First-time poster here, hope it all works ok with LaTex etc.
There's something here I'm missing. Would really appreciate any help.
The question states:
Suppose that $f$ is a real-valued function defined in an open set $E \subset \mathbb{R}^n$, and that the partial derivatives are bounded in $E$. Prove that $f$ is continuous in $E$.
I've found an answer on the internet that uses the Mean Value Theorem on the partial derivatives, together with the triangle inequality, to show that the function can be made less than epsilon by choosing for fix $x \in E$, a $y \in E$ such that $$ d(x,y) < \delta = \dfrac{\epsilon}{(n + 1)M},$$ where M is the maximum of the partial derivatives on E.
My question is: how do we know the partial derivatives are continuous in order to apply the Mean Value Theorem? Just because partial derivatives exist, it does not mean they have to be continuous, I thought? Is it something to do with that E is an open set or that the derivatives are bounded?
Thanks!
Edit: The solution I have found states:
By the Mean Value Theorem there is a number $c_k$ between $x_k^0$ and $y_k$ such that the last difference equals $$|{(D_kf)(x_1^0,x_2^0,…,x_{k-1}^0,c_k,y_{k+1},…,y_n)(y_k – x_k^0)}|,$$
where $D_kf$ is the k-th partial derivative of f.
Best Answer
Since you have not included the whole proof, I am not sure whether this helps. But note that the functions $$f_{k}(t)=f(x_{1},..,x_{k-1},t,x_{k+1},...,x_{n})$$ are all differentiable in $t$ for any collection of fixed $x_{1},..,x_{k-1},x_{k+1},...,x_{n}$ and thus continuous as a consequence. Hence we may apply the Mean Value Theorem on each of them seperately, i.e. in each axis direction. Then we may use the triangle inequality to get continuity of $f$ as a whole.