1
$\begingroup$

By Weierstrass's approximation theorem, every continuous function $f$ supported on an compact interval can be uniform approximated by polynomials. But, is it true that for every continuous $f$ on $[0,1]$, there exist a sequence of polynomials $\{f_n\}_{n=0}^{\infty}$ with $f_N=\sum_{n=0}^N a_n x^n$ ($a_N \neq 0$) such that $f_n$ converges to $f$?

My attempt:

For function $f=0$ , we can let $g_n=\frac{e^x}{n}$ and let $$f_N=\frac{\sum_0^N \frac{1}{n!}x^n}{N}$$ then we can find $f_n$ converges to $0$ uniformly. For each continuous function $g$ supported on $[0,1]$, if we can find a sequence of polynomials $\{h_n \}_{n=0}^{\infty}$ such that $h_n$ converges to $g$ uniformly and the degree of $g_n$ is less than $n$, then with $f_N$ defined above, let $g_n=h_n+f_n$. We can prove that $g_n$ converges to $g$ uniformly.

So, to prove the assertion above, it suffice to prove that following statement :
For every continuous function $f$ supported on $[0,1]$, we can find a sequence of polynomials $\{f_n \}_{n=0}^{\infty}$ with the degree of $f_n$ less than $n+1$, such that $f_n$ converges to $f$.

Let $\{g_n \}$ be polynomials converges to $f$ uniformly, and let $k(n)$ denote the degree of $g_n$. Then we can contrust $f_n$ such that $f_0=f_1=...=f_{k(0)-1}=0$, $f_{k(0)}=g_0$. For $n={1,2,3,...} $ , if $k(n)\le n$ then $f_{n}=g_{n}$. If $k(n) \gt n$ , let $f_n=f_{n+1}=...=f_{k(n)-1}=f_{n-1}$ and $f_{k(n)}=g_n$. Since $f_n$ conveges to $f$ uniformly, the proof is complete. Is my proof correct?

$\endgroup$
  • $\begingroup$ "closed interval" should be "closed bounded interval" $\endgroup$ – zhw. Jan 16 at 21:20
  • $\begingroup$ I changed your title to more accurately describe the result. If you don't like the change, feel free to go back to the originaL. $\endgroup$ – zhw. Jan 17 at 19:27
  • $\begingroup$ Very appreciate for you help , yours title is exactly what I want to ask . $\endgroup$ – J.Guo Jan 18 at 13:06
  • $\begingroup$ I upvoted your question; never thought about it before. $\endgroup$ – zhw. Jan 18 at 18:20
0
$\begingroup$

I think you have the right ideas, but I'm unsure about some of what you're doing. It may be just fine, but the notation is a little scrambled.

Here's what I did. Suppose f is continuous on $[a,b].$ Then by Weierstrass, there is a sequence of polynomials $p_n\to f$ uniformly on $[a,b].$ Suppose the degrees of the $p_n$ are unbounded. Then we can choose a subsequence $q_n$ of $p_n$ such that the degrees of the $q_n$ strictly increase to $\infty.$ Clearly $q_n\to f$ uniformly on $[a,b].$ Now add terms of the form $x^k/k!,$ as you were doing, to get a sequence of the desired form. Suppose for example $\text { deg } q_1=4, \text { deg } q_2=8,$ $ \text { deg } q_3=9, \text { deg } q_4=12,\dots$ Then we construct a new sequence as below:

$$1,x,x^2/2!,x^3/3!, q_1(x), q_1(x)+x^5/5!, q_1(x)+x^6/6!, q_1(x)+x^7/7!,$$ $$ q_2(x), q_3(x), q_3(x)+x^{10}/10!, q_3(x)+x^{11}/11!, q_4(x), \dots $$

This sequence does what you want.

If the degress of the $p_n$ are bounded, say by $d,$ then we can define

$$q_n(x) = p_n(x) + x^{d+n}/(d+n)!.$$

Then $\deg q_n = d+n$ and $q_n\to f$ uniformly on $[a,b].$ Since the degrees increase to $\infty,$ the above applies to $q_n$ and again we get the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.