Let $U \subset \mathbb{R}^n$ open set, $x_0 \in U$ and $f : U \rightarrow \mathbb{R}$ a function. Show that if $f$ is Lipschitz and Gâteaux-differentiable at point $x_0$, then $f$ is Fréchet-differentiable at point $x_0$.

For $\epsilon >0$ I tried to consider a finite set of vectors $v_1, ..., v_k \in S^{n-1}$ such that $\forall \, \, v \in S^{n-1} \, \, \exists \, i \in \{1, ..., k\}$ with $|x- v_i | < \epsilon $ and then using the Gâteaux-derivative in the directions $v_1, ..., v_k$, but I don't see how to proceed and to use it to get the result.


Let $v$ be the Gâteaux derivative of $f$ at $x_0$. To simplify inequalities, subtract off the linear term: that is, consider $g(x) = f(x) - f(x_0) - v\cdot(x-x_0)$.

Known: $g$ is Lipschitz continuous, and $\lim_{h\to 0} g(x_0+hv)/h = 0$ for every vector $v\in\mathbb{R}^n$.

Goal: $\lim_{x\to x_0} g(x)/\|x-x_0\| = 0$.

Idea of proof: since we control the behavior of $g$ on each line through $x_0$, take many such lines and use Lipschitz continuity to fill (small) gaps between them.

Proof: Given $\epsilon>0$, let $K$ be a finite subset of the unit sphere $S^{n-1}$ such that for every $u\in S^{n-1}$ there is $v\in K$ with $\|u-v\| \le \epsilon$. The existence of such $K$ follows from the compactness of $S^{n-1}$.

Since $K$ is finite, Gâteaux differentiability implies there is $\delta>0$ such that $|g(x_0+hv)| \le \epsilon h$ for all $v\in K$, whenever $|h|<\delta$.

Let $u$ be any unit vector. By the above, there exists $v\in K$ such that $\|u-v\| \le \epsilon$, hence $\|hu - hv\| \le \epsilon h$. The Lipschitz continuity of $g$ implies
$|g(x_0+hu)-g(x_0+hv)| \le L \epsilon h$ where $L$ is the Lipschitz constant. Combining the above, we get $$ |g(x_0+hu)| \le L \epsilon h + |g(x_0+hv)| \le (L+1) \epsilon h $$ whenever $|h| < \delta$. Writing $x = x_0+hu$ we can rephrase this as $$ \sup_{\|x-x_0\| < \delta} \frac{|g(x)|}{\|x-x_0\|} \le (L+1)\epsilon $$ Since $\epsilon$ can be arbitrarily small, it follows that $$\lim_{x\to x_0} \frac{|g(x)|}{\|x-x_0\|} = 0$$


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