Continuous partial derivatives implies continuous differential

Question

We have the well-known statement (Analysis I by Zorich, p.457):

Let $f: U(x) \to \mathbb{R}$ be a function defined in a neighbourhood $U(x) \subseteq \mathbb{R}^m$ of the point $x = (x^1,\dots,x^m)$. If the function $f$ has all partial derivatives $\frac{\partial f}{\partial x^1},\dots,\frac{\partial f}{\partial x^m}$ at each point of $U(x)$ and they are continuous at $x$, then $f$ is differentiable at $x$.

Now my question is, that somehow my lecture notes suggest, that when the partial derivatives are continuous at all points of $U(x)$, $f$ is then continuously differentiable, i.e. the map $U(x) \to \text{Hom}(\mathbb{R}^m,\mathbb{R})$ is continuous. How can this be seen?

Answer 1

The partial derivatives$\frac{\partial f}{\partial x^1},\dots,\frac{\partial f}{\partial x^m}$ are only continuous at $x$. Nothing is said about the continuouity of these derivatives in points of $U(x) \setminus \{x\}$. So $f$ need not to be continuously differentiable.

Answer 2

We have to show that $U(x) \to \text{Hom}(\mathbb{R}^m,\mathbb{R})$. Let $x_k$ be a sequence in $U(x)$ with $\lim_{k \to \infty} x_k = x$. Since $\text{Hom}(\mathbb{R}^m,\mathbb{R})$ is a finite dimensional real vector space, we have that all norms are equivalent. Thus in the Frobenius norm we get $$\|Df(x_k) - Df(x)\|^2_F = \sum_{i = 1}^m|Df(x_k) - Df(x)|^2 = \sum_i^n \left\vert \frac{\partial f}{\partial x_i}(x_k) - \frac{\partial f}{\partial x_i}(x)\right\vert^2 \to 0$$ by the continuity of the partial derivatives.