Prove that a product of continuous functions is continuous

In summary, we need to prove that for all continuous functions f and g at a given point a, the product function fg is also continuous at a. To do so, we must show that the key inequality |f(x)g(x)-f(a)g(a)| can be made smaller than any given epsilon. By breaking down this inequality and considering different cases for the terms, we can find suitable deltas to limit each term and ultimately prove the continuity of fg at a.
  • #1
docnet
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complex analysis
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##f## is continuou on ##\mathbb{C}##, so for al ##\epsilon>0##, there is a ##\delta>0## such that $$|\tilde{z}-z|\leq \delta \Rightarrow |f(\tilde{z})-f(z)|\leq \epsilon$$ for all ##\tilde{z}## and ##z## in ##\mathbb{C}##.

Complex conjugation is a norm preserving operation on ##\mathbb{C}##, so

$$|f(\tilde{z})-f(z)|=|\bar{f}(\tilde{z})-\bar{f}(z)|\leq \epsilon$$

The same ##\delta## satisfies the definition of continuity for ##\bar{f}##, so ##\bar{f}## is continuous on ##\mathbb{C}##.

The product of two continuous functions is continuous, so ##g=f\cdot \bar{f}## is continuous.

The conclusion feels shaky or like I'm not doing the real work of the problem somehow.
 
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  • #2
docnet said:
The absolute value function is a norm preserving operation on ##\mathbb{C}##,
I think you mean complex conjugation is norm preserving.

docnet said:
The product of two continuous functions is continuous, so ##g=f\cdot \bar{f}## is continuous.
If you are allowed to invoke that theorem, then this is fine.
 
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  • #3
PeroK said:
I think you mean complex conjugation is norm preserving.
Yes! I edited my post to fix my mistake.
 
  • #4
Not using the theorem:

$$|\bar{f}(\tilde{z})f(\tilde{z})-\bar{f}(z)f(z)|=|\bar{f}(\tilde{z})f(\tilde{z})-\bar{f}(z)f(z)+\bar{f}(\tilde{z})f(z)-\bar{f}(\tilde{z})f(z)|$$
$$\leq |\bar{f}(\tilde{z})||f(\tilde{z})-f(z)|+|f(z)||\bar{f}(\tilde{z})-\bar{f}(z)|= |\bar{f}(\tilde{z})|\epsilon +|f(z)|\epsilon$$Let ##\delta=\frac{\epsilon }{|\bar{f}(\tilde{z})|+|f(z)|}## then ##\bar{f}f## is continuous by the epsilon delta definition of continuity.
 
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  • #7
docnet said:
Thank you, I'll look into this and come up with a solution soon.
It's trickier than it looks. If you really want to master the epslion-delta technique then you have to really knuckle down and battle through something like this. Every step has to be justified and every nuance covered. For example, what happens to your proof if ##f(z) = f(\tilde z) = 0##?

The best plan, I find, is to get the strategy/outline first before diving in with the technical details.

Let me post an outline to show you what I mean ...
 
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  • #8
Outline proof (I'll do it for real ##f, g## just to simplify the notation a bit):

1) The first thing we must do is state clearly what we are going to prove:

Let ##f, g## be continuous at some point ##a##. Then the product function ##fg## is continuous at ##a##.

2) First technicality: what happens if ##f## and ##g## have different domains? What if ##f(x) = \sqrt x##, with domain ##[0, \infty)##; and, ##g(x) = \sqrt {-x}##, with domain ##(-\infty, 0]##. The product function is only defined at ##x = 0##. That's okay, because then ##fg## is continuous vacuously at ##x = 0##.

3) Do we mention domains further or just gloss over that point? For simplicity, let's assume that we have restricted our attention to a shared domain that inlcudes ##a## and not say any more about this.

4) The key inequality is: $$|f(x)g(x) - f(a)g(a)| \le |f(x)||g(x) - g(a)| + |g(a)||f(x) - f(a)|$$How do we deal with this?

4a) Note that ##|g(a)|## is some fixed number (but it might equal ##0##, so we need to be careful about that). The term ##|f(x) - f(a)|## can be made arbitrarily small. We're okay there.

4b) We can make ##|f(x)|## close to ##|f(a)|##, so we must be able to limit the size of this term. And, the term ##|g(x) - g(a)|## can be made arbitrarily small. We're okay here as well.

4c) We have the sum of two expressions, so we should try to make each ##< \frac \epsilon 2##.

5) Let's look at each of those points in more detail.

5a) If ##|g(a)| \le 1##, then the first term is easy. We can find ##\delta_1## such that ##|f(x) - f(a)| < \frac \epsilon 2##. And, if ##|g(a)| > 1##, then we need to find ##\delta_2## such that ##|f(x) - f(a)| < \frac \epsilon {2|g(a)|}##.

Alternatively, if ##|g(a)| = 0##, then the first term vanishes and otherwise we can find ##\delta_2## such that ##|f(x) - f(a)| < \frac \epsilon {2|g(a)|}##.

Another alternative, we could find ##\delta_3## such that ##|f(x) - f(a)| < \frac \epsilon {2(1+ |g(a)|)}##. That takes care of all cases.

We need to decide on one of those approaches. Pick whatever looks best.

5b) The term ##|f(x)|## is trickier, because ##x## is a variable. But, we can find ##\delta_4## such that ##|f(x)| < |f(a)| + 1##. We should be able to show that from the triangle inequality.

That just leaves ##\delta_5## such that ##|g(x) - g(a)| < \frac \epsilon{2(|f(a)| + 1)}##.

5c) We just need to decide on the best approach, rationalise our deltas, and take ##\delta = min(\delta_1, \delta_2)## and we are good to go.

6) How do all epsilon-delta proofs start?

Let ##\epsilon > 0 \dots##
 
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  • #9
PeroK said:
Outline proof (I'll do it for real ##f, g## just to simplify the notation a bit):

1) The first thing we must do is state clearly what we are going to prove:

Let ##f, g## be continuous at some point ##a##. Then the product function ##fg## is continuous at ##a##.

2) First technicality: what happens if ##f## and ##g## have different domains? What if ##f(x) = \sqrt x##, with domain ##[0, \infty)##; and, ##g(x) = \sqrt {-x}##, with domain ##(-\infty, 0]##. The product function is only defined at ##x = 0##. That's okay, because then ##fg## is continuous vacuously at ##x = 0##.

3) Do we mention domains further or just gloss over that point? For simplicity, let's assume that we have restricted our attention to a shared domain that inlcudes ##a## and not say any more about this.

4) The key inequality is: $$|f(x)g(x) - f(a)g(a)| \le |f(x)||g(x) - g(a)| + |g(a)||f(x) - f(a)|$$How do we deal with this?

4a) Note that ##|g(a)|## is some fixed number (but it might equal ##0##, so we need to be careful about that). The term ##|f(x) - f(a)|## can be made arbitrarily small. We're okay there.

4b) We can make ##|f(x)|## close to ##|f(a)|##, so we must be able to limit the size of this term. And, the term ##|g(x) - g(a)|## can be made arbitrarily small. We're okay here as well.

4c) We have the sum of two expressions, so we should try to make each ##< \frac \epsilon 2##.

5) Let's look at each of those points in more detail.

5a) If ##|g(a)| \le 1##, then the first term is easy. We can find ##\delta_1## such that ##|f(x) - f(a)| < \frac \epsilon 2##. And, if ##|g(a)| > 1##, then we need to find ##\delta_2## such that ##|f(x) - f(a)| < \frac \epsilon {2|g(a)|}##.

Alternatively, if ##|g(a)| = 0##, then the first term vanishes and otherwise we can find ##\delta_2## such that ##|f(x) - f(a)| < \frac \epsilon {2|g(a)|}##.

Another alternative, we could find ##\delta_3## such that ##|f(x) - f(a)| < \frac \epsilon {2(1+ |g(a)|)}##. That takes care of all cases.

We need to decide on one of those approaches. Pick whatever looks best.

5b) The term ##|f(x)|## is trickier, because ##x## is a variable. But, we can find ##\delta_4## such that ##|f(x)| < |f(a)| + 1##. We should be able to show that from the triangle inequality.

That just leaves ##\delta_5## such that ##|g(x) - g(a)| < \frac \epsilon{2(|f(a)| + 1)}##.

5c) We just need to decide on the best approach, rationalise our deltas, and take ##\delta = min(\delta_1, \delta_2)## and we are good to go.

6) How do all epsilon-delta proofs start?

Let ##\epsilon > 0 \dots##
please allow me plenty of time to digest this.. and Merry Christmas!
 
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FAQ: Prove that a product of continuous functions is continuous

What does it mean for a function to be continuous?

A function is continuous if it has no sudden jumps or breaks in its graph. This means that the function is defined and has a value at every point along its domain.

How do you prove that a product of continuous functions is continuous?

To prove that a product of continuous functions is continuous, we can use the product rule for continuity. This states that the product of two continuous functions is also continuous.

Can you give an example of a product of continuous functions that is not continuous?

Yes, an example would be the product of the functions f(x) = 1/x and g(x) = x. Individually, both functions are continuous, but their product f(x)g(x) = 1 is not continuous at x = 0.

Are there any exceptions to the product rule for continuity?

Yes, the product rule for continuity only applies when both functions are continuous at the same point. If one or both functions have a discontinuity at a certain point, the product rule does not hold.

Why is it important to prove that a product of continuous functions is continuous?

Proving that a product of continuous functions is continuous is important because it allows us to use the properties of continuous functions to analyze and solve problems involving the product. This is especially useful in real-world applications, such as in physics and engineering.

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