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Doing an exercise for exam preparation, I stumbled across the following function:

$f_n(x)= n^2x(1-nx), \quad \text{if }0 \leq x \leq \frac{1}{n} $

$f_n(x)= 0, \quad \text{if } \frac{1}{n} < x \leq 1$

The task is to find the limit of this function series and to determine whether this function converges uniformly in $[0,1]$

On the one hand $\frac{1}{n}$ approaches $0$ for $n \to \infty$. So one would just have to insert $0$ in $n^2x(1-nx)$. Thereby gaining $f_n(x) = 0$ for $n \to \infty$.

On the other hand the function has a maximum for $x=\frac{n}{2}$. Putting this into $n^2x(1-nx)$ and calculating $f_n(x)$ for $n \to \infty$ afterwards one gets $f_n(x) = \infty$.

So whats correct? How does one approach such a problem?

Thanks in advance

ftiaronsem

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We are talking about a function $sequence$ here, not a series. The notion of series comes into play when the $f_n(x)$ of a sequence of functions are $added up$ to form $\sum_{n=0}^\infty$ f_n(x)$ and one tries to make sense of such an expression. –  Christian Blatter Jan 7 '11 at 19:11
    
ohh, year right. Thanks, since english is not my native tongue, if have quite some difficulties, choosing the right terms. –  ftiaronsem Jan 7 '11 at 20:54
    
By the way: have you tried modifying the sequence in question? Try \[g_{n}(x) = \begin{cases} nx(1-nx) & \text{if $0 \leq x \leq \frac{1}{n}$} \\ 0 & \text{if $\frac{1}{n} \leq x \leq 1$} \end{cases}\] and \[h_{n}(x) = \begin{cases} x(1-nx) & \text{if $0 \leq x \leq \frac{1}{n}$} \\ 0 & \text{if $\frac{1}{n} \leq x \leq 1$} \end{cases}\] for example. –  t.b. Jan 8 '11 at 1:50
    
Ohh, nice. Thanks for providing this comment. $g_n(x)$ and $h_n(x)$ have both $\frac{1}{2n}$ as Maximum and both are converging pointwise to 0. $\lim_{n \to \infty}g_n(\frac{1}{2n}) = \frac{1}{4}$, which is why $g_n(x)$ is not converging uniformly.$\lim_{n \to \infty}h_n(\frac{1}{2n}) = \lim_{n \to \infty}\frac{1}{4n} = 0 $, which is why $h_n(x)$ is converging uniformly against $0$. Hope this was correct ^^ –  ftiaronsem Jan 8 '11 at 15:29
    
Right on! That was exactly the point of these two examples. Have you drawn a picture? Do you see the difference? –  t.b. Jan 8 '11 at 16:08

3 Answers 3

up vote 6 down vote accepted

I'd strongly encourage you to draw a picture of the graph of $f_{n}$.

Your argument that $f_{n} \to 0$ is not quite correct. I'd argue as follows:

We have $f_{n}(x) \to 0$ as $(n \to \infty)$ for all $x \in [0,1]$. This is clear for $x = 0$ and for $x > 0$ we have $f_{n}(x) = 0$ for all $n$ so large that $\frac{1}{n} < x$.

If $f_{n} \to f$ uniformly on $[0,1]$ then $f_{n} \to f$ pointwise, hence we must have $f = 0$.

On the other hand, the function $f_{n}$ has a maximum at $\frac{1}{2n}$ (not $\frac{2}{n}$ as you've written in your question) as can be found by differentiation, for example. Evaluation gives \[ f_{n}(\frac{1}{2n}) = n^{2} \frac{1}{2n} ( 1 - n\frac{1}{2n}) = \frac{n}{4} \] Therefore \[ \sup_{x \in [0,1]} |f_{n}(x) - f(x)| = \sup_{x \in [0,1]} |f_{n}(x)| = \frac{n}{4} \xrightarrow{n \to \infty} \infty \] and hence $f_{n}$ does not converge uniformly to $f = 0$.

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Thanks for your answer.Is the following correct? $f_n$ is converging pointwise, which is what we gain from our first argument. So we have showed that $\forall y \in \mathbb{R} \lim_{x \to y} \lim_{n\to\infty} f_n(x) = 0$ In order to show uniform convergence we have to show that $\forall y \in \mathbb{R} \lim_{n\to\infty} \lim_{x \to y} f_n(x) = 0$, which fails, by chosing $y=\frac{1}{2n}$. So both statements are true on their own. There is no contradiction, isn't it? The first is just the definition of pointwise convergence, whereas the second is the defintion fo uniform convergence? –  ftiaronsem Jan 7 '11 at 17:28
    
Not quite. Pointwise convergence $f_{n} \to f$ means: For all $x \in [0,1]$ we have $\lim_{n \to \infty} f_{n}(x) = f(x)$, or, in other words, $\lim_{n \to \infty}|f_{n}(x) - f(x)| = 0$. Indeed, that's what I've shown in the first few lines. Uniform convergence $f_{n} \to f$ means $\lim_{n \to \infty} \sup_{x \in [0,1]}|f_{n}(x) - f(x)| = 0$. Since $0 \leq |f_{n}(x) - f(x)| \leq \sup_{x \in [0,1]} |f_{n}(x) - f(x)|$ we have that uniform convergence implies pointwise convergence. We've seen that $\sup_{x \in [0,1]} |f_{n}(x) - f(x)| = \frac{n}{4}$, so $f_{n}$ doesn't converge uniformly. –  t.b. Jan 7 '11 at 17:51
    
Again: $f_{n} \to f$ uniformly implies $f_{n} \to f$ pointwise (same $f$!). On the other hand, we know already that in our example $f_{n} \to 0$ pointwise, so the only candidate for a uniform limit is $f = 0$. The second part of the argument shows that $f_{n} \not\to 0$ uniformly. No contradiction, of course. –  t.b. Jan 7 '11 at 17:55
    
Thank you very much for this explanations. However I am a little bit puzzled because you wrote "not quite" under my last comment ;-). Isn't $\forall y \in [0,1] \lim_{x \to y} \lim_{n\to\infty} f_n(x) = 0 = f(x)$ equivalent to "For all $x \in [0,1]$ we have $\lim_{n \to \infty} f_{n}(x) = f(x)$"? Presumably not, but why? –  ftiaronsem Jan 8 '11 at 16:03
    
That's exactly the point of contention. The additional $\lim_{x \to y} f(x) = f(y)$ just means that the pointwise limit $f$ is continuous. But a pointwise convergent sequence need not converge to a continuous function: Take $k_{n}(x) = x^{n}$ on $[0,1]$, then $k_{n}$ converges pointwise to \[k(x) = \begin{cases} 0 & \text{if $0 \leq x < 1$} \\\ 1 & \text{if $x = 1$} \end{cases}\] but $\lim_{x \to 1} \lim_{n \to \infty} k_{n}(x)$ does not exist. Take the sequence $x_{j} = 1 - \frac{1}{j}$ if $j$ is odd and $x_{j} = 1$ if $j$ is even. Then $x_{j} \to 1$ ..... –  t.b. Jan 8 '11 at 16:28

The limit function is $f(x) = 0 \quad \forall x \in [0,1]$ because, as $n \rightarrow \infty$ you have that the region where the sequence is $n^2 x(1-nx)$ is always smaller and smaller (this is a nice way of approaching sequences with boundary conditions that involve $n$).

As for uniform convergence, you should take the supremum for $x \in [0,1]$ and since the function has a maxmimum inside the interval, you can say that $\sup_{x \in [0,1]}\left| n^2 x(1- n x)\right| = \frac{n}{4}$, and if $n \rightarrow \infty$ it doesn't approach zero, so the convergence is not uniform.

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My answer is an intuitive argument (for pointwise convergence), while Theo Buehler gave a more rigrous one. I guess mine can be left here as a hint to get an idea and then start with a more rigorous argument. –  Andy Jan 7 '11 at 12:31
    
thanks too, for your answer. But I have to agree with Theo. What do you mean with "since the function is decreasing and negative"?. This function has a Maximum hasn't it? –  ftiaronsem Jan 7 '11 at 17:30
    
@Theo Buehler, @ftiaronsem : you're both right, I'll correct that remark (I was writing rather in a hurry, I'm sorry). I should have checked better when reading Theo's answer. –  Andy Jan 7 '11 at 18:26
    
Ok, now I agree with your description. Thanks for the fix. –  t.b. Jan 7 '11 at 19:01

Another way to proceed is to look at the area of the function in your domain $[0,1]$. (I usually use this as a first step to check if the function is not uniformly convergent, since it is relatively easy.)

The area of the function $f_n(x)$ in the domain is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$ irrespective of $n$.

Note that $f_n(0) = 0, \forall n$ and $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$, $\forall x \in (0,1]$.

This can be seen since for any $x$, $\exists N \in \mathbb{N}$ such that $\forall n > N$, $\frac{1}{n} < x \Rightarrow f_n(x) = 0$.

Hence, $f(x) = \displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$, $\forall x \in [0,1]$.

So we have $\displaystyle \int_{0}^{1} f_n(x) dx = \frac{1}{6}$, $\forall n \in \mathbb{N}$ and $\displaystyle \int_{0}^{1} f(x) dx = 0$. So we have $$\displaystyle \lim_{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx = \frac{1}{6} \neq 0 = \displaystyle \int_{0}^{1} f(x) dx$$

Hence, we have $\displaystyle \lim_{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx \neq \displaystyle \int_{0}^{1} \lim_{n \rightarrow \infty} f_n(x) dx$

And we know that if a sequence of functions converge uniformly, we can swap the limit and the integrals to get the same integral.

Hence, the function is not uniformly convergent.

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+1: I like it! It's quite easy in this case; very useful. –  Andy Jan 8 '11 at 10:19
1  
This is a cool and often useful argument. Two comments: 1. The limitations of this argument are shown by the sequence $g_{n}$ I posted in a comment to the question, where it doesn't work. 2. Could you please change ''And we know that if the function...'' to ''And we know that if the sequence...'' in the penultimate paragraph? –  t.b. Jan 8 '11 at 15:35
    
Thanks for this answer. This is indeed a nice test and good to know of. Thanks –  ftiaronsem Jan 8 '11 at 16:06
    
@Theo: Have changed it. –  user17762 Jan 8 '11 at 19:27
    
Yes at Theo stated, this only means one way implication. So this line of thought can be used to show that the functions is not uniformly convergent. –  user17762 Jan 8 '11 at 19:29

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