Prove functions f and g are continuous in the reals

[lep11]

Homework Statement

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

The Attempt at a Solution

[/B]
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?

 

[PeroK]

Homework Statement

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

The Attempt at a Solution

[/B]
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.

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

 

[lep11]

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

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.

 

[PeroK]

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

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

 

[lep11]

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

Please help me how to get started if we know nothing about g(0).

 

[PeroK]

Please help me how to get started if we know nothing about g(0).

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 $x_0$).

 

[PeroK]

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

 

[lep11]

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

|g(x)-g(x₀)|=...?

 

[PeroK]

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

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.

 

[lep11]

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

 

[PeroK]

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

No. Use some simple properties of limits.

 

[lep11]

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

 

[PeroK]

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

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.

 

[lep11]

What about f?

 

[lep11]

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

It's 0. But how to prove it?

 

[PeroK]

It's 0. But how to prove it?

Can you see how to use equation (iii)?

What about f?

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

 

[lep11]

Can you see how to use equation (iii)?

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?

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

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

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

 

[PeroK]

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

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, ...

 

[lep11]

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

if f(x)≠0 always, hmm...

 

[PeroK]

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

if f(x)≠0, hmm...

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 $f(y) \ne 0$ ...

 

[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

 

[PeroK]

ok, so I need to show again that $\lim_{x \rightarrow x_0} f(x)=f(x_0)$?now if f(y)≠0, $\lim_{x \rightarrow x_0} f(x)= \lim_{x \rightarrow x_0} [g(x-y)-g(x)g(y)]/f(y)]$

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.

 

[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!

 

[PeroK]

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 h(x)=f(y₀)

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.