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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.2011-01-07
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    ohh, year right. Thanks, since english is not my native tongue, if have quite some difficulties, choosing the right terms.2011-01-07
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    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.2011-01-08
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    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 ^^2011-01-08
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    Right on! That was exactly the point of these two examples. Have you drawn a picture? Do you see the difference?2011-01-08
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    Yeah, I believe so. The maximum of $f_n(x)$ is getting larger as $n \to \infty$. The area in which this part of the function is defined, gets smaller and smaller, so the graph of $f_n(x)$ is rising steeper. With $g_n(x)$ the maximum stays at $\frac{1}{4}$, the graph of $g_n(x)$ is also rising steeper and steeper as $n \to \infty$. The maximum of $h_n(x)$ is aproaching $0$ as $n \to \infty$ Therefore this graph converges uniformly.2011-01-09

4 Answers 4

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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?2011-01-07
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    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.2011-01-07
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    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.2011-01-07
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    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?2011-01-08
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    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 $x_{j} \to 1$ .....2011-01-08
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    ..... but $\lim_{n \to \infty} k_{n}(x_{j})$ is either zero or one depending on whether $j$ is odd or even.2011-01-08
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    Ahh. thank you so much, that was a very valuable example. I sum up as follows: $\forall x \in A: \lim_{n \to \infty} f_{n}(x) = f(x)$, means pointwise konvergence. $\forall y \in A \lim_{x \to y} \lim_{n\to\infty} f_n(x) = f(x)$ means continuous pointwise convergence and $\lim_{n \to \infty} \sup_{x \in A}|f_{n}(x) - f(x)| = f(x)$ means uniform convergence. So there is a difference in writing $\forall y \in A$ and $\sup_{x \in A}$. Let A now be $[0,1]$ and $f_n$ the function from our task. (next comment)2011-01-09
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    Lets say $x=\frac{1}{2n}$. Therefore $x$ is obviously not a member of $\forall x \in A$, since otherwise it would not converge pointwise to $0$. But I am using it in $\lim_{n \to \infty} \sup_{x \in A}|f_{n}(x) - f(x)| = f(x)$, where I am using Elements with $x \in A$. So what is the difference in writing $\forall y \in A$ and $\sup_{x \in A}$. Isn't the latter included in the former?2011-01-09
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    And one more question ;-). I think I remember that a uniformly convergent function implies: $\forall y \in \mathbb{R}: \quad \lim_{x \to y} \lim_{n\to\infty} f_n(x) = \lim_{n\to\infty} \lim_{x \to y} f_n(x)$, where $y$ is a Limit point. Is the reverse implication also true?2011-01-09
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    Concerning $\forall y \in A: f(y) = \lim_{x \to y} \lim_{n \to \infty} f_{n}(x)$: As I said, it says (implicitly) **1.** $\forall x \in A: f(x) = \lim_{n \to \infty} f_{n}(x)$ - the sequence $f_{n}$ converges pointwise to $f$ - *and* **2.** $\forall y \in A: f(y) =\lim_{x \to y} f(x)$ - the limit function $f$ is continuous.2011-01-09
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    Concerning your last comment: 1. *yes*, you remembered correctly. 2. *yes*, if $A$ is a closed and bounded set e.g. an interval $[a,b]$ - that's not entirely trivial2011-01-09
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    I don't quite understand the statement "Let's say $x = \frac{1}{2n}$. Therefore $x$ is obviously not a member of $\forall x \in A$, ...". First, $x = \frac{1}{2n}$ is a member of $A = [0,1]$ but it *depends* on $n$, that's why it's better to write $x_{n} = \frac{1}{2n}$ in order to indicate this dependence. Second, $\forall x \in A$ is the *statement* "for all $x \in A$" so it's *not* a set and thus cannot have members.2011-01-09
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    So what's the difference? Well $\forall y \in A$ means for each *individual* $y \in A$, while $s_{n} = \sup_{x \in A}|f_{n}(x) - f(x)|$ is simply a real number2011-01-09
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    Ahh, that made it clear. Thank you, this was great.2011-01-13
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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.2011-01-08
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    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?2011-01-08
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    Thanks for this answer. This is indeed a nice test and good to know of. Thanks2011-01-08
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    @Theo: Have changed it.2011-01-08
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    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.2011-01-08
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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.2011-01-07
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    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?2011-01-07
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    @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.2011-01-07
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    Ok, now I agree with your description. Thanks for the fix.2011-01-07
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If one has to justify the existence of the absolute maximum of $\:f_n,\:$ the following'd be non trivial. $$\text{For }\:n\in \mathbf N,\:]0,1[\:\ni u

$\varphi_n(x)\:$ being polynomial justifies its continuity so that $$\varphi_n(u)=\lim_{x\to u}\varphi_n(x)=\lim_{x\to u}{f_n(x)-f_n(u)\over x-u}=:f^{\large'}_n(u)=n^2-2n^3u,\:\:\forall n\in\mathbf N.$$

Thus, $\:f_n\:$ is differentiable on $\:[0,1].$

Now, $\:f^{\large'}_n(x)=0\iff x_n=1/{2n}\:\:\:\&\:\:f_n(x_n)=n/4.$

We hope that $\:f_n\:$ attains its absolute maximum at $\:x_n:$ $$\forall\delta>0\:\:\exists\mathcal{\:N}_\delta(x_n)=\left]\frac{1}{2n}-\delta,\frac{1}{2n}+\delta\right[:f^{\large'}_n(y)\ge0\:\:\text{ for }\:\frac{1}{2n}-\delta

For the increasing part, we must have that $\:\left|y+\frac{\delta}{2}\right|<\frac{1-n\delta}{2n}.$

Let's choose $\:y=\large{1-n\delta\over 2n}\normalsize\implies f^{\large'}_n(y)=n^3\delta>0$

For the decreasing part, we must have that $\:\left|y-\frac{\delta}{2}\right|<\frac{1+n\delta}{2n}.$

Let's choose $\:y=\large{1+n\delta\over 2n}\normalsize\implies f^{\large'}_n(y)=-n^3\delta<0.$

So that $\:f_n\:$ attains $\text{a relative maximum}\:$ at $\:x_n=1/{2n}.\:$

But since $\:\delta\:$ must be arbitrarily chosen small, whenever $\:\delta=1/{2n},\:$ our neighbourhood $\:\mathcal N_\delta(x_n)\:$ becomes asymptotically equivalent to $\:]0,1/n[\:\subset[0,1],\:$ for some $\:n\ge N\ge1$.

Therefore $\:f_n\:$ attains $\text{its absolute maximum}\:$ at $\:x_n=1/{2n}.\:$