Two-variable functions - If $f$ is a continuous function in $(x_0,y_0)$ and $g$ is a one-variable function continuous in $f(x_0,y_0)$ so...

Question

In a given exercise I am asking the following:

If $f$ is a continuous function in $(x_0,y_0)$ and $g$ is a one-variable function continuous in $f(x_0,y_0)$ so the composite function $h$, defined by $h(x,y)=g(f(x,y))$ is continuous in $(x_0,y_0)$. True or false?

Since the textbook has no solution so I would like to check my reasoning here on this forum.

For me it seems true since $f$ is a multivariable function continuous on the given point $(x_0, y_0)$ and $h(x,y) = g(f(x,y))$ is basically a level curve of $h(x,y)$.

Answer 1

Yes of course ! Let $((a_n,b_n))_n$ a sequence that converge to $(x_0,y_0)$. Since $f$ is continuous at $(x_0,y_0)$, then $(f(a_n,b_n))_n$ is a sequence that converge to $f(x_0,y_0)$. Since $g$ is continuous, one finally get $$\lim_{n\to \infty }h(a_n,b_n)=\lim_{n\to \infty }g(f(a_n,b_n))=g(f(x_0,y_0))=h(x_0,y_0)$$

Answer 2

If $X,Y,Z$ are topological spaces, $f:X\to Y$ is a function continuous at $x_0$ and $g:Y\to Z$ is a function continuous at $y_0=f(x_0)$ then composition $h=g\circ f:X\to Z$ is continuous at $x_0$.

Let $U$ be an open set of $Z$ with $h(x_0)=g(f(x_0))\in U$.

In order to prove that $h$ is continuous at $x_0$ it is sufficient to prove that some open set $W\subseteq X$ exists with $x_0\in W$ and $h(W)\subseteq U$.

Since $g$ is continuous at $f(x_0)$ an open subset $V\subseteq Y$ exists with $f(x_0)\in V$ and $g(V)\subseteq U$.

Then - since $f$ is continuous at $x_0$ - an open subset $W\subseteq X$ exists with $x_0\in W$ and $f(W)\subseteq V$.

Then consequently $h(W)=g(f(W))\subseteq g(V)\subseteq U$.

Proved is now that $h$ is continuous at $x_0$.

Any further structure of the functions/spaces is irrelevant.

Answer 3

Yes, this is true, even if we do not restrict to one variable functions. First, suppose $f$ is continuous on $(x_0, y_0)$, and suppose that $g$ is continuous on $(f(x_0),f(y_0))$. Since $g$ is continuous, then for any $\epsilon > 0$, we can find a $\delta_1 > 0$ such that $$ \left| x' - y' \right| < \delta_1 \Rightarrow \left| g(x') - g(y') \right| < \epsilon $$ for any $x',y' \in (f(x_0), f(y_0))$. Now, since $f$ is continuous, then for this $\delta_1$, we can find $\delta_2$ such that $$ \left| x - y \right| < \delta_2 \Rightarrow \left| f(x) - f(y) \right| < \delta_1. $$ But since $f(x), f(y) \in (f(x_0), f(y_0))$, this gives us that $\left| g(f(x)) - g(f(y)) \right| < \epsilon$ (from the first statement). But this tells us that $$ \left| x - y \right| \Rightarrow \left| g(f(x)) - g(f(y)) \right| < \epsilon $$ This tell us that $h := g \circ f$ is continuous on $(f(x_0), f(y_0))$.

If we replace our intervals $(x_0, y_0)$ and $(f(x_0), f(y_0))$ with an arbitrary set $S$ and its image under $f$, $f(S)$, and instead use an arbitrary distance metric, $d(\cdot, \cdot)$, then we have the same result by choosing $\delta_1$ such that $$ d(x',y') < \delta_1 \Rightarrow d(g(x'), g(y')) < \epsilon, $$ for $x',y' \in f(S)$, and then choosing $\delta_2$ such that $$ d(x,y) < \delta_2 \Rightarrow d(f(x), f(y)) < \delta_1 $$ for $x, y \in S$, and then we have that $$ d(x,y) < \delta_2 \Rightarrow d(g(f(x)), g(f(x))) < \epsilon $$ Thus, we see that the same result is true even if we do not restrict ourselves to one variable functions.