Let $f: \mathbb{R}^n \to \mathbb{R}^n$ satisfy the following: ($|| \cdot || = || \cdot ||_2$)

$$f(x+y) = f(x) + f(y) + ||x||^2y + ||y||^2x, \quad \forall x, y \in \mathbb{R}^n,$$ and $$\lim_{\delta \to 0} \frac{f(\delta x)}{\delta} = x, \quad \forall x \in \mathbb{R}^n.$$ Show that $f(\cdot)$ admits a gradient at any $x \in \mathbb{R}^n$. Find the gradient. Is $f(\cdot)$ Fréchet differentiable? Why?

First, I will show that $f(\cdot)$ admits a gradient. Let us find the directional derivative at $x_0$ in the direction of $u$.

\begin{align} f'(x_0; u) &= \lim_{\delta \ \downarrow \ 0} \frac{f(x_0 + \delta u) - f(x_0)}{\delta}\\ &= \lim_{\delta \ \downarrow \ 0} \frac{f(x_0) + f(\delta u) + ||x_0||^2\delta u + ||\delta u||^2 x_0 - f(x_0)}{\delta}\\ &= \lim_{\delta \ \downarrow \ 0} \frac{f(\delta u)}{\delta} + \frac{||x_0||^2\delta u + ||\delta u||^2 x_0}{\delta}\\ &= u + ||x_0||^2u \end{align}

In order for $f(\cdot)$ to admit a gradient, we must find a vector $(\Delta f)(x_0)$ such that $f'(x_0; u) = (\Delta f)(x_0) \cdot u$ for all unit vectors $u$.

Well, $f'(x_0; u) = u + ||x_0||^2u = (1 + ||x_0||^2) \cdot u$, so I think the gradient is $(1 + ||x_0||^2)$.

Next we check if $f(\cdot)$ is Fréchet differentiable. We say that $f(\cdot)$ is Fréchet differentiable at $x_0$ if there exists a vector denoted by $f_x(x_0)$ such that $$ \lim_{||y|| \to 0} \frac{f(x_0 + y) - f(x_0) - f_x(x_0) \cdot y}{||y||} = 0$$

We have learned that if a function is Fréchet differentiable, then $f(\cdot)$ admits a gradient $(\Delta f)(x_0)$ and that $(\Delta f)(x_0) = f_x(x_0).$ I believe this means that the only candidate I must check for $f_x(x_0)$ is the gradient found above, is this right? In that case, we must determine whether the limit below is zero.

\begin{align} &\lim_{||y|| \to 0} \frac{f(x_0 + y) - f(x_0) - f_x(x_0) \cdot y}{||y||} \\ = &\lim_{||y|| \to 0} \frac{f(y) + ||x_0||^2y + ||y||^2x_0 - (1+ ||x_0||^2)\cdot y}{||y||} \\ = &\lim_{||y|| \to 0} \frac{f(y) + ||y||^2x_0 - y}{||y||} \\ = &\lim_{||y|| \to 0} \frac{f(y) - y}{||y||} \\ = &\lim_{||y|| \to 0} \frac{f(||y||\frac{y}{||y||}) - ||y||\frac{y}{||y||}}{||y||} = 0. \end{align}

Thus, $f(\cdot)$ is Fréchet differentiable.

  • $\begingroup$ Thank you, I will fix this and see if it helps with the limit. $\endgroup$ – user389056 Feb 12 '17 at 21:47
  • $\begingroup$ Still stuck, but now I'm not confused about the random $u$ that was in my limit. $\endgroup$ – user389056 Feb 12 '17 at 21:55
  • $\begingroup$ You need to use the assumptions, esp. $f(y)=y+o(\|y\|)$. However, the second derivatives do not commute, $f_{1;12}$ is different from $f_{1;21}=0$ as $f_{1;1}=1+\|x\|^2$ while $f_{1;2}=0$. (Using some old derivative notation, $f_{j;k}=\frac{\partial f_j}{\partial x_k}$ etc.) $\endgroup$ – LutzL Feb 12 '17 at 21:55
  • $\begingroup$ What is $o(||y||)$? $\endgroup$ – user389056 Feb 12 '17 at 21:59
  • $\begingroup$ See Landau notation, the big and small $o$, theta and omega classes. It just repeats that $\lim_{\|y\|\to 0}\frac{f(y)-y}{\|y\|}=0$. $\endgroup$ – LutzL Feb 12 '17 at 23:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.