calculusdefinitionreal-analysis
I'm reading about the definition of exponential function on $\Bbb R$ and I came across a definition of pointwise convergence which I don't understand:
We say that a sequence of functions $(f_n)_n$ where $f_n:I\rightarrow \Bbb R, \ I\subseteq \Bbb R$, converges pointwise to function $f:I\rightarrow \Bbb R$ on the interval $I$ if a sequence of numbers $(f_n(x))_n$ converges to $f(x), \forall x\in I.$
Can someone please explain this definition and provide an example?
For every irrational $x \in (0,1)$, $$\lim_{n \to \infty} f_n(x) = 0$$ is obvious. For a rational $x = \frac{p}{q}$, its again easy to see that $$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{p}{q^n} = 0$$
Thus the pointwise limit is $0$ at all $x \in (0,1)$.
To show uniform convergence, note that $f_1(x) \leq 1$ for all $x$. Also $f_2(x) \leq \frac{1}{2}$ and so on, its easy to see (since for any rational $\frac{p}{q}$, $q \geq 2$) that $$f_n(x) \leq \frac{1}{2^n}$$
Now given $\epsilon > 0$, choose $N$ such that $$\frac{1}{2^N} < \epsilon$$ and you're done.
For your third question, JohnD's answer in the page you are quoting, contains a trick that's often used in Analysis courses. That is, try to find:
$$\sup_{x\in[0,1]}|f_n(x)-f(x)|$$
The supmemum occurs at $x$ such that:
$$\frac{df_n(x)}{dx}=0$$
Solving the above for $x$, you get:
$$x=\pm\frac{1}{n}$$
$\frac{1}{n}\in[0,1]$, so substitute back to the function to get:
$$\sup_{x\in[0,1]}|f_n(x)|=f_n\left(\frac{1}{n}\right)=\frac{1}{2}$$
And this is fixed and does not vanish, therefore convergence in $[0,1]$ is not uniform.
Addendum for comment:
I am adding a graphic, so you can see what's happening as a response to your second question.
This is the graph of $f_1(x)$, $f_2(x)$,..., $f_5(x)$, from right to left. Note that the supremum is given by $\left(\frac{1}{n},f_n\left(\frac{1}{n}\right)\right)=\left(\frac{1}{n},\frac{1}{2}\right)$ and is moved to the left on each iteration, but always stays at 1/2.
Best Answer
Pointwise convergence of $(f_n)_{n\in\mathbb{N}}$ to $f$ means that for each point $x \in I$, we have $\lim_{n \rightarrow \infty}{f_n(x)}=f(x)$.
Essentially we take a point $x \in I$ and look at $f_n(x)$ as $n \rightarrow \infty$. If this converges to a limit and does so for all $x \in I$, then it makes sense to say $(f_n)_{n\in\mathbb{N}}$ converges to the function $f(x)=\lim_{n\rightarrow\infty}f_n(x)$.
We call this pointwise convergence because it only looks at individual points rather than the functions as a whole, which is in contrast to something stronger like uniform convergence.
An example would be $$f_n(x)=\frac{x^2}{n} (x \in \mathbb{R})$$
For any $x \in \mathbb{R}$ we then have $$ \lim_{n \rightarrow \infty}f_n(x) = \lim_{n \rightarrow \infty}\frac{x^2}{n} = x^2\lim_{n \rightarrow \infty}\frac{1}{n}=x^2 \cdot0 = 0$$
So $(f_n)_{n\in\mathbb{N}}$ converges pointwise to the null function $f(x)=0$