# Prove functions f and g are continuous in the reals

1. Apr 9, 2016

### lep11

1. The problem statement, all variables and given/known data
Prove functions f and g are continuous in ℝ. It's known that:
i) lim g(x)=1, when x approaches 0
ii)g(x-y)=g(x)g(y)+f(x)f(y)
iii)f2(x)+g2(x)=1

3. The attempt at a solution

g(0) has to be equal to 1 because it's known that lim g(x)=1, when x approaches 0. Otherwise g won't be continuous at x=0.

g(0)=1 implies that f(0)=0 (iii)
And if g(0)=1 and f(0)=0 then g(-y)=g(y) (ii) so g is even function.

So it looks like g(x)=sin (x) and f(x)=cos(x). However, g and f may not be unique.
f+g is continuous if f and g are continuous but not vice versa
The definition of continuity is ∀ε>0∃δ>0: |x-x0|<δ ⇒|f(x)-f(x0|<ε ,x0∈ℝ

How to proceed?

Last edited: Apr 9, 2016
2. Apr 9, 2016

### PeroK

You are trying to prove that f and g are continuous, so you can't assume that g is continuous.

3. Apr 9, 2016

### lep11

If g(0)≠1 then g is not continuous at x=0 so g is not continuous ∀x∈ℝ and that's it. g(0)=1 is reasonable assumption.

4. Apr 9, 2016

### PeroK

You might as well go the whole way: we know (because we've been asked to prove it) that f and g are continuous, therefore they must be continuous. QED

5. Apr 9, 2016

6. Apr 9, 2016

### PeroK

I would choose a point $x_0$ and write down the definition of continuity of $g$ at $x_0$. That seems like a good place to start.

PS I would say "to show that $g$ is continuous at $x_0$, we need to show that ..." (not assume $g$ is continuous at $x_0$).

7. Apr 9, 2016

### PeroK

PPS You could additionally try to find $lim_{x \rightarrow 0} f(x)$

8. Apr 9, 2016

### lep11

|g(x)-g(x0)|<epsilon when |x-x0|<delta hmm

|g(x)-g(x0)|=...?

9. Apr 9, 2016

### PeroK

That isn't a definition of anything. As an aside, if you're doing analysis you have to been more precise.

In this case, you don't need to use epsilon-delta. You just need the properties of limits. Let me help you:

$g$ is continuous at $x_o$ if $lim_{x \rightarrow x_0} g(x) = g(x_0)$

Now, I haven't given too much away, because you have to write that definition in a slightly different form. Hint: look at equation (ii) that you were given.

10. Apr 9, 2016

### lep11

So I won't need epsilon-delta definition at all?

11. Apr 9, 2016

### PeroK

No. Use some simple properties of limits.

12. Apr 9, 2016

### lep11

$g$ is continuous at $x_o$ if $lim_{x \rightarrow x_0} g(x) = g(x_0)$⇔ $lim_{x \rightarrow x_0} (g(x)g(0)+f(x)f(0)) = g(x_0)$

13. Apr 9, 2016

### PeroK

Let me give you one more helping hand. Then I'll be offline anyway:

$g$ is continuous at $x_0$ if $lim_{y \rightarrow 0} g(x_0 + y) = g(x_0)$

Having $y \rightarrow 0$ is just another way to have $x \rightarrow x_0$

It might be worth looking at these two variations of the limit we have now:

$lim_{x \rightarrow x_0} g(x) = lim_{y \rightarrow 0} g(x_0 + y)$

And really understand why they are equivalent.

14. Apr 9, 2016

### lep11

Last edited: Apr 9, 2016
15. Apr 9, 2016

### lep11

It's 0. But how to prove it?

16. Apr 9, 2016

### PeroK

Can you see how to use equation (iii)?

Let's worry about proving $g$ is continuous first, then we can have a think about $f$.

17. Apr 9, 2016

### lep11

iii)f2(x)+g2(x)=1

$lim_{x \rightarrow 0} g(x) =1$ ⇒ $lim_{x \rightarrow 0} g^2(x) =1$

when x--->0 g2(x) approaches 1 so f2(x) has to approach 0 for the equation (iii) to hold. Right?

$lim_{x \rightarrow x_0} g(x) = lim_{h \rightarrow 0} g(x_0 - h)= lim_{h \rightarrow 0} [g(x_0)g(h)+f(x_0)f(h)]=...=g(x_0)*1+lim_{h \rightarrow 0} [f(x_0)f(h)]=g(x_0)*1+0$

∴ $lim_{x \rightarrow x_0} g(x) =g(x_0)$ ∀x0∈ℝ so g is always continuous

Last edited: Apr 9, 2016
18. Apr 9, 2016

### PeroK

Yes on both counts!

Now, you might want to wheel out the epsilons and deltas to show that:

$\lim_{x \rightarrow 0} f(x)^2 = 0 \ \Rightarrow \lim_{x \rightarrow 0} f(x) = 0$

Or, you might want to just state that as "obvious" and move quickly on.

Now, to show the continuity of $f$ follows from the continuity of $g$.

Hint: you said that you thought $g(x) = cos(x)$ and $f(x) = sin(x)$ but in fact $g(x) = 1$ and $f(x) = 0$ for all $x$ is also a solution.

If $f(x) = 0$ for all $x$, then it's continuous. Otherwise, ...

19. Apr 9, 2016

### lep11

if f(x)=0, then f is constant function and therefore always continuous

if f(x)≠0 always, hmm...

20. Apr 9, 2016

### PeroK

Let me explain how I thought about it.

First, I thought, what about equation (iii). If $g$ is continuous, then $f^2$ must be continuous. But, does $f^2$ being continuous imply $f$ is continuous? No, because $f$ could jump from positive to negative. So, perhaps I have to use equation (ii). If I picked a fixed value for $y$ then I could express $f(x)$ in terms of some combination of $g$ divided by $f(y)$ and that would do it. But, then, to do that I need $f(y) \ne 0$. That's when I noticed that there was the constant value solution. So, ...

If $f$ is not the zero function then $\exists y$ such that $f(y) \ne 0$ ...