# Homework Help: 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 $f(y) \ne 0$ ...

21. Apr 9, 2016

### lep11

ok, so I need to show again that $\lim_{x \rightarrow x_0} f(x)=f(x_0)$?

now if f(y)≠0, $\lim_{x \rightarrow x_0} f(x)= \lim_{x \rightarrow x_0} [g(x-y)-g(x)g(y)]/f(y)]= ...$im stuck here

22. Apr 9, 2016

### PeroK

You could do it that way. But, you can use the properties of continuous functions:

Let's use $y_0$ so that we know what is fixed and what is a variable:

$g(x-y_0)$ is a continuous function, so is $g(x)g(y_0)$ hence the sum of these is continuous etc.

23. Apr 9, 2016

### lep11

now if f(y0)≠0 and g is continuous, f(x)= [g(x-y0)-g(x)g(y0)]/f(y0)] is continuous

f(y0) is a real number, costant function (let h(x)=f(y0)) and y0 is fixed.

I really appreciate your help. Thanks!

24. Apr 9, 2016

### PeroK

Yes, and that's it done! With a little help here and there!

Seriously though, there is a lot of good stuff in this problem. It's worth not just reviewing it, but also looking at the overall strategic approach we took. As problems get harder, it becomes more difficult to hit on a solution. You need to work out a plan for each step.