MATH 414 Lecture 3

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Distance

Difference in length in a vector space:

$\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(\overrightarrow{u},\overrightarrow{v})=\Vert \overrightarrow{u}-\overrightarrow{v}\Vert$

In signal space $L^2$, recall

$\Vert f\Vert =\sqrt{\underset{a}{\overset{b}{\int }}{|f(t)|}^2\mathrm{d}t}$

Hence $\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(f,g)=\sqrt{\underset{a}{\overset{b}{\int }}{|f(t)-g(t)|}^2\mathrm{d}t}$

Convergence

Approximation

If we have a sequence of functions ${\{f_n(t)\}}_{n=1}^{\mathrm{\infty }}$

1. Naïve convergence: Fix $t$. then $\underset{n\to \mathrm{\infty }}{\lim }{f}_n(t)=f(t)$
2. Uniform convergence: (Visual convergence): A sequence $f_n$ converges uniformly to $f$ on $[a,b]$ if and only if for all $ϵ>0$ there exists a $N$ depending only on $ϵ$ such that $|f_n(t)-f(t)|<ϵ$ for all $n\ge N$ and $t\in [a,b]$.
3. Convergence in the mean or in $L^2$:

Example

$f_n(t)=t^n$ $0\le t<1$.

Fix $t$. Let $n\to \mathrm{\infty }$ and $f_n(t)$ converges to $0$.

Naïve convergence

observe $t^n={\mathrm{e}}^{-n\log \left(\frac{1}{t}\right)}$. For $t^N=ϵ$, we have $N=\frac{\log ϵ}{\log t}$.

for $n\ge N$, $t^n<t^N<ϵ$

Uniform Convergence

$t^n$ does not converge uniformly since for any $ϵ<1$ and $N$, $t^n\to 1$ as $t\to 1$

$L^2$

$\underset{a}{\overset{b}{\int }}{\left|t^n\right|}^2\mathrm{d}t=\underset{a}{\overset{b}{\int }}{t}^{2n}\mathrm{d}t={\frac{1}{2n+1}{t}^{2n+1}|}_a^b=\frac{1}{2n+1}$, which converges to $0$ as $n\to \mathrm{\infty }$. Therefore $f$ converges in $L^2$.

Note: Uniform convergence implies both Naïve convergence and $L^2$ convergence, but neither implies uniform, and they do not imply each other.

Orthogonality and Subspaces

We say that $\overrightarrow{u}$ and $\overrightarrow{v}$ are orthogonal if and only if $⟨\overrightarrow{u},\overrightarrow{v}⟩=0$.

We say that $U\subset V$ is a subspace (not empty, but $\overrightarrow{0}\in U$), then $\overrightarrow{v}$ is orthogonal to $U$ when $\overrightarrow{v}\perp \overrightarrow{u}$ for all $\overrightarrow{u}\in U$.

We say that $\{u_1,u_2,\dots ,u_n\}$ is an orthonormal set iff $⟨{\overrightarrow{u}}_j,{\overrightarrow{u}}_k⟩={\delta }_{j,k}$, where ${\delta }_{j,k}$ is the $j,k$-th entry in the identity matrix; that is,

${\delta }_{j,k}=\{\begin{array}{ll}1& j=k\\ 0& \text{otherwise}\end{array}$

1. An orthonormal set is always linearly independent
2. If $U=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\{{\overrightarrow{u}}_1,\dots ,{\overrightarrow{u}}_n\}\subseteq V$, then $\overrightarrow{u}\in U$ can be written as $\overrightarrow{u}=\sum _{j=1}^n⟨\overrightarrow{u},{\overrightarrow{u}}_j⟩{\overrightarrow{u}}_j$ (think $\hat{ı}$, $\hat{ȷ}$, and Failed to parse (unknown function "\math"): {\displaystyle \math{k}} in ${\mathbb{ℝ}}^3$)
3. Orthonormal projection onto $U$ of a vector $\overrightarrow{v}\in V$ is defined as $\overrightarrow{p}={\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{j}}_U\left(\overrightarrow{v}\right)=\sum _{j=1}^n⟨\overrightarrow{v},{\overrightarrow{u}}_j⟩{\overrightarrow{u}}_j$, where $\overrightarrow{v}-\overrightarrow{p}\perp U=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\{{\overrightarrow{u}}_1,\dots ,{\overrightarrow{u}}_n\}$.