# Prove $\exists$ a sequence of polynomials which converges to $f$ on any compact subset of $\mathbb{R}$. [duplicate]

If $f$ is continuous function ($f:\mathbb{R}\rightarrow\mathbb{R}$), then there exists a sequence of polynomials which converges to $f$ on any compact subset of $\mathbb{R}$.

Question: Since $K$ is compact, there exists some interval $[a,b]$, such that $K$ is contained in. By Weierstrass, for all $\epsilon>0$, there exits a sequence of polynomials of $P_n$, such that $P_n$ converges to $f$ uniformly, i.e. $\exists N\in\mathbb{N}$ $\forall n\geq N$, $|P_n(x)-f(x)|<\epsilon$, for all $x\in [a,b]$, therefore, certainly in $K$. So we are done. Correct?

## marked as duplicate by José Carlos Santos, Nosrati, mechanodroid, Namaste, mlcSep 27 '17 at 20:41

• @RobertZ my question is on my proof. I'm not looking for A answer. – 2ndaccount Sep 26 '17 at 16:48
• Your proof is incorrect, because your polynomials depend on $K$. – Martin R Sep 26 '17 at 16:50
• You are not correct. You need a sequence of polynomials which converges on ANY compact. – Robert Z Sep 26 '17 at 16:50
• @RobertZ But K is given right? – 2ndaccount Sep 26 '17 at 16:52
• @RobertZ I see. My bad. K is arbitrary. Thank you. – 2ndaccount Sep 26 '17 at 16:53

We can find continuous real function $f_{0}$ such that $f_{0}$ not differentiable everywhere!
But as $f_{0}$ be the limit of a real polynomials at such interval $[a,b]$ then $f_{0}$ will be expanded by Taylor expansion, so $f_{0}$ is infinitely differentiable over $[a,b]$. Which contradict with the first property of $f_{0}$ which is: $f_{0}$ not differentiable everywhere.

Now let us tray to set a continuous real function $f_{0}$, such that $f_{0}$ not differentiable everywhere:

let us define $g$ as following: $$g(x)=x-E(x) \text{ if } E(x) \text{ is even}\\ g(x)=E(x)-x+1 \text{ if } E(x) \text{ is odd}\\ \text{where } E \text{ is the integer part function}$$

We can easily show that $g$ is continuous, periodic with period 2, and bounded, moreover norm of $g$ less than $1$.

Take the following function sequence: $$g_{n}(x) =g(2^{n}x+1/2)/2^{n} \text{ for all } x\text{ in } \mathbb{R} \\ \text{for any positive integer } n$$ We can easily show that $g_{n}$ is continuous, periodic with period 2^{-n+1}, and bounded, moreover norm of $g_{n}$ less than $1/2^{n}$, for every positive integer $n$.

Take the following function series: $$S_{n}(x) =\sum_{i=0}^{i=n}g_{i}(x) \text{ for all } x\text{ in } \mathbb{R} \\ \text{for any positive integer } n$$ We can easily show that $S_{n}$ is continuous for every positive integer $n$.

Therefore this series is converges by norm to a real continuous function S.

Now we just need to show that S is not differentiable every where.

Let us find $B$ the set of all points where S is not differentiable Using the following steps:

$\mathbb{Z}$ is the set of all points where $g$ is not differentiable,see the definition of $g$.

let $A_{n}$ be the set of all points where $g_{n}$ is not differentiable. $g_{n}$ is not differentiable at the points $x$ such that $2^{n}x+1/2$ in $\mathbb{Z}$, which means $A_{n}=2^{-n}\mathbb{Z}-2^{-n-1}$. The must important think is $A_{i}\cap A_{j}=\phi$ for each two different positive integers $i,j$.

let $B_{n}$ be the set of all points where $S_{n}$ is not differentiable. $$\text{let } x\in \bigcup_{u=0}^{u=n}A_{u}\\ x\in \bigcup_{u=0}^{u=n}A_{u} \Rightarrow x\in A_{u} \text{ for some } 0\leq u\leq n\\ \Rightarrow g_{u}\text{ is not differentiable at } x \text{ for some } 0\leq u\leq n$$ since $A_{i}\cap A_{j}=\phi$ for each two different positive integers $i,j$: so we have $x$ doesn't belong to every $A_{v}$ where $0\leq v\leq n$ and $v\neq u$, that means $g_{v}$ differentiable at $x$ for every $0\leq v\leq n$ and $v\neq u$. Which gives: $$\sum_{i=0}^{i=n}g_{i} \text{ is not differentiable at } x \text{, hence } S_{n} \text{ is not differentiable at } x.\\ S_{n} \text{ is not differentiable at } x \Rightarrow x\in B_{n}$$ therefore: $B_{n}=\bigcup_{u=0}^{u=n}A_{u}$ ($\bigcup_{u=0}^{u=n}A_{u}\subseteq B_{n}$ has been proved; $B_{n}\subseteq \bigcup_{u=0}^{u=n}A_{u}$ is trivial inclusion). $$\text{Then } B=\bigcup_{u=0}^{u=\infty}A_{u}.$$

In order to show that S is not differentiable every where we will show that: $$(\forall \epsilon>0)(\forall a \in \mathbb{R})(S \text{ is not diff. over } ]a-\epsilon,a+\epsilon[)$$

Let us choose $a \in \mathbb{R}$. for each fixed $\epsilon>0$ the following is true:

We will choose $n_{0} \in \mathbb{N}^{*}$ such that $1/2^{n_{0}}<\epsilon$. Suppose $g_{n_{0}}$ is differentiable over $]a-\epsilon,a+\epsilon[$, then $g_{n_{0}}$ is differentiable over $[a-\epsilon/2,a+\epsilon/2]$, then $g_{n_{0}}$ is differentiable over $[a-\epsilon/2,a+\epsilon/2]+m/2^{n_{0}}$ for all integer m (since $1/2^{n_{0}}$ is the period of $g_{n_{0}}$), then $g_{n_{0}}$ is differentiable over $\bigcup_{m=0}^{m=\infty} ([a-\epsilon/2,a+\epsilon/2]+m/2^{n_{0}})$, then $g_{n_{0}}$ is differentiable over $\mathbb{R}$ which isn't true. Therefore $g_{n_{0}}$ isn't differentiable over $]a-\epsilon,a+\epsilon[$.

but: $g_{n_{0}}$ is not differentiable over $]a-\epsilon,a+\epsilon[$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $g_{n_{0}}$ is not differentiable at $b$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $b \in A_{n_{0}}$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $b \in B$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $S$ is not differentiable at $b$, equivalent to say $S$ is not differentiable over $]a-\epsilon,a+\epsilon[$.

Finally $S$ isn't differentiable every where, $S$ is an example for $f_{0}$ continuous function not differentiable every where.