MHB (f_n) converges pointwise to a continuous f

[evinda]

Hey again! (Blush)

I am looking at the following exercise:
Let $$f_n(x)= \begin{cases}
0,x< \frac{1}{n+1} \text{ or } \frac{1}{n}<x \\
\sin^2( \frac{ \pi}{x}), \frac{1}{n+1} \leq x \leq \frac{1}{n}
\end{cases}.$$
Prove that $(f_n)$ converges pointwise to a continuous $f$ in $ \mathbb{R}$.
Which is this $f$ ? Does $f_n \to f$ uniformly in $\mathbb{R}$ ?

Can I show that $(f_n)$ converges pointwise to $f=0$ like that?
$ \displaystyle \forall x<\frac{1}{n+1}$ or $x>\dfrac{1}{n}$: $f_n(x)=0 \to 0$
$ \displaystyle \forall \frac{1}{n+1} \leq x \leq \frac{1}{n} \Rightarrow n \leq \frac{1}{x} \leq n+1 \Rightarrow \pi n \leq \frac{\pi}{x} \leq \pi(n+1) \Rightarrow $
$ \displaystyle \sin^2( \pi n) \leq \sin^2 ( \frac{ \pi}{x}) \leq \sin^2( \pi(n+1)) \Rightarrow$
$\displaystyle \lim_{n \to + \infty} \sin^2( \pi n) \leq \lim_{n \to +\infty} \sin^2 ( \frac{ \pi}{x}) \leq \lim_{n \to +\infty} \sin^2( \pi(n+1)) \Rightarrow 0 \leq f \leq 0 \Rightarrow f=0$

 

[Deveno]

Let $x$ be any real number. We wish to show that for any $\epsilon > 0$, there is some $N \in \Bbb N$ such that for all $n > N$:

$|f_n(x)| < \epsilon$.

If $x > 1$, we can choose $N = 1$, and similarly for $x \leq 0$. If $x = 1$, we can pick $N = 2$.

So we just need to focus on $0 < x < 1$.

Choose $N = \lfloor \frac{1}{x} \rfloor + 1$. This tells us that:

$\dfrac{1}{x} < N$ so that for $n > N$:

$\dfrac{1}{n} < \dfrac{1}{N} < x$

in which case we have $f_n(x) = 0$ for all $n > N$.

Note that our choice of $N$ depends on $x$.

Can we pick SOME $N$ that will work for EVERY $x \in (0,1)$?

Suppose we could.

Suppose we have some $n > N$, and consider $x = \dfrac{2}{2n+1}$.

For such $x$, we have $f_n(x) = \sin^2(\frac{\pi}{x}) = \sin^2(\frac{(2n+1)\pi}{2})$

$= \sin^2(\pm \frac{\pi}{2}) = (\pm 1)^2 = 1$, so if:

$1 > \epsilon > 0$, we have, for this $x$:

$|f_n(x)| = 1 > \epsilon$, contradicting our choice of $N$.

So $f_n$ does not converge uniformly.

 

[evinda]

Let $x$ be any real number. We wish to show that for any $\epsilon > 0$, there is some $N \in \Bbb N$ such that for all $n > N$:

$|f_n(x)| < \epsilon$.

If $x > 1$, we can choose $N = 1$, and similarly for $x \leq 0$. If $x = 1$, we can pick $N = 2$.

So we just need to focus on $0 < x < 1$.

Choose $N = \lfloor \frac{1}{x} \rfloor + 1$. This tells us that:

$\dfrac{1}{x} < N$ so that for $n > N$:

$\dfrac{1}{n} < \dfrac{1}{N} < x$

in which case we have $f_n(x) = 0$ for all $n > N$.

Note that our choice of $N$ depends on $x$.

Can we pick SOME $N$ that will work for EVERY $x \in (0,1)$?

Suppose we could.

Suppose we have some $n > N$, and consider $x = \dfrac{2}{2n+1}$.

For such $x$, we have $f_n(x) = \sin^2(\frac{\pi}{x}) = \sin^2(\frac{(2n+1)\pi}{2})$

$= \sin^2(\pm \frac{\pi}{2}) = (\pm 1)^2 = 1$, so if:

$1 > \epsilon > 0$, we have, for this $x$:

$|f_n(x)| = 1 > \epsilon$, contradicting our choice of $N$.

So $f_n$ does not converge uniformly.

Could you explain me further how you concluded that $f_n$ converges pointwise to $0$,when $ \frac{1}{n+1} \leq x \leq \frac{1}{n} $ ? (Blush)

 

[Deveno]

Put somewhat broadly, the $n$'s we need to look at depend on our $x$.

As $n$ increases, the $x$'s that are in the interval $\left(\dfrac{1}{n+1},\dfrac{1}{n}\right)$ keep getting closer to 0.

So for any $x$, we just need to pick an $N$ big enough so that $x$ is to the right of all those intervals for $n > N$.

Of course, as $x$ gets really close to 0, we keep having to pick enormous $N$'s, and no single $N$ will work for ALL $x$.

If it weren't for the fact that $f_n(0) = 0$ for all $n$, we WOULD have a problem at 0, because we get an "infinitely small area with infinitely fast oscillation" for $f_n$ as $n \to \infty$.

In essence, we "sweep the problem under the rug" by pushing the "problem area" to the left. Since the real numbers are "infinitely deep", we never run out of room, even though we keep having to squeeze the "bad part" in-between $x$ and 0.

The point being, for any given $x$ (at that "point"), we can arrange it so that $x$ is NOT IN the interval you speak of. So the situation you mention does not occur.

 

[evinda]

Put somewhat broadly, the $n$'s we need to look at depend on our $x$.

As $n$ increases, the $x$'s that are in the interval $\left(\dfrac{1}{n+1},\dfrac{1}{n}\right)$ keep getting closer to 0.

So for any $x$, we just need to pick an $N$ big enough so that $x$ is to the right of all those intervals for $n > N$.

Of course, as $x$ gets really close to 0, we keep having to pick enormous $N$'s, and no single $N$ will work for ALL $x$.

If it weren't for the fact that $f_n(0) = 0$ for all $n$, we WOULD have a problem at 0, because we get an "infinitely small area with infinitely fast oscillation" for $f_n$ as $n \to \infty$.

In essence, we "sweep the problem under the rug" by pushing the "problem area" to the left. Since the real numbers are "infinitely deep", we never run out of room, even though we keep having to squeeze the "bad part" in-between $x$ and 0.

The point being, for any given $x$ (at that "point"), we can arrange it so that $x$ is NOT IN the interval you speak of. So the situation you mention does not occur.

So,you mean that, as $\frac{1}{n+1} \leq x \leq \frac{1}{n}$, $x \to 0$ ?
Why do we know that $f_n(0)=0$ ? (Wondering)

 

[Deveno]

No matter how big we pick $n$, we still have:

$0 < \dfrac{1}{n+1}$, so from the definition of $f_n$, we have $f_n(0) = 0$, for ALL $n$.

That's one of the things about natural numbers: no matter how huge they are, every one of them is still finite.

You see to be missing a fundamental point: to show $f_n$ converges at the point $x$, we first FIX $x$. THEN we look for "big enough $n$". We don't "move $x$" along with the $n$'s: the point $x$ stays put.

All we need for "pointwise convergence" is for $f_n$ to "settle down" LOCALLY, which in this case it DOES: eventually, for every $x > 0$, there is a (perhaps very small) neighborhood (interval) around $x$ so that $f_n$ is CONSTANT on that tiny interval, for all $n > N$ (and the $N$ we pick depends on an essentially way on the point $x$).

For "uniform convergence" we need the $f_n$ to settle down "everywhere all at once" to some function, in other words the rate of convergence is "uniform", whereas for pointwise convergence the rate of convergence can vary (in this case, the closer to 0 we get, the slower the convergence gets, and it takes MANY $n$'s for it to converge within a small $\epsilon$ if $x$ is very close to 0).

You can look at it this way:

For $f_1$ everything to the right of 1 is 0.
For $f_2$ everything to the right of 1/2 is 0.
For $f_3$ everything to the right of 1/3 is 0.
For $f_4$ everything to the right of 1/4 is 0.
...
For $f_n$ everything to the right of 1/n is 0.

So if $x > 0$, surely we can pick $N$ SO BIG, that $x > \dfrac{1}{N}$. This gives us some "wiggle room" (namely, if we set:

$d = x - \dfrac{1}{N}$ then on the entire interval $(x-d,x+d)$ we have $f_n = 0$ for all $n > N$).

We aren't concerned with $f_n(x)$ FOR ALL $x$, we're just interested in $f_n(x)$ at the PARTICULAR $x$ we are trying to prove pointwise convergence at.

The entire point of this problem is to SHOW you that pointwise convergence isn't really the "best" kind of convergence, that sequences of functions that converge pointwise can behave in unexpected ways:

In this case, the range of EVERY $f_n$ is [-1,1], but the range of the function they converge to pointwise is just {0}.

There are other "bad" sequences like this: one can construct a sequence like:

$f_n(x) = \dfrac{1}{x},\ 0 < x < \dfrac{1}{n}$

$f_n(x) = 0$, otherwise.

This likewise converges pointwise to the 0 function, but none of the $f_n$ are continuous, or even bounded.