# Weierstrass's approximation theorem with polynomials $p_n$ of degree $n$ for all $n.$

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?

• "closed interval" should be "closed bounded interval" – zhw. Jan 16 at 21:20
• 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. – zhw. Jan 17 at 19:27
• Very appreciate for you help , yours title is exactly what I want to ask . – J.Guo Jan 18 at 13:06
• I upvoted your question; never thought about it before. – zhw. Jan 18 at 18:20

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.