No matrix, and continuity is no longer free
In Vol I every linear transformation between finite-dimensional spaces was a matrix once you chose bases, and every such map was automatically continuous. Both facts evaporate in infinite dimensions: there is no finite matrix, and a linear map can be wildly discontinuous, sending a convergent input sequence to a divergent output.
An operator T is a bounded operator when there is a constant C with ||Tx|| <= C ||x|| for all x. "Bounded" here does not mean its values stay in a box — it means it does not stretch lengths by more than a fixed factor. The decisive theorem: for linear maps, bounded is equivalent to continuous. So bounded operators are exactly the linear maps that respect limits.
Worked examples: shift, multiplication, differentiation
Three operators on sequence/function spaces
1) RIGHT SHIFT on ell^2:
S(x_1, x_2, x_3, ...) = (0, x_1, x_2, ...)
||Sx|| = ||x|| exactly -> bounded, ||S|| = 1 (an isometry).
Note: S is injective but NOT surjective (nothing maps to e_1).
In finite dim, injective => surjective. Here it FAILS.
2) MULTIPLICATION on L^2[0,1]: (M_g f)(t) = g(t) f(t)
if |g(t)| <= K everywhere, ||M_g f|| <= K ||f||
-> bounded, ||M_g|| = sup |g| = ess-sup of g.
3) DIFFERENTIATION D f = f' on L^2:
take f_n(t) = sin(n t): ||f_n|| stays ~ constant,
but D f_n = n cos(n t): ||D f_n|| grows like n -> infinity.
No constant C bounds it -> D is UNBOUNDED (discontinuous).The shift operator already breaks a load-bearing Vol I intuition: in finite dimensions an injective operator is automatically surjective (the rank-nullity theorem guarantees it). The right shift is injective yet misses e_1 entirely — left versus right inverses come apart, and one-sided invertibility becomes a real phenomenon. This single example will return to haunt the spectrum in Guide 5.
Riesz representation and the adjoint
How do we recover the transpose without a matrix? Through the Riesz representation theorem: in a Hilbert space, *every* bounded linear functional f(x) is just an inner product f(x) = <x, v> against one fixed vector v, and ||f|| = ||v||. Functionals and vectors are the same data. This is the engine that turns abstract dual statements back into concrete geometry.
Riesz lets us define the adjoint T* of a bounded operator by the rule <Tx, y> = <x, T* y> for all x, y. For a matrix this is exactly the conjugate transpose, so T* generalizes A^T (or A^* over C). An operator is self-adjoint when T = T* — the infinite-dimensional twin of a symmetric matrix — and self-adjointness is what will make the spectral theorem work in Guide 4.
Adjoint via <Tx, y> = <x, T*y>
Right shift S on ell^2: S(x_1, x_2, ...) = (0, x_1, x_2, ...)
<Sx, y> = x_1 y_2 + x_2 y_3 + ...
= <x, (y_2, y_3, y_4, ...)>
=> S*(y_1, y_2, y_3, ...) = (y_2, y_3, ...) = the LEFT shift.
So (right shift)* = left shift. Check the asymmetry:
S* S = I (left-then-right undoes -> identity)
S S* != I (S S* kills the first coordinate)
Left and right inverses differ -> a purely infinite-dim effect.