# What's the reason why this sequence of function doesn't converge uniformly to $f$?

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 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. – sizz Jan 5 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 at 2:56
What about pointwise divergence? Does that imply $f_n$ cannot uniformly converge? – sizz Jan 6 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 at 4:21
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## 3 Answers

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 $f_{10}(x) = \sqrt[10]{x}$:

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 $|f_N(x) - 1| = 1/2$? – sizz Jan 5 at 23:52
@sizz Because we picked $x$ so that $f_N(x) = 1/2$. – Ayman Hourieh Jan 5 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 – sizz Jan 5 at 23:55
@sizz I edited my question with a plot and more text to help you visualize it. – Ayman Hourieh Jan 6 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 at 23:44
Thanks! I kept thinking it was $x^{n}$. – Haskell Curry Jan 5 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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