Proving continuity of f(x,y) = g(x)p(y)

I know this must be easy, but...

Say real functions g(x) and p(y) are continuous and f(x,y) = g(x)p(y). How to proof rigorously the continuity of f in a point (x1,y1)?

In other words, how to obtain l g(x)p(y) - g(x1)p(y1) l < epsilon (for any epsilon).

I can prove that l g(x)p(y1) - g(x1)p(y) l < any epsilon, but I cant see how to go from here to there. I am trying all variations of the triangular inequality, to no avail.

We want to prove
$$\forall \varepsilon>0\quad\exists\delta>0\quad\forall (x,y)\in B_{\delta}(x_0,y_0):f(x,y)\in B_{\varepsilon}(f(x_0,y_0))$$

Let fix epsilon and we want to find such delta. From definition of ball B, we compute

$$\begin{array}{ll}|f(x,y)-f(x_0,y_0)|:&=|g(x)p(y)-g(x_0)p(y_0)|\\&=|g(x)p(y)-g(x)p(y_0)+g(x)p(y_0)-g(x_0)p(y_0)|\\&\le|g(x)p(y)-g(x)p(y_0)|+|g(x)p(y_0)-g(x_0)p(y_0)|\\&=|p(y)-p(y_0)||g(x)|+|g(x)-g(x_0)||p(y_0)|\end{array}$$

Now, I think you are able to complete this proof. Use the continuity of g and p.

Hi Castilla!

$$|g(x)p(y)-g(x_1)p(y_1)|=|(g(x)p(y)-g(x_1)p(y))+(g(x_1)p(y)-g(x_1)p(y_1))|\leq |g(x)-g(x_1)||p(y)|+|g(x_1)||p(y)-p(y_1)|$$

I'll let you continue from there.

Thank you, stanley and micromass. You enlightened my mind.

Just one question. Looking at both answers, basically the same, it seems that I need that the functions g and p to be bounded.

So I would need not only that g and p are continuous, but also that they're continuous in a compact set?

thanks again!

disregardthat
You don't need that, but good catch!

You have

|g(x)p(y)-g(x_0)p(y_0)| = |(g(x)-g(x_0))(p(y)-p(y_0)) +p(y_0)(g(x)-g(x_0)) + g(x_0)(p(y)-p(y_0))| <= |g(x)-g(x_0)||p(y)-p(y_0)| +|p(y_0)||g(x)-g(x_0)| + |g(x_0)|p(y)-p(y_0)|

Last edited:
Excellent explanation. Thank you.

Hurkyl
Staff Emeritus