Let $f(x_1', ..., x_n')$ and $g_1(x_1), ..., g_n(x_n)$ be continuous functions. Is $f(g_1(x_1), ..., g_n(x_n))$ continuous?

Question

Let $f:X_1' \times...\times X_n' \rightarrow Y$, and $g_1:X_1 \rightarrow X_1'$, ..., $g_n:X_n \rightarrow X_n'$ be continuous functions on their respectives metric spaces $X_1', ..., X_n', Y, X_1, ..., X_n$.

Are the conditions above sufficient to conclude that $f(g_1(x_1), ..., g_n(x_n))$ is a continuous function $X_1 \times ... \times X_n \rightarrow Y$?

If so, how could one prove it? If not, what would be a counterexample?

Answer 1

The proof of that the composition is a continuous function $X_1 \times \cdots \times X_n \to Y$ should be fairly easy to understand in terms of the "engineering" interpretation of the $\epsilon$-$\delta$ definition of continuity. In this interpretation, $\epsilon$ is interpreted as a maximum allowed error in the output of the function, and $\delta$ is interpreted as a corresponding imposed tolerance on the input of the function which guarantees the requirements of the application.

So, suppose you have designed a process which in theory takes inputs $x_1 \in X_1, \ldots, x_n \in X_n$ and produces a desired output $y \in Y$. This process is a composition of subprocesses as follows: you apply $g_1$ to $x_1$ to get $x_1' \in X_1'$, $\ldots$, $g_n$ to $x_n$ to get $x_n' \in X_n'$. Then, you feed these intermediate results into $f$ to get $y$.

Now, you cannot guarantee the inputs will be exactly $x_1, \ldots, x_n$; but the engineering requirements are for the resulting output to be within $\epsilon$ of $y$. Is there some set of deviations from $x_1, \ldots, x_n$ within which you can guarantee the output will satisfy the requirements? If $f, g_1, \ldots, g_n$ are continuous, then the answer is yes: first, find some $\delta_1', \ldots, \delta_n' > 0$ such that when you have inputs to $f$ which are within $\delta_1'$ of $x_1'$, $\ldots$, $\delta_n'$ of $x_n'$, then the output is within $\epsilon$ of $y$. Now, for each $i \in \{ 1, 2, \ldots, n \}$, using the continuity of $g_i$, find a $\delta_i$ such that the input of $g_i$ being within $\delta_i$ of $x_i$ guarantees the output is within $\delta_i'$ of $x_i'$. It is now easy to see that if the inputs to the composed process are within $\delta_1$ of $x_1$, $\ldots$, within $\delta_n$ of $x_n$, then the output of the composed process will be within $\epsilon$ of $y$ and so the customer will be satisfied. (And no matter how demanding the customer is, we can still satisfy their requirements, as long as our suppliers of $x_1, \ldots, x_n$ can satisfy any of our derived requirements on them.)

Answer 2

For a "yes" it is enough that $X_1\times\cdots\times X_n$ and $X'_1\times\cdots\times X'_n$ are both equipped with product topology.

For $i=1,\dots,n$ let $p_i':X_1'\times\cdots\times X'_n\to X_i'$ and $p_i:X_1\times\cdots\times X_n\to X_i$ denote projections.

These functions are continuous (characteristic for product spaces).

Then $g_i\circ p_i:X_1\times\cdots\times X_n\to X'_i$ is continuous for $i=1,\dots,n$ so that a unique continuous function $h:X_1\times\cdots\times X_n\to X_1'\times\cdots\times X_n'$ exists with $p_i'\circ h=g_i\circ p_i$ for $i=1,\dots,n$.

This is characteristic for the product space $X_1'\times\cdots\times X_n'$.

This function is prescribed by:$$(x_1,\dots,x_n)\mapsto(g_1(x_1),\dots,g_n(x_n))$$

Then $f\circ h:X_1\times\cdots\times X_n\to Y$ is a composition of continuous functions, hence is continuous itself.

This function is prescribed by:$$(x_1,\dots,x_n)\mapsto f((g_1(x_1),\dots,g_n(x_n)))$$

It is not necessary that the spaces are metric (or metrizable).