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(a) Define a sequence of functions on $\mathbb{R}$ by $f_n(x) = 1$ if $x\in\{\dfrac{1}{2}, \dfrac{1}{3},\cdots ,\dfrac{1}{n}\},$ $f=0$ otherwise. Let $f$ be the pointwise limit of $f_n$

Does $f_n \rightarrow f$ uniformly on $\mathbb{R}$? Is $f$ continuous at $0$?

(b) Repeat this exercise using the sequence of functions $g_n(x) = x$ if $\{ x=1,\dfrac{1}{2}, \dfrac{1}{3},...\dfrac{1}{n}$ or $0$ otherwise. Let $f$ be the pointwise limit of $f_n$

For part (a) I see I can get an $N \geq \dfrac{1}{\epsilon}$ so that only a finite many points are outside the epsilon, thus when $n > N$, $|f_n(x)-f(x)|<\epsilon$. This means that it converges uniformly to $0$? What happens at 0, that is what is confusing me.

For part (b) I'm at a loss as to where to start. Thank you in advance for any help.

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    A similar question there http://math.stackexchange.com/questions/2135199/having-trouble-showing-continuity-and-pointwise-limit-of-this-function/2135347#2135347 – 2017-02-25

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Let's do (a) first. I recommend you to first study the pointwise convergence. So, pick $x_0\in \mathbb R$ and study the limit $\lim_{n\to +\infty} f_n(x_0)$. We have:

  • If $x_0\notin\{\frac1m, m\in\mathbb N\}$, then $f_n(x_0)=0$ $\forall n\in \mathbb N$, so $\lim_{n\to +\infty} f_n(x_0)=0$.
  • If $x_0=\frac{1}{m}$ for $m\in \mathbb N$, then we have $f_n(x_0)=1$ if we pick $n>m$. So $\lim_{n\to +\infty} f_n(x_0$)=1$.

So the pointwise limit would be $$f(x)=\begin{cases} 1 \text{ if $x=\frac{1}{n}$ for $n\in\mathbb N$} \\ 0 \text{ otherwise}\end{cases}$$

Now, to prove uniform convergence, you have to see that when $n\to+\infty$ $$\sup_{x\in\mathbb R}|f_n(x)-f(x)|\to 0$$

So, let's study the quantities $|f_n(x)-f(x)|$. Note that $f_n(\frac1{n+1})=0$, but $f(\frac{1}{n+1})=1$. So $\sup_{x\in\mathbb R}|f_n(x)-f(x)|>1$, and that means it cannot converge to $0$. So $f_n$ does not converge uniformly to $f$.

To study if $f$ is continuous at $0$, note that if there wasn't for the points $\frac{1}{n}$, $\lim_{x\to 0} f(x)=0$. That's the only possibility. But if you try to apply the definition of limit: $$\forall \varepsilon>0, \ \exists \delta>0 : |x|<\delta \Rightarrow |f(x)|<\varepsilon$$ you quickly know that if you take $\varepsilon<1$, this can't be satisfied, because for every $\delta>0$ you can find a point $\frac1n$ so that $\frac1n<\delta$ and $|f(\frac1n)|=1\nless \varepsilon$.

Now, for part $(b)$, let's study pointwise convergence. If $x_0\in\mathbb R$, we have:

  • If $x_0\notin\{\frac1m, m\in\mathbb N\}$, then $g_n(x_0)=0$ $\forall n\in \mathbb N$, so $\lim_{n\to +\infty} g_n(x_0)=0$.
  • If $x_0=\frac{1}{m}$ for $m\in \mathbb N$, then we have $g_n()=\frac{1}{m}$ if we pick $n>m$. So $\lim_{n\to +\infty} g_n(x_0)=\frac1m$.

so the pointwise limit is, in this case, $$g(x)=\begin{cases} \frac1n \text{ if $x=\frac{1}{n}$ for $n\in\mathbb N$} \\ 0 \text{ otherwise}\end{cases}$$

Note that $\sup_{x\in \mathbb R} |f_n(x)-f(x)| = \frac{1}{n+1}$, because these two functions only differ in the points $\{\frac{1}{n+1}, \frac{1}{n+2}, ...\}$ and the maximum difference is reached at $\frac{1}{n+1}$. So: $$\sup_{x\in \mathbb R} |f_n(x)-f(x)| = \frac{1}{n+1} \to 0$$ when $n\to +\infty$. There is uniform convergence to the function $f(x)=0$.

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    It's the way the problem is set up, as a piecewise function that is throwing me off. So the value of $f(x)$ will oscillate between $0$ and $1$ as $x$ takes on values of the sequence? – 2017-02-25
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    No, $f(x)$ would take the value $1$ at the points $\frac1n$, and $0$ in the other points. Try making a sketch of $f_1$, $f_2$, $f_3$... and see how that functions behave. The difference between $f_n$ and $f_{n+1}$ is that you are "moving" the image of $\frac{1}{n+1}$ from $0$ to $1$. – 2017-02-25
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    That was my intended meaning, I could have phrased it better, thanks. So if it's bouncing between 0 and 1, it can't be uniform, then. Because at $1/n$, $f(x) =1$ and for the infinite number of points between that and $1/(n+1)$, $f(x) = 0$. Am I getting this now? – 2017-02-25
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    Quite well. Now, try to do part $(b)$, using the same technique. Note that this functions still "bounce", but the situation is slightly different now. Please ask if you get stuck. – 2017-02-25
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    Yes, I do appreciate that. In the case of $(b)$ then the pointwise limit of $f_n$ is $x$ for the points in the sequence, but $0$, otherwise. I can at least see some sort of convergence there. – 2017-02-25
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    So, to put it more specifically, $$f(x)=\begin{cases} \frac1n \text{ if $x=\frac1n$} \\ 0 \text{ otherwise } \end{cases}$$ Now, what can you say about $|f_n(x)-f(x)|$? – 2017-02-25
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    When $x=1$, $f(x)=1$ in $(1, 1/2)$, $f(x) = 0$. At $x=1/2$, $f(x)=1/2$. On $(1/2, 1,3)$, $f(x)=0$, etc. It looks like $|f_n(x) - f(x)| $ is either $0$ or approaching $0$. At any rate, there has to be a certain $N$ for where $|f_n(x) - f(x)| < \epsilon$. – 2017-02-25
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    Yes, that's it. I've edited my answer with some indications. – 2017-02-25