Let $f: \Omega \subseteq \mathbb{R}^{n} \longrightarrow \mathbb{R}$ be a continuous function in the closed segment $[x,y] \subset \Omega $, such that the partial derivative with respect to the j-th variable $( \frac{\partial f(x)}{ \partial x_{j}})$ is defined in the segment $ (x,y) $. Prove that $ \exists z \in (x,y) $ such that:
$$f(y) – f(x) = \frac{\partial f(z)}{ \partial x_{j}} (y_{j} – x_{j})$$
Honestly I'm really surprised that I ended up having to ask this, because at first I thought the prove would just closely follow the same structure from similar theorems. But my main problem is that the conditions I'm given restrict me from using the theorems that I am comfortable with. Most similar questions I've found (like this or this) refer to different MVTs (which I actually already know), but can't seemingly be applied here. The proof for the first one (it's Theorem 36, just using this as a reference) doesn't work here because it only proves the existence of a directional derivative, where the direction is the one from the segment, so for example in my case it could only prove the directional derivative in the direction $ \frac{y-x}{||y-x||} $. In fact, the proof does require that the limit $ \lim_{t \rightarrow t_{0}} g(t)$ is only evaluated for points that are in the segment, because otherwise you can't guarantee that the composition of f and g are continuous, and can't apply the single-variable MVT.
The second one does imply existence of all partial derivatives, but it requires differentiability so that's out of the question.
What I gathered from both proofs is that they are usually revolved around reducing the multivariable functions down to functions in $ \mathbb{R} $, where we can use the MVT for the single-variable case. However, I don't know how to do that in this case. I think that my main problem comes from the fact that I don't know what is the direction of the segment $[x,y]$.
Could anyone please give me a hint on how to build a function that lets me reduce this problem to a single-variable case? Or should I take a completely different approach?
Best Answer
We may have $x_j=y_j$ but $f(x) \neq f(y)$. So the assertion is false.