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.

  • $\begingroup$ 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) $\endgroup$ – GeauxMath Nov 13 '18 at 23:33
  • $\begingroup$ Also note a vector valued function is continuous if it is continuous in all its arguments $\endgroup$ – GeauxMath Nov 13 '18 at 23:39
  • $\begingroup$ @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. $\endgroup$ – kyriakosSt Nov 14 '18 at 0:24
  • 1
    $\begingroup$ yes, you are correct $\endgroup$ – GeauxMath Nov 14 '18 at 1:18
  • $\begingroup$ Cheers. If you would like to modify your comment to an answer, I am willing to accept it $\endgroup$ – 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".

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.