Prove that a product of continuous functions is continuous

[docnet]

Homework Statement

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Relevant Equations

complex analysis

$f$ 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.

 

[PeroK]

The absolute value function is a norm preserving operation on $\mathbb{C}$,

I think you mean complex conjugation is norm preserving.

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.

 

[docnet]

I think you mean complex conjugation is norm preserving.

Yes! I edited my post to fix my mistake.

 

[docnet]

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.

 

[PeroK]

The main problem is the logic is not right. There is also a technicality in establishing a bound for $|\bar f(\tilde z)|$, assuming that $\tilde z$ is a variable.

E.g.
https://math.stackexchange.com/ques...uct-of-two-continuous-functions-is-continuous

 

[docnet]

The main problem is the logic is not right. There is also a technicality in establishing a bound for $|\bar f(\tilde z)|$, assuming that $\tilde z$ is a variable.

E.g.
https://math.stackexchange.com/ques...uct-of-two-continuous-functions-is-continuous

Thank you, I'll look into this and come up with a solution soon.

 

[PeroK]

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 ...

 

[PeroK]

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 $\epsilon > 0 \dots$

 

[docnet]

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!