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?



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