Show Pointwise Convergence of g_n to Zero Function

[Carl140]

Homework Statement

Let f: [0,1] -> R (R-real numbers) be a continuous non constant
function such that f(0)=f(1)=0. Let g_n be the function: x-> f(x^n)
for each x in [0,1]. I'm trying to show that g_n converges pointwise to
the zero function but NOT uniformly to the zero function.

The Attempt at a Solution

I thought: sup |g_n(x) | = sup |f(x^n)| now if I can only show that this doesn't
tends to zero then we are done but I can't. I also have no idea how to show
it converges pointwise to zero. Can you please help?

 

[Dick]

Pick a definite point, say x=1/2. What's lim x^n? What's lim f(x^n)?

 

[Carl140]

Pick a definite point, say x=1/2. What's lim x^n? What's lim f(x^n)?

Ok here's what I think it's a "proof".

To show pointwise convergence:

If x<1 then x^n converges pointwise to 0 , then because f is continuous f(x^n) -> f(0) =0.
But f(x^n) = g_n(x) so g_n(x) ->0 if n->infinity and x<1.
Now if x=1 then g_n(1) = f(1^n ) = f(1) =0 so g_n(1) =0.
Conclusion: g converges pointwise everywhere to 0.

Is this proof correct?

Now to show it is NOT uniformly convergent.
Let x=1/2 then x^n = (1/2)^n = 1/2^n which goes to 0.
Now f(x^n) = f(1/2^n) but then I don't know how to compute this limit.

 

[Dick]

That's the idea for pointwise convergence alright. To show it's not uniform go back to sup |f(x^n)| over x in [0,1]. How does that compare with sup |f(x)| over x in [0,1]?

 

[Carl140]

That's the idea for pointwise convergence alright. To show it's not uniform go back to sup |f(x^n)| over x in [0,1]. How does that compare with sup |f(x)| over x in [0,1]?

Thank you for helping, really appreciated.
I think you suggest to show that sup |f(x^n) | >= |sup(f(x)| but I don't know how
to prove this inequality. Any hints you can please provide?

 

[Office_Shredder]

Thank you for helping, really appreciated.
I think you suggest to show that sup |f(x^n) | >= |sup(f(x)| but I don't know how
to prove this inequality. Any hints you can please provide?

For every n, find an x such that xⁿ=1/2. Since f is non-constant, you know somewhere f is non-zero, so do the same for that point

 

[Dick]

I, in fact, think to show sup |f(x^n)|=sup |f(x)|. x goes from 0 to 1, x^n also goes from 0 to 1 for each n. f is going to range over the same values in both cases, isn't it?

 

[Carl140]

I, in fact, think to show sup |f(x^n)|=sup |f(x)|. x goes from 0 to 1, x^n also goes from 0 to 1 for each n. f is going to range over the same values in both cases, isn't it?

Thanks guys, is this correct?
We want to show that sup | f(x^n) | = sup |f(x)|.
Let M = sup |f(x)| which exists because f is continuous and [0,1] is compact.

Now observe that no matter for each x in [0,1] x^n is also in [0,1].
For simplicity let y_n = x^n. Then each y_n is an element of [0,1].
Then by assumption M = sup | f(x) | x in [0,1], so simply "replace" x by y_n.
Then M = sup | f(x^n) | x in [0,1] and we are done (I think)

Is this correct?


One more thing:
To show x^n converges pointwise to zero we shall prove that for all e>0 there
exists natural n>=N such that x^n< e.
n< log(e)/log(x).

So what natural N can we take?

 

[Dick]

That's coming out a little confusing. I think you just say that range(f)=f([0,1]) for both f(x) and f(x^n). Since as x goes from 0 to 1, x^n also goes from 0 to 1. But for purposes of the proof, all you really need is that sup |f(x^n)| is greater than some positive constant. You could follow Office-Shredder's suggestion and just concentrate on a single nonzero point, if that makes it easier to write down. You might be getting confused in your epsilon proof because log(x) and log(e) are negative. Multiply both sides of n*log(x)<log(e) by -1 to get n>|log(e)|/|log(x)|.

 

[Carl140]

Thanks very much! Got it now.