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Duplicate: [[pointwise convergence]]
- Suppose $X$ is a set and $Y$ a [[topological space]].
> [!definition] Definition. ([[pointwise converge]])
> A [[sequence]] of functions $(f_{n}:X \to Y)$ is said to **converge pointwise** to a given function $f:X \to Y$ often written as $\lim_{ n \to \infty } f_{n} = f \text{ pointwise}$
> if $\lim_{ n \to \infty } f_{n}(x)=f(x) \ \ \forall x \in \dom f$.
>
>
From a [[topological space|topological perspective]], pointwise convergence is equivalent to [[sequence|convergence]] wrt the [[product topology]] on the set $Y^{X}$ of functions $X \to Y$.
> [!basicexample]
>
> Suppose $f_{k}:[-1,1]\to \mathbb{R}$ is the function whose graph is shown below (courtersy of Axler's MIRA)
> ![[CleanShot 2025-09-19 at
[email protected]]]
>
> and $f:[-1,1]\to \mathbb{R}$ is the function $f(x)=\begin{cases}
> 1 & x \neq 0 \\
> 2 & x = 0.
> \end{cases}$
> Then $f_{1},f_{2},\dots$ [[pointwise converge|converges pointwise]] on $[-1,1]$ to $f$.
>
>
>
>
> Note that this example shows that the pointwise limit of [[continuous]] functions need not be [[continuous]].
>
>
> >[!proof]- Proof.
> >
> > Explicitly, $f_{k}(x)=\begin{cases}
> > 1 & x \leq -\frac{1}{k} \\
> > k\left( x+\frac{1}{k} \right) + 1 & -\frac{1}{k} \leq x \leq 0 \\
> > -k\left( x - \frac{1}{k} \right) + 1 & 0 \leq x \leq \frac{1}{k} \\
> > 1 & x \geq \frac{1}{k},
> > \end{cases}$
> > though the exact definition matters less than just the fact $f_{k}(x)$ equals $1$ outside of $\left[ -\frac{1}{k}, \frac{1}{k} \right]$ and $f_{k}(0)=0$.
> >
> >
> > Now, fix $\varepsilon>0$ and $x \in [-1,1]$. If $x=0$ the result is immediate since $f_{1}(0),f_{2}(0),\dots$ is the constant sequence $(2)$. So suppose $x \neq 0$. In this case we let $N=\lfloor\frac{1}{|x|}+1\rfloor$ so that $f_{k}(x)=1$ for all $k \geq N$. Then for any $k \geq N$, we have $|f_{k}(x)-f(x)|=|1-1|=0<\varepsilon$.
^c6bab8
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#### References
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