Functions Convergence

Convergence

Def'n. Pointwise convergence

Given $\{h_k\}\subset C(S,\mathbb R)$. We say that they converge pointwise if $\forall x\in S. h_k(x)\rightarrow h(x)$.

\[\forall \epsilon > 0. \forall x \in S. \exists N > 0. \forall k \geq N. |f_k(x)-f(x)|<\epsilon\]

Def'n. Convergence Uniformly

$f_k\rightarrow^{u.c.}f$ if $\|f_k - f\|_{\infty}\rightarrow 0$.

\[\forall \epsilon > 0. \exists N > 0. \forall k\geq N. \sup_{x\in S}|f_k(x)-f(x)|\leq \epsilon\]

Example

consider $f_k(x):= \begin{cases} k^2 x & 0\leq x\leq 1/k \\ k^2 x (\frac{2}{k}-x) & 1/2 \leq x \leq 2/k \\ 0 &2/k \leq x \leq 1 \end{cases}$

Claim $f_k \rightarrow^{p.w.} 0$

proof. Let $x\in[0,1]$. take $N > 0$ s.t. $2/N \leq x$. Therefore, $\forall k \geq N, |f_k(x)-0|<\epsilon$

Claim $f_k\not\rightarrow^{u.c.} 0$

proof. $\sup_{x\in[0,1]}|f_k(x)-0|=k\rightarrow\infty$

Claim $\lim_{k\rightarrow\infty}\int_0^1 f_k(x)dx \neq \int_0^1 \lim_{k\rightarrow\infty}f_k(x)dx$

proof. $LHS=\lim_{k\rightarrow\infty}\int_0^{1/k} k^2 x dx + \int_{1/k}^{2/k} k^2(\frac{2}{k}-x)dx + 0 = \lim_{k\rightarrow\infty}1=1$
$RHS = 0 \neq 1$

Example

Let $f_n:[0,A]\rightarrow \mathbb R, f(x):= \frac{\sin(nx)}{n}\rightarrow^{u.c.}0$.

proof. $\sup_{x\in[0,A]}|\frac{\sin(nx)}{x} - 0| \leq 1/n \rightarrow 0$

Thrm 1.

Claim u.c. does not preserve derivatives

proof. $f'(x)=\frac{n\cos(nx)}{n} = cos(nx)$ doesn't converge to any function.

Thrm 2. Dini's Theorem

If $f,f_n\in C([a,b]). (\forall n \geq 1. f_n \leq f_{n+1} \land f_n(x)\rightarrow f(x))\Rightarrow f_n\rightarrow^{u.c.} f$

proof. Since $f_n \leq f_{n+1}\land f_n \rightarrow f \Rightarrow g(x):=f(x)-f_n(x)\geq 0$

(i) By continuity at $x_0$, take $\delta_1 > 0, \forall y\in\mathbb R. |x_0 - y|\Rightarrow |g(x_0) - g(y)|\leq \epsilon/2$

(ii) By $f_n(x_0)\rightarrow^{p.w.}f(x_0)$, take $N>0, \forall n > N. |f_n(x_0)-f(x_0)|\leq \epsilon/2$

\[\begin{align*} |g(y)| &= |f(y)-f_n(y)| \\ &\leq |f(x_0)-f_n(x_0)| + |g(x_0) - g(y)| \\ &\leq \epsilon /2 + \epsilon /2 = \epsilon. &\forall y\in (x_0-\delta, x_0 + \delta) \end{align*}\]

Then, suppose that we don't have u.c.
$\Rightarrow \sup_{x\in[a,b]}|g(x)|=\sup_{x\in[a,b]} |f(x)-f_n(x)|\geq d, d > 0$.
$\|g\|_{\infty}>d\Rightarrow \forall n > 1. \exists x_n \in [a,b]. g_n(x_n)\geq d$
By $\{x_n\}\in[a,b]$, with bolzano weierstrass theorem, $x_{n_k}\rightarrow x_0 \in [a,b]$.
But we show for $\epsilon < d. \exists \delta > 0. |x_0 - y|<\delta\Rightarrow |g(y)|< \epsilon < d$.
For $\delta, \exists N>0$ so that $\forall k\geq N.|x_{n_k}-x_0|\leq \delta$. Take $y = x_{n_k}\Rightarrow |g(x_{n_k})|<\epsilon < d$ causes contradiction

Uniform convergence and completeness

Thrm 3.

If $f_k\in C([a,b]),f_k \rightarrow^{u.c.}f$, then $f\in C([a,b])$

proof. Let $\epsilon > 0$, and $x_0\in [a,b]$.
By uniform continuous, take $N_{\epsilon/3} > 0, \forall k > N_{\epsilon/3}. \sup_{x\in[a,b]}|f_k(k)-f(x)|\leq \epsilon/3$
By continuity of $f_k$, take $\delta_{\epsilon/3} > 0, |x_0 - y|\leq \delta \Rightarrow |f_{N_{\epsilon/3} }(x_0) - f_{N_{\epsilon/3} }(y)|\leq \epsilon/3$
Take $\delta = \delta_{\epsilon/3}$,

\[\begin{align*} |f(x_0) - f(y)| &\leq |f(x_0) - f_{N_{\epsilon/3} }(x_0)| &\text{by uniform continuous} \\ &\quad+ |f_{N_{\epsilon/3} }(x_0) - f_{N_{\epsilon/3} }(y)|&\text{by continuity} \\ &\quad+ |f_{N_{\epsilon/3} }(y) - f(y)|&\text{by uniform continuous} \\ &= \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon \end{align*}\]

Example

$f_n(x) = (1 + \frac{x}{n})^n\rightarrow^{u.c.} e^x := \sum^\infty \frac{x^k}{k!}$ over any interval $[a,b]$

proof.

\[\begin{align*} \sup_{x\in [a,b]} |(1+\frac{x}{n})^n - e^x| &=\|\sum_{k=1}^n {n\choose k}\frac{x^k}{n^k} - \sum\frac{x^k}{k!}\|_\infty \\ &\leq |\sum^\infty \frac{x^k}{k!}| &{n\choose k}/n^k \leq \frac{n}{k!}\\ &\leq |\sum \frac{b^k}{k!}| &x\in [a,b] \end{align*}\]

By big-oh, Take $N_\epsilon > 0, \forall n \geq N. \sum \frac{b^k}{k!} \leq \epsilon$

Thrm 4.

Claim $\lim_{n\rightarrow\infty}\int_0^1 f_n(x)dx = \int_0^1 \lim_{n\rightarrow\infty}f_n(x)dx = \int_0^1 e^x dx = e-1$

proof.

\[\begin{align*} \int_0^1 (1 + \frac{x}{n})^n dx &= \big(\frac{ {n+x}^{n+1} }{(n+1)n^n}\big)\big|_0^1 \\ &= \frac{(n+1)^{n+1} }{(n+1)n^n} - \frac{n^{n+1} }{(n+1)n^n}\\ &= (1 + \frac{1}{n})^n - \frac{n}{n+1}\\ \lim_{n\rightarrow \infty} (1 + \frac{1}{n})^n - \frac{n}{n+1} &= e - 1\\ \end{align*}\]

Example

Point-wise convergence does not imply uniform continuous

proof. Take $f_n:[0,1]\rightarrow \mathbb{R} : = x^n$.
$f_n(x_0) \rightarrow^{p.w.} \mathbb I (x_0 = 1)$
But $\sup_{x\in[0,1]}|x_n - \mathbb I(x = 1)| = \sup_{x\in(0,1)}|x^n|$.
Therefore, for $x_0 = 0.5^{1/n}, f_n(x_0) = 1/2 > 0$.
You will lose u.c. when you arbitrarily close to 1.

Example

$\forall f,g\in C(K). K$ compact, if $f_n \rightarrow^{u.c.} f, g_n \rightarrow^{u.c.} g$, then $f_n g_n \rightarrow^{u.c.} f g$

proof. Let $\epsilon > 0$, for all $x\in K$

\[\begin{align*} |f_n(x)g_n(x) - f(x)g(x)| &= |f_n(x)g_n(x) - g(x)f_n(x) + g(x)f_n(x) - f(x)g(x)| \\ &\leq |f_n(x)||g_n(x)-g(x)| + |g(x)||f_n(x)-f(x)| \\ &< B_1 |g_n(x)-g(x)| + B_2 |f_n(x)-f(x)|&\text{By EVT on $f_n, g_n$, since $K$ compact} \\ &\leq B_1 \|g_n - g\|_\infty + B_2 \|f_n-f\|_{\infty}&\text{By u.c. of $f,g$} \\ \text{take } N_1&\implies \forall n \geq N_1 . \|g_n-g\|_\infty \leq |\frac{\epsilon}{2B_1}|, N_2\\&\implies \forall n \geq N_2 . \|f_n-f\|_\infty \leq |\frac{\epsilon}{2B_2}| \\ &\leq \epsilon \end{align*}\]

Integral Convergence Theorem

If $f_n \in C([a,b])\land f_n \rightarrow^{u.c.}f$,
then $\forall [c,d]\subseteq [a,b]. \lim_{n\rightarrow\infty}\int_c^d f_n(t)dt = \int_c^d \lim_{n\rightarrow\infty}f_n(t)dt$

proof.

\[\begin{align*} \Big|\int_c^d f_n(t) - f(t)dt\Big| &\leq \int_c^d |f_n(t)-f(t)|dt \\ &\leq \int_c^d \sup_{s\in[a,b]}|f_n(s)-f(s)|dt \\ &= \sup_{s\in[a,b]} |f_n(s)-f(s)| \int_c^d d_t \\ &= \|f_n-f\|_\infty(d-c) \\ &\text{by u.c., take }\tilde\epsilon = \frac{\epsilon}{b-a} s.t. \exists N_{\tilde \epsilon} \geq 0. \forall n\geq N_{\tilde\epsilon}, \|f_n-f\|_\infty \leq \tilde\epsilon \\ &\leq \frac{d-c}{a-b}\epsilon \end{align*}\]

Note that if $[a,b]=[0,\infty)$, then the proof breaks down.
because $\int_c^d dt$ can be arbitrarily large, so one fix is to have a density $\rho(x)$ so that $\int_c^d \rho(s)ds\leq B$

Thrm 5. Leibriz's Rule (differentiation under the integral)

$\forall f(x,t), d_x f(x,t) \in C([a,b]\times[c,d])$. If $f(x) = \int_c^d f(x,t)dt$, then $\frac{dF}{dx} = \int_c^d d_x f(x,t)dt$

proof. Fix $x_0\in[a,b]$,

\[F'(x_0) = \lim_{h\rightarrow 0}\frac{F(x_0 + h) - F(x_0)}{h}\]

Consider the RHS,

\[\begin{align*} \lim_{h\rightarrow 0}\frac{F(x_0 + h) - F(x_0)}{h} &= \frac{1}{h}\int_c^d f(x_0+h,t)-f(x_0, t)dt \\ \text{by MVT, take }\xi(t,h)\in [x_0, x_0+h], &\frac{F(x_0 + h) - F(x_0)}{h} = d_x f(\xi(t,h), t) \\ &= \int_c^d d_xf(\xi(t,h),t)dt \end{align*}\]

WTS given $\epsilon$, can find $h$ small enough so that $|\int_c^d d_xf(\xi(t,h), t) - d_xf(x_0, t)dt|\leq \epsilon$
Using continuity of $d_xf(x_0, t)$, take $\delta > 0. |x_0 - y|<\delta \Rightarrow |d_xf(y,t) - d_xf(x_0,t)|\leq \frac{\epsilon}{d-c}$

Since $|\xi(t,h)-x_0|<h$, take $\delta < h$, then $|d_xf(\xi(t,h), t) - d_xf(x_0, t)|\leq \frac{\epsilon}{d-c}$,

\[\begin{align*} |\int_c^d d_xf(\xi(t,h), t)-d_xf(x_0, t)dt| &\leq \int_c^d |d_xf(\xi(t,h), t)-d_xf(x_0, t)|dt \\ &\leq \frac{\epsilon}{d-c}\int_c^d dt = \epsilon \end{align*}\]

Therefore, $\epsilon>0, \exists h$ small enough so that

\[|\frac{F(x_0 + h, t) - F(x_0)}{h} - \int_c^d d_xf(x_0,t)dt|\leq \epsilon\]

Therefore,

\[F'(x_0)=\lim_{h\rightarrow 0} \frac{F(x_0+h)-F(x)}{h} = \int_c^d d_xf(x_0, t)dt\]