# Let $f_n(x) = (x-\frac{1}{n})^2$ for $x\in[0,1]$.

pointwise-convergencereal-analysis

Does the sequence $$\{f_n\}$$ converge pointwise in the set $$[0,1]$$?

And the question asks to give the limit function as well.

So far I have:

$$f(x)=\lim_{n\rightarrow\infty}f_n(x)$$
$$f(x)=\lim_{n\rightarrow\infty}(x^2-\frac{2x}{n}+\frac{1}{n^2})$$
$$f(x)=x^2 \;\forall x\in[0,1]$$
I am stuck from here

Your approach is correct and indeed $$f_{n}\to x^{2}$$ pointwise on $$[0,1]$$.

The definition of pointwise convergence of a sequence of functions $$(f_{n}(x))_{n\in \mathbb{N}}$$ on a set $$I\subseteq \mathbb{R}$$ says, $$f_{n}\to f$$ pointwise on $$I$$ if, and only if, $$(\color{red}{\forall x\in I})(\forall \varepsilon >0)(\color{red}{\exists N\in \mathbb{N}})(\forall n\in \mathbb{N})(n>N\implies |f_{n}(x)-f(x)|<\varepsilon.$$

Let the sequence of functions $$f_{n}(x)=\left(x-\frac{1}{n}\right)^{2}$$ and let's to show that $$f_{n}\to f$$ pointwise on $$[0,1]$$, with $$f(x)=x^{2}$$.

Let $$x\in [0,1]$$ and let $$\varepsilon >0$$. By the Archimedean property there exists a natural number $$N>\frac{\varepsilon}{3}$$. If $$n>N$$ we have \begin{align*} |f_{n}(x)-f(x)|&=\left|\left(x-\frac{1}{n}\right)^{2}-x^{2}\right|\\&=\left|x^{2}-\frac{2x}{n}+\frac{1}{n^{2}}-x^{2}\right|\\ &\leqslant \frac{|2x|}{n}+\frac{1}{n^{2}},\quad 0\leqslant x\leqslant 1,\\&\leqslant \frac{2}{n}+\frac{1}{n^{2}}\\&\leqslant \frac{3}{n},\\&<3\left(\frac{\varepsilon}{3}\right),\\&=\varepsilon. \end{align*} Therefore $$f_{n}\to f$$ pointwise on $$[0,1]$$.

Moreover $$f_{n}\rightrightarrows f$$ uniformly on $$[0,1]$$.

The definition of uniform convegence of a sequence of functions $$(f_{n}(x))_{n\in \mathbb{N}}$$ on a set $$I\subseteq \mathbb{R}$$ says, $$f_{n}\rightrightarrows f$$ uniformly on $$I$$ if, and only if, $$(\forall \varepsilon >0)(\color{red}{\exists N\in \mathbb{N}})(\forall n\in \mathbb{N})(\color{red}{\forall x\in I})(n>N\implies |f_{n}(x)-f(x)|<\varepsilon.$$

Let the sequence of functions $$f_{n}(x)=\left(x-\frac{1}{n}\right)^{2}$$ and let's to show that $$f_{n}\rightrightarrows f$$ uniformly on $$[0,1]$$, with $$f(x)=x^{2}$$.

Let $$\varepsilon >0$$, thus choose $$N\in\mathbb{N}$$ such that $$\frac{1}{N^{2}}<\frac{\varepsilon}{2}$$ and $$\frac{2x}{N}<\frac{\varepsilon}{2}$$. Thus for all $$n>N$$ we have \begin{align*} |f_{n}(x)-f(x)|&=\left|\left(x-\frac{1}{n}\right)^{2}-x^{2}\right|,\\&=\left|-\frac{2x}{n}+\frac{1}{n^{2}}\right|,\\&\leqslant \frac{2x}{n}+\frac{1}{n^{2}},\\&<\frac{2x}{N}+\frac{1}{N^{2}},\\&=\frac{\varepsilon}{2}+\frac{\varepsilon}{2},\\&=\varepsilon. \end{align*}

Therefore $$f_{n}\rightrightarrows f$$ uniformly on $$[0,1]$$.

Alternatively notice that $$\sup_{x\in [0,1]}\left|f_{n}(x)-f(x)\right|\to 0,\quad n\to +\infty$$ Therefore $$f_{n}\rightrightarrows f$$ uniformly on $$[0,1]$$. Then also $$(f_{n}(x))_{n\in \mathbb{N}}$$ is uniform Cauchy.