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Math Help - Pointwise convergence to uniform convergence

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    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!
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    Quote Originally Posted by tttcomrader View Post
    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.
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    Okay, here is my formal proof.

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

    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!!!
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    Quote Originally Posted by tttcomrader View Post
    Okay, here is my formal proof.

    Define f_n(x): [0,1] \rightarrow \mathbb {R} by <br />
f_n(x) = nx^n(1-x) \ \ \ \forall n \in \mathbb {N}<br />
    Both parts of this are actually quite tricky.

    Quote Originally Posted by tttcomrader View Post
    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 View Post
    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.
    Last edited by Opalg; November 9th 2008 at 11:42 PM. Reason: corrected mistake
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    Quote Originally Posted by Opalg View Post

    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.
    I'm a bit unclear about how <br />
\bigl(\tfrac n{n+1}\bigr)^{n+1}<br />
 \rightarrow \frac {1}{e} Did we use the natural log here?
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    Quote Originally Posted by tttcomrader View Post
    I'm a bit unclear about how <br />
\bigl(\tfrac n{n+1}\bigr)^{n+1}<br />
 \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}<br />
{{n + b}}} \right)^{cn}  = e^{ac} (with the correct restrictions on b)
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    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} ?
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    Quote Originally Posted by tttcomrader View Post
    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.
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  9. #9
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    Quote Originally Posted by tttcomrader View Post
    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.)
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    Quote Originally Posted by Opalg View Post
    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.
    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?
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    Quote Originally Posted by anlys View Post
    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.
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    Quote Originally Posted by Opalg View Post
    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?
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    Quote Originally Posted by anlys View Post
    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'H˘pital'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.
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    Quote Originally Posted by Opalg View Post
    There is certainly more than one way to show that \lim_{n\to\infty}nx^n=0 when |x|<1. L'H˘pital'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.
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