Duality

Vector Space

If V is a vector space over the real numbers \bm{R} then its dual, V^*, is the set of all linear functionals from V to \bm{R}. 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\bm{R}, 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\bm{R} 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\bm{R} 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 Transformation

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.

This is an example of a natural transformation in category theory.

Integration

Given a function on a finite interval f\colon [a,b]\to\bm{R}, a,b\in\bm{R}, 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\bm{R} 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.