Locally Lipschitz and Gâteaux Derivative if and only if Frechet Derivative

Question

Consider $f$ locally Lipschitz. So $f$ is Gâteaux Derivative if and only if $f$ is Frechet Derivative.

PS.: the converse is trivial.

Answer 1

I assume the Gâteaux derivative is required to be linear: this is often, but not always, a part of its definition. Without linearity the claim would be false.

Let's consider differentiability at $a\in\mathbb{R}^n$; we may assume $a=0$ by translation. As often, it helps to subtract off a linear function to achieve $f(0)=0$ and $Df(0)=0$ (where $Df$ is the Gâteaux derivative). The goal is to prove that $$\frac{1}{r}\max_{|x| = r} |f(x)| \to 0,\quad r\to0$$ which gives Fréchet derivative.

Let $L$ be the Lipschitz constant of $f$ within the unit ball. Given $\epsilon>0$, pick a finite $(\epsilon/L)$-net $N$ within the unit sphere of $\mathbb{R}^n$. By the Gâteaux differentiability, there is $\rho>0$ such that $$|f(r\xi )|/r <\epsilon$$ whenever $r<\rho$ and $\xi\in N$ (the key point is that $N$ is finite).

Observe that every point $x$ with $|x|=r$ lies within distance $(\epsilon/L) r$ of some point of the form $r\xi$ with $\xi\in N$. Conclude by the Lipschitz property that $$|f(x)|\le |f(x)-f(r\xi)|+|f(r\xi)|\le 2 \epsilon r$$

Answer 2

Seems to me the result is false.

Say $S$ is the unit circle in $\mathbb R^2$. Say $\phi:S\to\mathbb R$ is any smooth function with $\phi(-x)=\phi(x)$. There's a function $f:\mathbb R^2\to\mathbb R$ such that

1. $f(x)=\phi(x)\quad(x\in S),$

2. $f(tx)=tf(x)\quad(x\in\mathbb R^2,t\in\mathbb R).$

This seems like a counterexample to me, unless it just happens that $\phi$ is the restriction to $S$ of a linear functional. What am I missing?

Answer 3

Given $x \in U$, let us suppose that there exists $g \in \mathbb R^n$ (the Gâteaux derivative of $f$ at $x$) such that $$\lim_{t \rightarrow 0}\frac {f(x+th)-f(x)-t\langle g,h \rangle}{t}=0$$ for every $h \in \mathbb R^n$.

We want to show that $$\lim_{h\rightarrow 0} \frac {f(x+h)-f(x)-\langle g,h \rangle}{\|h\|}=0$$ if $f$ is $L$-Lipschitz at $x$.

If it is not true, there exist $\varepsilon>0$ and a sequence $\{h_n\}$ with $h_n \rightarrow 0$ and $h_n \neq 0$ for every $n$ such that $$\frac {|f(x+h_n)-f(x)-\langle g,h_n \rangle|}{\|h_n\|} \ge \varepsilon$$ (remember the sequential criterium for functional limits)

Since the unit sphere in $\mathbb R^n$ is compact, $\{h_n\}$ admits a subsequence $\{u_n\}$ such that the sequence $\left\{\frac {u_n}{\|u_n\|}\right\}$ converges. Let $t_n=\|u_n\|$, $v_n=\frac {u_n}{\|u_n\|}$ and $v_n \rightarrow v\,$. Note that $t_n \rightarrow 0\,$ and $\|v\|=1$.

The compactness of the unit sphere in $\mathbb R^n$ is essential in this proof and characterizes the Banach spaces of finite dimension.

Then, for $n$ sufficiently large $$\varepsilon \le \frac {|f(x+u_n)-f(x)-\langle g,u_n \rangle|}{\|u_n\|}=$$$$\frac {|f(x+t_nv_n)-f(x)-t_n\langle g,v_n \rangle|}{t_n}=$$$$\frac {|f(x+t_nv_n)-f(x+t_nv)+t_n \langle g,v \rangle-t_n \langle g,v_n \rangle +f(x+t_nv)-f(x)-t_n \langle g,v \rangle|}{t_n} \le $$$$L \cdot \|v_n-v\|+\|g\| \cdot \|v_n-v\|+\frac {|f(x+t_nv)-f(x)-t_n \langle g,v \rangle |}{t_n} \rightarrow 0$$ so $\varepsilon \le 0$, a contradiction.