Uniform Convergence of Power Series

Question

Let $\sum_{n=0}^\infty a_n \left( x-x_0 \right)^n$ be a power series that converges uniformly over all $x\in \mathbb R$. Prove there exists $N\in\mathbb N$ such that for all $n>N,\; a_n = 0$.

I fail to see how this is true. From the radius of convergence formula $$\frac{1}{R}=0=\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_n\right|}$$ Now for all $n$ choose $a_n=\frac{1}{n^n}\neq0$ and we get $a_n>a_{n+1}$ so that $$\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_n\right|}=\lim_{n\rightarrow\infty}\frac{1}{n}=0$$ We therefore got a convergent series over all the reals, thus uniformly convergent in every $[a,b]\in \mathbb R$ contradicting the above statement...

Answer 1

Suppose that a sequence $f_n(x)$ converges pointwise to the function $f(x)$ for all $x\in S$.

NEGATION OF UNIFORM CONVERGENCE

The sequence $f_n(x)$ fails to converge uniformly to $f(x)$ for $x\in S$ if there exists a number $\epsilon>0$ such that for all $N$, there exists an $n_0>N$ and a number $x\in S$ such that $|f_{n_0}(x)-f(x)|\ge \epsilon$.

Now, let $f_n(x)=a_nx^n$ with $\lim_{n\to \infty}f_n(x)=0$ for all $x\in \mathbb{R}$. Certainly, either $a_n=0$ for all $n$ sufficiently large or for any number $N$ there exists a number $n_0>N$ such that $a_{n_0}\ne 0$.

Suppose that the latter case holds. Now, taking $\epsilon=1$, we find that

$$|a_{n_0}x^{n_0}|\ge \epsilon$$

whenever $|x|\ge |a_{n_0}|^{-1/n_0}$. And this negates the uniform convergence of $f_n(x)$.

And inasmuch as the sequence $f_n(x)$ fails to uniformly converge to zero, then the series $\sum_{n=0}^\infty f_n(x)$ fails to uniformly converge.

Note for the example for which $a_n=n^n$ we can take $x>1/n$.

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NOTE:

If $a_n=0$ for all $n>N+1$, then we have

$$\sum_{n=0}^\infty a_nx^n = \sum_{n=0}^N a_nx^n$$

which is a finite sum and there is no issue regarding convergence.

Answer 2

Let $f$ denote this power series and $(P_n)$ denote the sequence of partial sums. Uniform convergence on $\Bbb R$ implies that $(P_n)$ is uniformly Cauchy. I'll use this in two places. First, there must be some $N$ such that for all $n,m \ge N$,

$$\|P_n - P_m\|_{\infty} < 1$$

In particular, $\|P_n - P_N\|_{\infty} < 1$ for all $n \ge N$, and since non-constant polynomials are unbounded, $P_n - P_N$ must be constant in $x$, say $:=b_n$. Now,

$$|b_n - b_m| \le \|P_n - P_m\|_{\infty} \to 0 \text{ as $n,m \to \infty$}$$

Where again I used that $(P_n)$ is uniformly Cauchy. Hence $(b_n)$ is Cauchy and so convergent, say to $b$. Thus, $f = b + P_N$ is a polynomial.

Answer 3

It appears that you're under the impression that "uniformly convergent" means convergent everywhere in $\mathbb R$.

Consider the sequence of functions $x\mapsto x^n$ on the interval $0<x<1$.

For every value of $x$ in the interval you have $x^n\to0$ as $n\to\infty$. That is what it means to say this sequence of functions converges pointwise on the interval $0<x<1$.

But the uniform distance from $x\mapsto x^n$ to $x\mapsto0$ is $\displaystyle\sup_{0<x<1} |x^n - 0| = 1$, and that does not approach $0$ as $n\to\infty$. That is what it means to say this sequence of functions does not converge uniformly on the interval $0<x<1$.

Answer 4

Uniform convergence of the power series of $f$ is equivalent to the fact that the sequence $$ f_n(x)=\sum_{k=0}^na_kx^k, \quad n\in\mathbb N, $$ is uniformly convergent and equivalently uniformly Cauchy in $\mathbb R$, i.e., for every $\varepsilon>0$, there is an $N$, such that, $$m,n\ge N\quad\Longrightarrow\quad \sup_{x\in\mathbb R}\lvert\, f_m(x)-f_n(x)\rvert<\varepsilon.\tag{1}$$

Assume that the power series contains infinitely many terms; say $a_{k_n}\ne 0$, $n\in\mathbb N$, let $\varepsilon>0$ and $N$, such that $(1)$ holds. Then, for $k_n\ge N$ we have $$ \sup_{x\in\mathbb R}\lvert\, f_{k_n}(x)-f_{k_n-1}(x)\rvert=\sup_{x\in\mathbb R}\lvert a_{k_n}x^{k_n}\rvert=\infty. $$ Contradiction. Hence, the power series contains only finitely many terms.