Sequences & Series of Functions
Sequences & Series of Functions
A sequence of functions $(f_n)$ may converge pointwise ($f_n(x)\to f(x)$ for each $x$) or uniformly (the convergence is simultaneous, uniform in $x$). Uniform convergence preserves continuity, integrability, and (with additional hypotheses) differentiability. The Weierstrass M-test provides a practical sufficient condition for uniform convergence of series.
Uniform vs Pointwise Convergence
$f_n\to f$ pointwise on $D$: $\forall x\in D,\forall\epsilon>0,\exists N(x,\epsilon)$ with $n>N\Rightarrow|f_n(x)-f(x)|<\epsilon$. Uniform convergence: $\forall\epsilon>0,\exists N(\epsilon)$ (independent of $x$) with $n>N\Rightarrow|f_n(x)-f(x)|<\epsilon$ for all $x\in D$. Equivalently, $\|f_n-f\|_\infty=\sup_{x\in D}|f_n(x)-f(x)|\to 0$. Key theorem: if $f_n\to f$ uniformly on $[a,b]$ and each $f_n$ is continuous, then $f$ is continuous and $\int_a^b f_n\to\int_a^b f$ (can interchange limit and integral).
Weierstrass M-test
If $\sum M_n<\infty$ and $|f_n(x)|\leq M_n$ for all $x\in D$ and all $n$, then $\sum f_n$ converges uniformly and absolutely on $D$. Proof: $\sum|f_n(x)|\leq\sum M_n<\infty$ gives absolute convergence; for uniform convergence, the tail $|\sum_{n>N}f_n(x)|\leq\sum_{n>N}M_n\to 0$ independently of $x$.
Example 1
Show $f_n(x)=x^n$ converges pointwise on $[0,1]$ but not uniformly.
Solution: Pointwise: $f_n(0)=0\to 0$; $f_n(1)=1\to 1$; for $x\in(0,1)$, $x^n\to 0$ since $x<1$. So $f(x)=0$ for $x\in[0,1)$ and $f(1)=1$. Not uniform: $\|f_n-f\|_\infty=\sup_{x\in[0,1]}|x^n-f(x)|=\sup_{x\in[0,1)}x^n=1$ for all $n$ (not $\to 0$). Alternatively: each $f_n$ is continuous but the limit $f$ is discontinuous at 1 — uniform convergence would preserve continuity.
Example 2
Use the Weierstrass M-test to show $\sum_{n=1}^\infty\frac{\cos(nx)}{n^2}$ converges uniformly on $\mathbb{R}$.
Solution: Set $M_n=1/n^2$. Since $|\cos(nx)/n^2|\leq 1/n^2$ and $\sum 1/n^2=\pi^2/6<\infty$, the M-test gives uniform convergence. The limit function $f(x)=\sum\cos(nx)/n^2$ is continuous everywhere (uniform limit of continuous functions).
Practice
- Prove that if $f_n\to f$ uniformly and each $f_n$ is integrable on $[a,b]$, then $f$ is integrable and $\int f_n\to\int f$.
- Show that pointwise convergence plus bounded convergence does not imply $\int f_n\to\int f$ (give an example).
- Prove the differentiation theorem: if $f_n'\to g$ uniformly and $f_n(x_0)\to L$ for some $x_0$, then $f_n\to f$ uniformly and $f'=g$.
- Show that the power series $\sum a_n x^n$ converges uniformly on $[-r,r]$ for any $r
Show Answer Key
1. Given $\epsilon>0$, $\exists N$: $\|f_n-f\|_\infty<\epsilon/(b-a+1)$ for $n\ge N$. Then $|\int f_n-\int f|\le\int|f_n-f|\le(b-a)\|f_n-f\|_\infty<\epsilon$. Integrability of $f$ follows since $f$ is the uniform limit of measurable functions.
2. Let $f_n=n\cdot\mathbf{1}_{[0,1/n]}$ on $[0,1]$. Then $f_n\to0$ pointwise, $|f_n|\le n$ (unbounded), and $\int f_n=1\neq0=\int0$.
3. Let $F_n(x)=f_n(x_0)+\int_{x_0}^x f_n'$. Since $f_n'\to g$ uniformly, $\int f_n'\to\int g$ uniformly (by the integration result above). Since $f_n(x_0)\to L$, $F_n\to f$ uniformly where $f(x)=L+\int_{x_0}^x g$. Then $f'=g$ by FTC.
4. For $|x|\le r