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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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    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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    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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    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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    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 $\:x_n=1/{2n}.\:$