# Pointwise convergence to uniform convergence

Printable View

• Nov 6th 2008, 03:06 PM
tttcomrader
Pointwise convergence to uniform convergence
Does pointwise convergence of continuous functions on a compact set to a continuous limit imply uniform convergence on that set?

I think yes.

Proof.

Suppose that $(M,d)$ is a metric space, let K be a compact subset of M, and define $f_n:K \rightarrow \mathbb {R} \ \ \ \forall n \in N$ such that there exists a continuous function $f:K \rightarrow \mathbb {R}$ with $f_n \rightarrow f$ pointwise.

By definitions, $\forall \epsilon > 0, \exists M_{( \epsilon ,x)}$ such that $d(f_n,f)< \epsilon \ \ \ n \geq M$

Now, what I want to show is that $\exists M_{ \epsilon }$ such that $d(f_n,f) < \epsilon \ \ \ n \geq M$, so an epsilon that doesn't depend on x.

But I find it difficult to work through this, perhaps my understanding of uniform convergence is still poor. Am I on the right track thou?

Thanks!
• Nov 7th 2008, 01:04 AM
Opalg
Quote:

Originally Posted by tttcomrader
Does pointwise convergence of continuous functions on a compact set to a continuous limit imply uniform convergence on that set?

I think yes.

The answer is No. For example, the sequence $f_n(x) = nx^n(1-x)$ converges pointwise but not uniformly to 0 on the unit interval.
• Nov 7th 2008, 11:53 AM
tttcomrader
Okay, here is my formal proof.

Define $f_n(x): [0,1] \rightarrow \mathbb {R}$ by $
f_n(x) = nx^n(1-x) \ \ \ \forall n \in \mathbb {N}
$

Claim: This sequence of functions converges pointwise to $f(x)=0$

Proof.

For $x=0$ and $x=1$, we would have $f_n(x)=0$.

Given $\epsilon > 0$, pick $N = ?$
For $x \in (0,1)$, we would have $|f_n(x)-0|=|nx^n(1-x)|<|nx^n|$ I'm stuck here, I know that this would converge to 0, but how would I prove that? What N I should pick to ensure this distance is less than epsilon?

Claim: This sequence of functions do not converge uniformly to 0.

Proof.

Pick $\epsilon = 1$, then $\forall N \in \mathbb {N}$, whenever $n \geq N$ and $x = \frac {1}{n}$ for each index n, we will have $|f_n(x)-0|=|n( \frac {1}{n} ) ^n(1- \frac {1}{n} )| = | \frac {1}{n^{n-1}}(1- \frac {1}{n})|=| \frac {1}{n^{n-1}}- \frac {1}{n^n}| = | \frac {n-1}{n^n}|$ But doesn't this still converges to 0? Did I pick the wrong x?

Thanks!!!
• Nov 7th 2008, 12:29 PM
Opalg
Quote:

Originally Posted by tttcomrader
Okay, here is my formal proof.

Define $f_n(x): [0,1] \rightarrow \mathbb {R}$ by $
f_n(x) = nx^n(1-x) \ \ \ \forall n \in \mathbb {N}
$

Both parts of this are actually quite tricky.

Quote:

Originally Posted by tttcomrader
Claim: This sequence of functions converges pointwise to $f(x)=0$

Proof.

For $x=0$ and $x=1$, we would have $f_n(x)=0$.

Good start. So we now want to show that if 0<x<1 then $nx^n\to0$ as $n\to\infty$. As you can see, this is not obvious. One sneaky way to prove it is to notice that the series $\sum_{n=0}^\infty nx^n$ converges if |x|<1 (easily shown by using the ratio test), and therefore the n'th term of the series must tend to 0.

Quote:

Originally Posted by tttcomrader
Claim: This sequence of functions do not converge uniformly to 0.[/tex]

To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of $x^n(1-x)$ in the unit interval occurs when the derivative is 0. This occurs when $x=1-\tfrac 1{n+1}$, at which point the value of the function is $\bigl(\tfrac n{n+1}\bigr)^{n+1}$, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.

As I said, it's a tricky question. (Wink)
• Nov 9th 2008, 10:37 AM
tttcomrader
Quote:

Originally Posted by Opalg

To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of $x^n(1-x)$ in the unit interval occurs when the derivative is 0. This occurs when $x=1-\tfrac n{n+1}$, at which point the value of the function is $\bigl(\tfrac n{n+1}\bigr)^{n+1}$, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.

As I said, it's a tricky question. (Wink)

I'm a bit unclear about how $
\bigl(\tfrac n{n+1}\bigr)^{n+1}
\rightarrow \frac {1}{e}$
Did we use the natural log here?
• Nov 9th 2008, 10:56 AM
Plato
Quote:

Originally Posted by tttcomrader
I'm a bit unclear about how $
\bigl(\tfrac n{n+1}\bigr)^{n+1}
\rightarrow \frac {1}{e}$
Did we use the natural log here?

Just 'break it up': $\left( {\frac{n}{{n + 1}}} \right)^{n + 1} = \left( {1 - \frac{1}{{n + 1}}} \right)^n \left( {\frac{n}{{n + 1}}} \right)$
$\lim _{n \to \infty } \left( {1 - \frac{1}{{n + 1}}} \right)^n = e^{ - 1} \,\& \,\lim _{n \to \infty } \left( {\frac{n}{{n + 1}}} \right) = 1$.

In general $\lim _{n \to \infty } \left( {1 + \frac{a}
{{n + b}}} \right)^{cn} = e^{ac}$
(with the correct restrictions on b)
• Nov 9th 2008, 05:39 PM
tttcomrader
I'm just really confuse about here:

So take $x= 1- \frac {n}{n+1}$, then $f_n(x)= | n(1- \frac {n}{n+1})^n[1-(1- \frac {n}{n+1}] |$ $=(1- \frac {n}{n+1})^n( \frac {n^2}{n+1} )$, now, why does this mess equals to $( \frac {n}{n+1})^{n+1}$?
• Nov 9th 2008, 06:02 PM
Plato
Quote:

Originally Posted by tttcomrader
I'm just really confuse about here:

Now I do agree with that!
Could it be a matter of simple algebra?
I do think you are over doing all of this.
Stop! Think about a simple approach to this problem.
• Nov 10th 2008, 12:51 AM
Opalg
Quote:

Originally Posted by tttcomrader
I'm just really confuse about here:

So take $x= 1- \frac {n}{n+1}$, then $f_n(x)= | n(1- \frac {n}{n+1})^n[1-(1- \frac {n}{n+1}] |$ $=(1- \frac {n}{n+1})^n( \frac {n^2}{n+1} )$, now, why does this mess equals to $( \frac {n}{n+1})^{n+1}$?

Sorry, that's my mistake. In one of the comments above, I wrote $x= 1- \tfrac {n}{n+1}$ when it should have been $x= 1- \tfrac 1{n+1}$. (But you should have spotted that for yourself, because I explained that this is the value of x where the function $nx^n(1-x)$ has its maximum value. If you take the trouble to differentiate, you soon see that this happens when $x=\tfrac n{n+1}$, which can equivalently be written $x=1- \tfrac 1{n+1}$. What I did was to write one of those expressions and then edit it to write the other one, but forgot to alter the n in the numerator.)
• Nov 27th 2009, 09:11 PM
anlys
Quote:

Originally Posted by Opalg
Both parts of this are actually quite tricky.

Good start. So we now want to show that if 0<x<1 then $nx^n\to0$ as $n\to\infty$. As you can see, this is not obvious. One sneaky way to prove it is to notice that the series $\sum_{n=0}^\infty nx^n$ converges if |x|<1 (easily shown by using the ratio test), and therefore the n'th term of the series must tend to 0.

To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of $x^n(1-x)$ in the unit interval occurs when the derivative is 0. This occurs when $x=1-\tfrac 1{n+1}$, at which point the value of the function is $\bigl(\tfrac n{n+1}\bigr)^{n+1}$, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.

As I said, it's a tricky question. (Wink)

I have a question regarding this problem. By using the ratio test, we get the sum of infinite series to be convergent if |x| <1. But how do we know it converges to 0?
• Nov 29th 2009, 06:13 AM
Opalg
Quote:

Originally Posted by anlys
I have a question regarding this problem. By using the ratio test, we get the sum of infinite series to be convergent if |x| <1. But how do we know it converges to 0?

It is not true that the series converges to 0. The series converges (in other words, the terms add up to a finite sum), but its sum is not zero. What is true is that the individual terms of the series converge to 0. That is always the case if a series converges.
• Nov 29th 2009, 07:58 AM
anlys
Quote:

Originally Posted by Opalg
It is not true that the series converges to 0. The series converges (in other words, the terms add up to a finite sum), but its sum is not zero. What is true is that the individual terms of the series converge to 0. That is always the case if a series converges.

Wouldn't it be easier to apply the L'Hospital rule to show that nx^n converges to 0 as n approaches to infinity? For 0<x<1, we can always write x = 1/(1+a) for some positive number a. So, the limit of nx^n as n goes to infinity equals to the limit of n/(1+a)^n, which is in the infinity/infinity form. So we can use the L'Hospital's rule in this case, which gives us limit of 1/(n*(1+a)^n-1), by differentiating both the numerator and denominator. Thus, as n approaches to infinity, this will approach to 0. So, we get nx^n converges to 0. Is this correct?
• Nov 29th 2009, 09:16 AM
Opalg
Quote:

Originally Posted by anlys
Wouldn't it be easier to apply the L'Hospital rule to show that nx^n converges to 0 as n approaches to infinity? For 0<x<1, we can always write x = 1/(1+a) for some positive number a. So, the limit of nx^n as n goes to infinity equals to the limit of n/(1+a)^n, which is in the infinity/infinity form. So we can use the L'Hospital's rule in this case, which gives us limit of 1/(n*(1+a)^n-1), by differentiating both the numerator and denominator. Thus, as n approaches to infinity, this will approach to 0. So, we get nx^n converges to 0. Is this correct?

There is certainly more than one way to show that $\lim_{n\to\infty}nx^n=0$ when |x|<1. L'Hôpital's rule is one way to approach it, but you need to be more careful in applying it. When you say $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{n(1+a)^{n-1}}$, you are differentiating the numerator as a function of n, and the denominator as a function of a. If you are using n as the variable, then $\tfrac d{dn}(1+a)^n = (1+a)^n\ln (1+a)$. So the calculation should go like this: $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{(1+a)^n\ln (1+a)} = 0$. That looks strange, because it's unusual to see n used as the name for a real variable rather than an integer. But the method is correct.
• Nov 29th 2009, 09:25 AM
anlys
Quote:

Originally Posted by Opalg
There is certainly more than one way to show that $\lim_{n\to\infty}nx^n=0$ when |x|<1. L'Hôpital's rule is one way to approach it, but you need to be more careful in applying it. When you say $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{n(1+a)^{n-1}}$, you are differentiating the numerator as a function of n, and the denominator as a function of a. If you are using n as the variable, then $\tfrac d{dn}(1+a)^n = (1+a)^n\ln (1+a)$. So the calculation should go like this: $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{(1+a)^n\ln (1+a)} = 0$. That looks strange, because it's unusual to see n used as the name for a real variable rather than an integer. But the method is correct.

Oh, you're right. Since I'm differentiating with respect to n, it should be like how you stated it. I should me more careful on determining which variables I am differentiating next time :) But, I'm glad the rest is correct. Thank you again for your feedback.