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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.

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    Thank you, I will fix this and see if it helps with the limit. – 2017-02-12
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    Still stuck, but now I'm not confused about the random $u$ that was in my limit. – 2017-02-12
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    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.) – 2017-02-12
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    What is $o(||y||)$? – 2017-02-12
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    See Landau notation, the big and small $o$, theta and omega classes. It just repeats that $\lim_{\|y\|\to 0}\frac{f(y)-y}{\|y\|}=0$. – 2017-02-12
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    I think the proof is right now? – 2017-02-13
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    Yes. If (big IF here) the functional equation has a solution, then it is differentiable by this proof. From $\frac{d}{dt}f(tx)=f'(tx;x)=(1+t^2\|x\|^2)x$ it follows by integration along the ray $f(x)=x+\frac13\|x\|^2·x$, which however has the derivative $f'(x;u)=u+\frac13\|x\|^2u+\frac23 xx^Tu$, which is obviously not what we started with. – 2017-02-13
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    Are you saying the function is not Fréchet differentiable? We haven't done anything with integration yet, so I do not quite follow what you're talking about. Perhaps you know of a more elementary way to see it is not Fréchet differentiable? (We also haven't done anything with second derivatives. The definition above and checking if the partials are continuous are all that we have covered so far.) – 2017-02-13

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