# Meeting the Continuity Condition for the Use of Mean Value Theorem in Multivariable Differentiability Proof

My question concerns the use of the Mean Value Theorem in the proof of the following theorem:

Theorem(1): Suppose $$f : D \rightarrow \mathbb{R}^p$$ (where $$D \subset \mathbb{R}^n$$ is open) is a function with continuous partial derivatives $$D_kf$$ for $$\mathbf{x} \in D$$. Then $$f$$ is differentiable.

A typical proof of Theorem(1) invokes the Mean Value Theorem to derive equality. I walk through a proof here for clarity/consistency of notation.

Given the properties of $$f$$, we must prove that there exists a linear function $$L : \mathbb{R}^n \rightarrow \mathbb{R}^p$$ such that:

$$\lim\limits_{\mathbf{h} \rightarrow \mathbf{0}} \dfrac{f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - L(\mathbf{h})}{\|\mathbf{h}\|} = \mathbf{0}$$

Then, given $$\mathbf{h} = (h_1,\dotsc,h_n)$$, we define $$\mathbf{z}_k = (h_1,\dotsc,h_k,0,\dotsc,0)$$ and replace the difference $$f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x})$$ with the telescoping sum

$$f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) = \sum\limits_{k=1}^n f(\mathbf{x} + \mathbf{z}_k) - f(\mathbf{x} + \mathbf{z}_{k-1})$$

Next, we define single-variable functions $$g_k$$

$$g_k : t \mapsto f(\mathbf{x} + \mathbf{z}_{k-1} + th_k\mathbf{e}_k)$$ for $$t \in [0,1]$$

such that

$$f(\mathbf{x} + \mathbf{z}_k) - f(\mathbf{x} + \mathbf{z}_{k-1}) = g_k(1) - g_k(0)$$

And at this point, we use MVT.

On first pass, I wrongly assumed that the continuity condition enabling use of the Mean Value Theorem was met by the fact that differentiability (of $$f$$) implies continuity (of $$f$$). (I eventually noted that this reasoning would be circular and that this surely could not be the justification because we are trying to prove differentiability!) So then I came to the conclusion that despite each of the constructed single variable functions $$g_k$$ being defined in terms of the original function $$f$$, to apply MVT, we need only continuity of $$g_k$$.

First, I'd appreciate it if someone could verify my rationale:

The use of the Mean Value Theorem here is justified because the definitions of single variable functions $$g_k$$, is such that $$g_k' = D_k f$$. Thus, existence of each $$D_k f$$ is equivalent to differentiability of each $$g_k$$, which implies continuity of each $$g_k$$, so the conditions for MVT are satisfied (never mind the particulars of closed set continuity and open set differentiability).

Second, upon going through this proof construction, I think, with some modification, there is some intuition to be had from my initial (incorrect) justification of MVT:

The existence of each partial derivative guarantees that each $$g_k$$ is continuous over its domain. Thus,

$$g(t_0) = \lim\limits_{t \rightarrow t_0} g(t) = \lim\limits_{t \rightarrow t_0} f(\mathbf{x} + \mathbf{z}_{k-1} + th_i\mathbf{e}_k) = f(\mathbf{x} + \mathbf{z}_{k-1} + t_0h_i\mathbf{e}_k)$$ for all $$t_0 \in [0,1]$$

As a similar relation would hold were $$\mathbf{e}_k$$ replaced with any other vector, we have that for any direction along which the directional derivative exists, $$f$$ is "continuous" along that path. I suppose this can be generalized to any arbitrary smooth path (as opposed to straight-line path) as well. Is this correct?