Prove Uniform Convergence of f^{(n)} to 0 on (-1,1)

[mattmns]

Here is the question from the book
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For each integer $n\geq 1$, let $f^{(n)}:(-1,1) \to \mathbb{R}$ be the function $f^{(n)}(x):= x^n$. Prove that $f^{(n)}$ converges pointwise to the zero function, but does not converge uniformly to any function $f:\mathbb{R} \to \mathbb{R}$.
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I already proved that it converges pointwise to the zero function, I am having trouble with the second part.

If $f^{(n)}$ is going to converge uniformly to any function, then it has to converge uniformly to the zero function (f(x)=0 for all x), since we know that it converges pointwise to the zero function.

So we want to show that it does not converge uniformly to the zero function. That is, we want to show that there exists $\epsilon > 0$ such that for all $N > 0$ there is an $n>N$ and $x \in (-1,1)$ such that $|x^n| > \epsilon$.

I first tried $\epsilon = 1/5, \ x =\frac{N}{N+1}, \ n = N+1$

If $x^n$ is increasing then we are done, but I can't quite prove it. Though I do think it is true.

Any ideas on how to prove that $x^n$ is increasing. Or of a better example? Thanks!-----

Definitions:

Uniform Convergence: Let $(f^{(n)})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_X)$ to another $(Y,d_Y)$, and let $f:X\to Y$ be another function. We say that $(f^{(n)})_{n=1}^{\infty}$ converges uniformly to $f$ on $X$ if for every $\epsilon > 0$ there exists $N>0$ such that $d_Y(f^{(n)}(x),f(x))< \epsilon$ for every $n>N$ and $x\in X.$

Pointwise Convergence: Let $(f^{(n)})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_X)$ to another $(Y,d_Y)$, and let $f:X\to Y$ be another function. We say that $(f^{(n)})_{n=1}^{\infty}$ converges uniformly to $f$ on $X$ if for every $x$ and $\epsilon>0$ there exists $N>0$ such that $d_Y(f^{(n)}(x),f(x))< \epsilon$ for every $n>N$.

 

[HallsofIvy]

Here is the question from the book
---------
For each integer $n\geq 1$, let $f^{(n)}:(-1,1) \to \mathbb{R}$ be the function $f^{(n)}(x):= x^n$. Prove that $f^{(n)}$ converges pointwise to the zero function, but does not converge uniformly to any function $f:\mathbb{R} \to \mathbb{R}$.
----------

I already proved that it converges pointwise to the zero function, I am having trouble with the second part.

If $f^{(n)}$ is going to converge uniformly to any function, then it has to converge uniformly to the zero function (f(x)=0 for all x), since we know that it converges pointwise to the zero function.

So we want to show that it does not converge uniformly to the zero function. That is, we want to show that there exists $\epsilon > 0$ such that for all $N > 0$ there is an $n>N$ and $x \in (-1,1)$ such that $|x^n| > \epsilon$.

I first tried $\epsilon = 1/5, \ x =\frac{N}{N+1}, \ n = N+1$

If $x^n$ is increasing then we are done, but I can't quite prove it. Though I do think it is true.

On the contrary, if 0< x< 1, then $x^n$ is a decreasing function of n! That's why[\b] xⁿ converges pointwise to 0!

Any ideas on how to prove that $x^n$ is increasing. Or of a better example? Thanks!


-----

Definitions:

Uniform Convergence: Let $(f^{(n)})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_X)$ to another $(Y,d_Y)$, and let $f:X\to Y$ be another function. We say that $(f^{(n)})_{n=1}^{\infty}$ converges uniformly to $f$ on $X$ if for every $\epsilon > 0$ there exists $N>0$ such that $d_Y(f^{(n)}(x),f(x))< \epsilon$ for every $n>N$ and $x\in X.$

Pointwise Convergence: Let $(f^{(n)})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_X)$ to another $(Y,d_Y)$, and let $f:X\to Y$ be another function. We say that $(f^{(n)})_{n=1}^{\infty}$ converges uniformly to $f$ on $X$ if for every $x$ and $\epsilon>0$ there exists $N>0$ such that $d_Y(f^{(n)}(x),f(x))< \epsilon$ for every $n>N$.

 

[AKG]

Note that by your definition of uniform convergence, you only need $|x^n| \geq \epsilon$, not $|x^n| > \epsilon$. This isn't a big deal, but finding an x to satisfy the first relation can be done neatly and cleanly. If you then want to use this answer to find an x satisfying the second relation, you have to increase your original x by some factor such that the new x is still < 1. This isn't hard, it just takes a few more seconds, and looks less elegant.

So do $n = N+1$, $x = \epsilon ^{1/n}$.

I don't know how your example is going to work. How does "xⁿ is increasing" mean you're done, i.e. how does it prove that $(\frac{N}{N+1})^{n} \geq \epsilon$? Anyways, as HallsofIvy pointed out, those functions aren't increasing on (-1,1), but they are increasing on [0,1) and maybe this is good enough for you. Proof: Suppose 0 < x < y < 1. Then

yⁿ-xⁿ = (y-x)(y^n-1 + y^n-2x + y^n-3x² + ... + y²x^n-3 + yx^n-2 + x^n-1)

This is a product of two factors. The factor on the right is a sum of non-negative numbers, at least one of which is strictly positive (y^n-1) so it is strictly positive. The factor on the left is y-x, which is positive because x < y. So the whole product is a positive number, meaning yⁿ > xⁿ, so f⁽ⁿ⁾ is increasing on [0,1).

 

[mattmns]

Wow, that is much nicer, thanks!
As for my "increasing" function, I guess I may have been unclear, and probably wrong as well.

Is the function

$$\left(\frac{n}{n+1}\right)^{n+1}$$

not increasing? I thought it was, and if it was then it would attain its minimum value at n = 1 (since we are assuming n>0), which would give us 1/4 which is greater than my epsilon of 1/5. That was what I was getting at, was I incorrect?

I do see that x^n is decreasing for 0<x<1, but I thought here things may be different since x depends on n. For large n, it seems that (n/(n+1))^(n+1) ~= .367879

 

[AKG]

xⁿ is decreasing on (-1,0] and increasing on [0,1). (n/n+1)ⁿ⁺¹ is increasing. Consider n as a real variable, and show that it is increasing on n > 1, by showing that the derivative is always positive. If we let f(n) be that expression above, then

f(n) = e^log(f(n)) = e^{(n+1)log(n/n+1)}

compute f'(n), and show that it is always positive. It will be helpful to know the Taylor expansion of log(1+1/x) for x > 1.

 

[Hurkyl]

For large n, it seems that (n/(n+1))^(n+1) ~= .367879

You can do better than that! After all, you're simply asking for the limit of

(1 - 1/(n+1))^(n+1),​

right?Incidentally, there's an easy way to find a sequence {x_n} such that (x_n)^n obviously doesn't converge to zero.But here's the easy way to tackle this problem: if the sequence f^(n) (x) were to converge uniformly to the zero function, then what do you know about the suprema of the f^(n) (x)?

 

[mattmns]

You can do better than that! After all, you're simply asking for the limit of

(1 - 1/(n+1))^(n+1),​

right?


Incidentally, there's an easy way to find a sequence {x_n} such that (x_n)^n obviously doesn't converge to zero.

I would think those limits were the same, and would have the same value. Perhaps my calculator is not quite correct (or I am misunderstanding something), but it evaluates both limits (as $n \to \infty$) as .367879.

But here's the easy way to tackle this problem: if the sequence f^(n) (x) were to converge uniformly to the zero function, then what do you know about the suprema of the f^(n) (x)?

I am guessing the supremum would be 0, and in which case we evaluate at some point (other than zero) and get a contradiction.