title: Duality
author: Keith A. Lewis
institution: KALX, LLC
email: [email protected]
classoption: fleqn
abstract: Mirror, mirror, ...
...
\newcommand\RR{\bm{R}}
\newcommand\CC{\bm{C}}
\newcommand\FF{\bm{F}}
\newcommand\ker{\operatorname{ker}}
\newcommand\ran{\operatorname{ran}}
Duality shows up in many mathematical guises. One of the simplest is in
the case of vector spaces. The dual of a vector space is the collection
of all linear functionals on the vector space. It too is a vector space
and has the same dimension as the original vector space if is finite dimensional.
Identifying when a space is a dual help clarifies mathematical concepts.
For example, if $F\colon V\to\RR$ is a smooth function then its derivative
is a function $DF\colon V\to V^$, where $V^$ denotes the dual of $V$.
There is no need for "row" or "column" vectors when using duals.
If $V$ is a vector space over the real numbers $\RR$ then its dual, $V^$,
is the set of all linear functionals from $V$ to $\RR$. The dual space
is also a vector space with scalar multiplication and vector addition
defined pointwise: $(tv^)(u) = t(v^(u))$ and $(v^ + w^)(u)
= v^(u) + w^(u)$, $t\in\RR$, $u\in V$, $v^,w^\in V^$. The dual
pairing is $\langle v,v^\rangle = v^(v)$, $v\in V$, $v^\in V^$.
Since $v^\colon V\to\RR$ is a function, if we know $\langle v,
v^\rangle$ for all $v\in V$ we know $v^*$.
Exercise. If $\langle v,v^\rangle = 0$ for $v^\in V^*$ show $v = 0$.
Exercise. If $\langle v,v^\rangle = \langle w,v^\rangle$ for $v^\in V^$ show $v = w$.
Hint: Use linearity and reduce to the first exercise.
If $V$ is finite dimensional with basis ${e_i}$ then define the dual basis
$e_j^\colon V\to\RR$ by $\langle e_i, e_j^\rangle = δ_{i,j}$, where
$δ_{i,j}$ is the Kronecker delta,
and extend to $V$ linearly to get $e_j^\in V^$.
This determines an isomorphism $T\colon V\to V^$ by $Te_i = e_i^$.
For any vector space $V$ define the canonical injection $ι_V\colon V\to V^{**}$ by
$\langle ι_Vv, v^\rangle = \langle v, v^\rangle$.
Exercise. Show $ι_V$ is a linear transformation.
Exercise. Show $ι_Vv = 0$ if and only if $v = 0$, $v\in V$.
The proof of this is trivial, unlike the proof that vector spaces of
the same dimension are isomorphic. That proof involves the Steinitz
replacement theorem and introducing an arbitrary basis for each vector
space.
Linear transformations on vector spaces $T\colon V\to W$ are also a vector
space with scalar multiplication and vector addition defined pointwise.
They have a notion of duality called adjoint. Define $T^\colon W^\to
V^$ by $\langle v, T^w^\rangle = \langle Tv, w^\rangle$, $v\in V$,
$w^\in W^$.
Exercise. If $S\colon U\to V$ and $T\colon V\to W$ show $(TS)^* = S^T^$.
Exercise. If $T\colon V\to W$ show $T^{**}ι_V = ι_WT$.
Solution
For $v\in V$, $w^*\in W^*$,
$\langle T^{**}ι_Vv, w^*\rangle
= \langle ι_Vv, T^*w^*\rangle
= \langle v, T^*w^*\rangle
= \langle Tv, w^*\rangle
= \langle ι_W Tv, w^*\rangle$.
This is an example of a natural transformation in category theory.
Given a function on a finite interval $f\colon [a,b]\to\RR$, $a,b\in\RR$,
and a partition of $[a,b]$, $Δ = (x_0,\ldots,x_n)$ with
$a = x_0 < x_1 < \cdots < x_n = b$ define
$\underline{R}{[a,b]}f = \sum{0\le i < n} \underline{f_i},Δx_i$
where $\underline{f_i} = \inf_{x_i \le x < x_{i+1}} f(x)$ and $Δx_i = x_{i + 1} - x_i$.
Likewise define $\overline{R}{[a,b]}f = \sum{0\le i < n} \overline{f_i},Δx_i$ where
$\overline{f_i} = \sup_{x_i \le x < x_{i+1}} f(x)$.
Define $Δ\preceq Δ'$ if every $x_i$ in $Δ$ is equal to some
$x_i'$ in $Δ'$. This ordering is a net so we can consider the limit
in both cases. Riemann showed that if $f$ is continuous on
$[a,b]$ the the lower and upper limits exist and are equal.
If $F\colon [a,b]\to\RR$ is any non-decreasing function a similar
argument goes through for sums of the for $\sum_{0\le i < n}
f(x_i^),(F(x_{i+1}) - F(x_i))$, where $x_i^\in [x_i, x_{i+1}]$,
which leads to Riemann-Stieljes integation. For example,
if $F(x) = 1_{[c,\infty)}$ is the Heavyside function it can be shown $\int_a^b f(x),dF(x) = f(a)$
when $a < c < b$.
Exercise. Prove this.
Hint: Only one term of $\sum f(x^*),ΔF(x)$ is non-zero.
There are also improper integrals that can be defined if $a = -\infty$
or $b = +\infty$ if limits exist. If $f$ tends to infinity at either endpoint then the
limits over $[a + ε,b]$ and $[a, b - ε]$ may exist as $ε\to 0$. If $f$
has a countable number of jumps it is also possible to fix things up.
Things become even more complicated when trying find conditions when limits can be exchanged:
$\lim_{n\to\infty}\int_a^b f_n(x),dx = \int_a^b \lim_{n\to\infty} f_n(x),dx$.
Lebesgue cleaned up this mess by noticing the continuous functons on the interval $[a,b]$,
$C[a,b]$, is a vector space and $f\mapsto \int_a^b f(x),dx$ is a linear functional.
This led him to the notion of set functions called measures that greatly
simplify the concept of integration.
Let $\Omega$ be any set and define the bounded functions
$B(\Omega) = {f\colon\Omega\to\RR:\sup_{\omega\in\Omega} |f(\omega)| < \infty}$.
Given a linear functional $L\in B(\Omega)^*$ and any subset $E\subseteq\Omega$ define
$λ(E) = L(1_E)$, where $1_E(ω) = 1$ if $ω\in E$ and $1_E(ω) = 0$ if $ω\not\in E$.
If $E$ and $F$ are disjoint subsets of $\Omega$ then $1_{E\cup F} = 1_E + 1_F$ so
$λ(E\cup F) = λ(E) + λ(F)$.
Exercise. Show $λ(E\cup F) = λ(E) + λ(F) - λ(E\cap F)$ for any $E,F\subseteq \Omega$.
While $f$ is a function on $\Omega$, $λ$ is a function on subsets of $\Omega$ that satifies
$λ(E\cup F) = λ(E) + λ(F) - λ(E\cap F)$ for any $E,F\subseteq \Omega$. Such functions
are called measures. Measures don't count things twice.
We also need $λ(\emptyset) = 0$, the measure of nothing is zero.
Exercise. If $λ(E\cup F) = λ(E) + λ(F) - λ(E\cap F)$, $E,F\subseteq\Omega$, then
$μ = λ + a$ also satisfies this condition.
Hint. The "measure" $μ + a$ is $(μ + a)(E) = μ(E) + a$.