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Consider $f_n = \sqrt[n]{x}$ on $[0,1]$

So it converges to the step function $f = 0$ if $x = 0$ and $f=1$ otherwise

I could see why it doesn't converge if i draw an epsilon rectangle over one part since for each $n$, $f_n$ lies completely outside of the function.

If I draw an epsilon rectangle over the whole $f$, I don't see why this isn't uniform convergence

EDIT: I got this from Spivak, so keep it at that level please...

Added question: if $f_n\nrightarrow f$ for some $x \in \mathbb{R}$, can I conclude that it is not uniformly convergent over $\mathbb{R}$?

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What do you know about the limit of a uniformly convergent sequence of continuous functions? –  hardmath Jan 5 '13 at 23:17
    
Oh then the limit $f$ must also be continuous, if the set is closed and bounded right? But in my notebook, I wrote down that pointwise convergence does not imply regular convergence. –  sidht Jan 5 '13 at 23:25
    
@sizz: You are right to say that pointwise convergence does not imply uniform convergence. The problem that you have described proves exactly this fact. However, uniform convergence implies pointwise convergence, as I have shown below. –  Haskell Curry Jan 6 '13 at 2:56
    
What about pointwise divergence? Does that imply $f_n$ cannot uniformly converge? –  sidht Jan 6 '13 at 4:12
    
@Sizz: I also answered this question below. You see, uniform convergence implies pointwise convergence. If you have pointwise divergence, then you cannot have uniform convergence, otherwise you would have pointwise convergence --- a contradiction. :) –  Haskell Curry Jan 6 '13 at 4:21

3 Answers 3

up vote 7 down vote accepted

I like @hardmath's approach in the comments above. But here is an approach using the definition of uniform convergence.

Assume $\{f_n\}$ converges uniformly to $f$. Pick $\varepsilon < 1/2$. We can find an $N$ for which $\forall x \in [0, 1] : |f_N(x) - f(x)| < \varepsilon$. Clearly, $f_N(1) = 1$ and $f_N(0) = 0$. Since $f_N$ is continuous, we can find $x \in (0, 1)$ for which $f_N(x) = 1/2$ (by the intermediate value theorem). This means that $|f_N(x) - 1| = 1/2 > \varepsilon$. This is a contradiction and $\{f_n\}$ doesn't converge uniformly to $f$.


Here is a plot of $f_{10}(x) = \sqrt[10]{x}$:

plot

Since $f_n$ is continuous, there will always be values of $x > 0$ for which $f_n(x)$ is too far away from $1$. Uniform convergence requires that after a certain $N$, $f_n(x)$ must be within a small distance $\varepsilon$ from $f(x)$ for all $x$ . This fails for the sequence we have.

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Why is $|f_N(x) - 1| = 1/2$? –  sidht Jan 5 '13 at 23:52
    
@sizz Because we picked $x$ so that $f_N(x) = 1/2$. –  Ayman Hourieh Jan 5 '13 at 23:53
    
Would it be too much to ask to draw me a picture? I am have trouble visualing $|f_N(x) - f(x)| < \epsilon$ part –  sidht Jan 5 '13 at 23:55
    
@sizz I edited my question with a plot and more text to help you visualize it. –  Ayman Hourieh Jan 6 '13 at 0:10

We will first prove two results.

Theorem 1 Let $ (f_{n})_{n \in \mathbb{N}} $ be a sequence of continuous functions on $ [0,1] $ that converges uniformly to some function $ f $ on $ [0,1] $. Then $ f $ must be continuous on $ [0,1] $.

Proof: Let $ \epsilon > 0 $. Then there exists an $ N \in \mathbb{N} $ such that for all integers $ n \geq N $, we have $$ \forall x \in [0,1]: \quad |{f_{n}}(x) - f(x)| < \frac{\epsilon}{3}. $$ To prove that $ f $ is continuous, pick an arbitrary $ x_{0} \in [0,1] $. As $ f_{N} $ is continuous by assumption, there exists a $ \delta > 0 $ such that $ |{f_{N}}(x_{0}) - {f_{N}}(x)| < \dfrac{\epsilon}{3} $ for all $ x \in (x_{0} - \delta,x_{0} + \delta) \cap [0,1] $. Hence, by the Triangle Inequality, we see that for all $ x \in (x_{0} - \delta,x_{0} + \delta) \cap [0,1] $, the following relations hold: \begin{align} |f(x_{0}) - f(x)| &= |[f(x_{0}) - {f_{N}}(x_{0})] + [{f_{N}}(x_{0}) - {f_{N}}(x)] + [{f_{N}}(x) - f(x)]| \\ &\leq |f(x_{0}) - {f_{N}}(x_{0})| + |{f_{N}}(x_{0}) - {f_{N}}(x)| + |{f_{N}}(x) - f(x)| \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\ &= \epsilon. \end{align} As $ x_{0} $ and $ \epsilon $ are arbitrary, we conclude that $ f $ is indeed continuous on $ [0,1] $. $ \quad \spadesuit $

Theorem 2 Let $ (f_{n})_{n \in \mathbb{N}} $ be a sequence of (not-necessarily-continuous) functions on $ [0,1] $ that converges uniformly to some function $ f $ on $ [0,1] $. Then $ (f_{n})_{n \in \mathbb{N}} $ converges pointwise to $ f $.

Proof: This follows directly from the definition of uniform convergence. For any $ \epsilon > 0 $, there exists an $ N \in \mathbb{N} $ such that for all integers $ n \geq N $, we have $$ \forall x \in [0,1]: \quad |{f_{n}}(x) - f(x)| < \epsilon. $$ Therefore, for all $ x \in [0,1] $, we get $ \displaystyle \lim_{n \to \infty} {f_{n}}(x) = f(x) $. $ \quad \spadesuit $

This corollary is in response to the OP's latest question.

Corollary Let $ (f_{n})_{n \in \mathbb{N}} $ be a sequence of (not-necessarily-continuous) functions on $ [0,1] $ and $ f $ a function on $ [0,1] $ also. If $ {f_{n}}(x) \nrightarrow f(x) $ for some $ x \in [0,1] $, then $ (f_{n})_{n \in \mathbb{N}} $ does not converge uniformly to $ f $.

Proof: By Theorem 2, uniform convergence implies pointwise convergence; if pointwise convergence fails, then uniform convergence fails. $ \quad \spadesuit $


Assume now, for the sake of contradiction, that the sequence $ (f_{n})_{n \in \mathbb{N}} := (\sqrt[n]{\bullet})_{n \in \mathbb{N}} $ of continuous functions on $ [0,1] $ converges uniformly to some function $ f $ on $ [0,1] $. By Theorem 1, $ f $ is continuous on $ [0,1] $. By Theorem 2, $ f $ can be computed as the pointwise limit of $ (f_{n})_{n \in \mathbb{N}} $. Hence, \begin{equation} f(x) = \left\{ \begin{array}{ll} 0 & \text{if $ x = 0 $}; \\ 1 & \text{if $ x \in (0,1] $}. \end{array} \right. \end{equation} However, $ f $ is clearly not continuous at $ 0 $, thus contradicting Theorem 1.

Conclusion: $ (f_{n})_{n \in \mathbb{N}} $ does not converge uniformly to any function on $ [0,1] $. However, it does converge pointwise to the piecewise-defined function $ f $ described above.

The main point here is that uniform convergence and pointwise convergence are two different concepts. Uniform convergence implies pointwise convergence, but not vice-versa.

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Is that really the correct pointwise limit? –  mrf Jan 5 '13 at 23:44
    
Thanks! I kept thinking it was $ x^{n} $. –  Haskell Curry Jan 5 '13 at 23:45

@sizz : It does look as if you don't know the difference between pointwise convergence and uniform convergence.

If for every $x\in\mathbb R$, $\lim_{n\to\infty} f_n(x)=f(x)$, that is pointwise convergence of $f_n$ to $f$. That's what you've got here (although in order for the statement to be true just as you've stated it, you ought to have $\sqrt[n]{|x|}$).

Now look at the supremum over all $x\in\mathbb R$ of the distance between $f_n(x)$ and $f(x)$. If that goes to $0$ as $n\to\infty$, then that is uniform convergence of $f_n$ to $f$. That doesn't happen here. You have $$ f(x) = \begin{cases} 1 & \text{if }x\ne 0, \\ 0 & \text{if }x=0 \end{cases} $$ and $f_n(x)=\sqrt[n]{|x|}$. As $x\to0$, you have the distance between $f_n(x)$ and $f(x)$ approaching $1$. So the "uniform distance" between $f_n$ and $f$ is $1$. And that doesn't go to $0$ as $n\to\infty$. So you don't have uniform convergence.

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