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For $X \subset \mathbb{R}^d$ open, we define $$ C^1(X) := \left \{ f : X \to \mathbb{C} : f \text{ is a function s.t. } \frac{\partial f}{\partial x_j} \text{ exists and is continuous for } j = 1, 2, \ldots, d \right \}. $$ Consider the following mean-value theorem :

Thm. : Let $X \subset \mathbb{R}^d$ be open, let $f \in C^1(X)$ and let $a,b \in X$ be s.t. $[a,b] \subset X$. Then $$ |f(b)-f(a)| \leq M|b-a| $$ where $M := \displaystyle \sum_{j=1}^d \sup_{x \in [a,b]} \left| \frac{\partial f}{\partial x_j} (x) \right|$.

My questions regarding this theorem are :

  1. Could someone give me a counterexample when $f$ is differentiable on $X$ but $f \notin C^1(X)$ ?

  2. Is it necessary that $f \in C^1(X)$ ? More precisely, does the theorem hold if we suppose that $\frac{\partial f}{\partial x_j}$ is continuous on $[a,b]$ only ?

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There is no such counterexample and the usual proof of the multidimensional MVT shows that continuity of the derivative is completely superfluous: One reduces to the onedimensional case by considering the function $h:=t\mapsto f(ta+(1-t)b)$ on the interval $[0,1]$ which is by construction differentiable. Without assuming continuity of the derivative the ondimensional MVT gives the existence of a $\tau\in [0,1]$ with $f(a)-f(b) = h'(\tau) = Df(\tau b + (1-\tau)b)\cdot(a-b)$ and thus $|f(a)-f(b)| \leq \|Df(\tau a+ (1-\tau b))\| \cdot \|a-b\|$. The operator norm of the differential is bounded by $M$.

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