My goal is to examine the continuity of a certain multivariable function. In that regard, (mainly because my math in the specific area is dusty) I find it somewhat tricky to either prove or disprove that property in the N-diensional space. However, I have come up with the following idea:
let $f : \mathbb{R}^2 \rightarrow \mathbb{R} $ be the function that I want to examine at some point $(x_0, y_0)$.
Now I am using the following two functions $g: \mathbb{R}^2 \rightarrow \mathbb{R} $ and $h: \mathbb{R} \rightarrow \mathbb{R} $ so that I can write function $f$ as: $$ f(x,y) = h(g(x,y)),\quad \forall (x,y) \in \mathbb{R}^2 $$ I also know that function $g$ is indeed continuous in $\mathbb{R}^2$.
My assumption is that I can examine the continuity of $h$ at $x' = g(x_0,y_0)$ in order to decide the continuity of $f$ at $(x_0,y_0)$ - a much easier task. In mathimatical notation, I guess what I mean is: $$ \lim_{(x,y)->(x_0,y_0)} f(x,y) = f(x_0, y_0) \quad \iff \\ \lim_{x->g(x_0,y_0)}h(x) = h(g(x_0,y_0)) $$
Actual Questions:
1. Is my assumption correct?
2. If not, maybe do either of $\Rightarrow$ or $\Leftarrow$ hold? (Because if one of those holds then I can use the above to only prove or only disprove accoridnlgy.)
3. Does the same hold for proving/disproving differentiability as well?
4. Under what conditions do any of those hold for a (finite) sum of $h,g$ functions, i.e. $f(x,y) = h_1(g_1(x,y)) + h_2(g_2(x,y)) + ... + h_k(g_k(x,y))$
Note: I can actually post my original function and the transformation if the question is too broad or the answers depend greatly on the actual $f,g$ and $h$ definitions. However I believe a more general question may help more people find it and also make it easier to be understood.
Just for closure, I am answering myself in case it helps anyone else. Merit goes to @GeauxMath and @Todor Markov for providing answers in comments.
We can answer mainly by using the following proposition:
In detail:
The $\Leftarrow$ direction holds by using the above proposition. The opposite is not true in every case. As a counter example
The corresponding property applies also to differentiable functions, so we can only prove that $f$ is differentiable that way, but not disprove it.