Proving continuity of a multivariable function by “reducing” it to a single variable function

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.

• We know that the composition of continuous functions is continuous. So it suffices to show that $g(x,y)$ is continuous at $\xi_0 = (x_0,y_0)$. That is that $\lim_{\xi \rightarrow \xi_0} g(\xi) = g(\xi_0)$. Also note that showing that the derivative exists at $x_0$ is a sufficient, but not necessary condition for continuity. That is differentiability implies continuity but not the converse. For an example see $f(x) = |x|$. This is continuous at $0$, and in fact Lipschitz continuous at $0$, but not differentiable there. (continued) – GeauxMath Nov 13 '18 at 23:33
• Also note a vector valued function is continuous if it is continuous in all its arguments – GeauxMath Nov 13 '18 at 23:39
• @GeauxMath First of all, thank you for responding. Now, I am fully aware about how differentiability implies continuity but not the other way around. Reading back my post, I think I was indeed looking for the property of compositions of continuous functions being continuous. I also take it, that in your first comment you meant to say that it suffices to show that $h(x)$ is continuous and not $g(x,y)$, as I already mentioned that $g$ is indeed continuous. – kyriakosSt Nov 14 '18 at 0:24
• yes, you are correct – GeauxMath Nov 14 '18 at 1:18
• Cheers. If you would like to modify your comment to an answer, I am willing to accept it – kyriakosSt Nov 14 '18 at 10:06

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:

A composition of continuous functions is continuous

In detail:

1. My assumption is incorrect as a whole
2. The $$\Leftarrow$$ direction holds by using the above proposition. The opposite is not true in every case. As a counter example

If $$g(x_0,y_0)=a$$ and $$f(x,y)≥a$$, then $$h$$ can have a discontinuity at a. If $$h(x)$$ is continuous for $$x>a$$, $$f$$ would still be continuous.

3. The corresponding property applies also to differentiable functions, so we can only prove that $$f$$ is differentiable that way, but not disprove it.

4. Again, by using that proposition we can extend our statements to that specific case, so we can prove continuity and differentiability of $$f$$ but not disprove. A counter example here would be a case where $$h_1$$ and $$h_2$$ are not continuous, but by adding them, their discontinuities "cancel out".