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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?

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  • $\begingroup$ $f_n(x)=\frac1x-\frac1n$, for an example. $I=\mathbb R\setminus \{0\}$. $\endgroup$
    – zoli
    Feb 5, 2017 at 13:18
  • $\begingroup$ What is unclear to you? The definition is itself evident. See personal.psu.edu/auw4/M401-notes1.pdf for some examples. $\endgroup$
    – Crostul
    Feb 5, 2017 at 13:22
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    $\begingroup$ What the definition says is that for each fixed $x \in I$, the sequence $(f_n(x))_n$ converges to the value $f(x)$. Example: let $f_n(x) = \frac{x}{n}$. For a fixed $x \in \Bbb R$, the sequence $(f_n(x))$ converges to $0$ because $\lim_{n \to\infty} f_n(x) = \lim_{n \to \infty} \frac{x}{n} = x \lim_{n \to \infty} \frac1n = x \cdot 0 = 0$. So if we define $f: \Bbb R \to \mathbb R$ by $f: x \mapsto 0$, we get that $(f_n)$ converges to $f$ pointwise on $\mathbb R$. $\endgroup$
    – user384138
    Feb 5, 2017 at 13:23

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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 $f(x)=0$

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To converge pointwise in this context means that you can take any $x\in I$ and then look at the function $f_n$ evaluated at $x$. This gives you a value in $\mathbb{R}$ and we call it $f_n(x)$. Now if you look at this with $x$ fixed you look at a sequence of numbers! For every $n$ you have a different number in $\mathbb{R}$ and this sequence converges to the number $f(x)$. If the same holds for any $x \in I$, i.e. any sequence $f_n(x)$ (now with $x$ arbitrary, not fixed) converges to its respective limit $f(x)$, we say the the sequence of functions $f_n$ converges to the function $f$.

An example is provided by the function $f\colon (0,1) \rightarrow \mathbb{R}, x \mapsto x^n$, which converges pointwise to the zero function on $(0,1)$. For any $x\in (0,1)$, the sequence $x^n$ converges to $0$.

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