{bmatrix} \begin{bmatrix} y_1\\ y_2 \end{bmatrix}. \end{equation*} Choosing $y = x$ (for purposes of comparison with the above Taylor expansion) we get \begin{equation*} \partial^2f(0)[x,x] =\sum_{i, j = 1}^2 \frac{\partial^2f}{\partial x_i\partial x_j}(0) x_ix_j. \end{equation*} Using these matrix representations of the first two Frechet derivatives of $f$ at $0$, the second-order Taylor approximation of $f$ at the origin becomes \begin{equation*} f(x) \sim f(0) + \sum_{i =1}^2\frac{\partial f}{\partial x_i}(0)x_i + \frac12\sum_{i,j = 1}^2 \frac{\partial^2 f}{\partial x_i\partial x_j}(0) x_i x_j \end{equation*} which is the classical Taylor expansion one encounters in multivariable calculus. For this reason, I interpret the $n$th-order Frechet derivatives of $f$ at $0$ as the $n$th-order approximation to $f$ at $0$.

To address item 2: Look at the matrix representation of $\partial^2f(0)$ relative to the standard basis. Letting this matrix act on, for example $e_1 = [1, 0]$ and $e_2 = [0, 1]$ we get \begin{equation*} [1,0] \begin{bmatrix} \frac{\partial^2f}{\partial x_1^2}(0)& \frac{\partial^2 f}{\partial x_1\partial x_2}(0)\\ \frac{\partial^2f}{\partial x_2\partial x_1}(0) & \frac{\partial^2 f}{\partial x_2^2}(0) \end{bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix} = [\frac{\partial^2 f}{\partial x_1^2}(0), \frac{\partial^2f}{\partial x_1\partial x_2}(0)] \begin{bmatrix} 0\\ 1 \end{bmatrix} = \frac{\partial^2f}{\partial x_1\partial x_2}(0). \end{equation*} Similarly, one can check that letting the matrix representation of $\partial^2f(0)$ act on $(e_i, e_j)$ for any $i,j\in \{1, 2\}$ gives $\frac{\partial f}{\partial x_i\partial x_j}(0)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .