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

• Castilla
In summary, You are trying to find the point (x1,y1) where f(x,y) is continuous. You are using the continuity of g and p to try to find this point. However, you do not need to use that property, as you have an excellent explanation.
Castilla
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 can't 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!

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.

You could have used the fact multiplication is continuous...

## 1. What is the definition of continuity in mathematics?

The mathematical definition of continuity is the property of a function where the values of the function change gradually as the input values change, without any sudden jumps or breaks. In other words, a continuous function can be drawn without lifting your pencil from the paper.

## 2. How do you prove the continuity of a function?

To prove the continuity of a function, you must show that it satisfies the three conditions of continuity: the function is defined at a given point, the limit of the function at that point exists, and the limit is equal to the value of the function at that point. This can be done using algebraic manipulation, the definition of a limit, or theorems such as the Intermediate Value Theorem.

## 3. What is the difference between uniform continuity and pointwise continuity?

Uniform continuity is a stronger version of pointwise continuity, where not only does the function have to be continuous at each point, but the rate of change of the function must also be continuous. In other words, the difference between the output values of the function at two points must approach zero as the difference between the input values approaches zero.

## 4. How does continuity relate to differentiability?

If a function is differentiable at a point, it must also be continuous at that point. However, the converse is not necessarily true; a function can be continuous at a point but not differentiable at that point. This is because differentiability requires not only continuity, but also a well-defined slope at that point.

## 5. Can a function be continuous at one point but not continuous on a larger interval?

Yes, it is possible for a function to be continuous at a single point but not on a larger interval. This would occur when the function has a discontinuity at one or more points within that interval, but is still continuous at the specific point in question.

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