Functional Analysis 18 | Compact Operators [dark version]

Q: How does boundedness of the integral operator follow from the supremum norm?

With ||f||∞=1, estimate |(Tf)(s)| = |∫_0^1 k(s,t)f(t)dt| ≤ ∫_0^1 |k(s,t)| |f(t)| dt ≤ ∫_0^1 |k(s,t)| dt. The transcript simplifies this by bounding with a constant derived from the supremum norm of k, showing the operator norm is finite and depends only on k, not on f.

Q: Why does Arzelà–Ascoli imply compactness here?

Arzelà–Ascoli says that a set of functions on a compact domain that is uniformly bounded and uniformly equi-continuous has compact closure in C([0,1]) with the supremum norm. Since the image of the unit ball under T satisfies both properties (from the kernel’s uniform continuity and the norm estimate), the closure of T(B_X) is compact, so T is compact.

TL;DR

In finite-dimensional normed spaces, every bounded linear operator is compact because bounded sets have compact closures.

Briefing Cornell Notes

Briefing

Compact operators extend the “finite-dimensional finiteness” idea to infinite-dimensional normed spaces by forcing the image of the unit ball to become compact after taking closure. In finite dimensions, every bounded linear operator automatically behaves compactly: the image of the unit ball is bounded, and its closure is compact. The key point is that this compactness property can fail badly in infinite dimensions—so compact operators are precisely the bounded operators whose unit-ball images (after closure) are compact.

The discussion starts by contrasting finite-dimensional and infinite-dimensional behavior. For a linear map between finite-dimensional normed spaces, continuity (equivalently boundedness for linear operators) is guaranteed, and bounded sets have compact closures. That motivates the definition: a bounded linear operator T: X → Y is compact if the closure of T applied to the unit ball in X is compact in Y. The “why it matters” becomes clear through a standard counterexample: the identity operator on L^p. Its unit-ball image is the unit ball itself, and the closed unit ball in L^p is closed and bounded but not compact. Since the identity operator does not shrink sets into something compact, it cannot be compact—showing that compactness is a genuine extra requirement beyond boundedness.

To make the definition concrete, the transcript builds a classic example: an integral operator on C([0,1]) with the supremum norm. The operator takes a continuous function f on [0,1] and produces a new function (Tf)(s) given by an integral of the form (Tf)(s) = ∫_0^1 k(s,t) f(t) dt, where k is a fixed continuous kernel on [0,1]×[0,1]. The first technical hurdle is ensuring Tf is continuous in s; the calculation compares (Tf)(s1) and (Tf)(s2) and uses uniform continuity of k on the compact domain [0,1]×[0,1]. Uniform continuity lets the difference in kernels k(s1,t) − k(s2,t) be made uniformly small for all t when s1 and s2 are close, which forces |(Tf)(s1) − (Tf)(s2)| to be small. This yields continuity and, more importantly, a uniform equi-continuity property for the entire image set of the unit ball.

Next comes boundedness. The operator norm is estimated by taking sup over s of |∫_0^1 k(s,t) f(t) dt| with ||f||∞ = 1. Pulling absolute values inside the integral and using the supremum bound on k shows the output is uniformly bounded by a constant depending only on k. With both ingredients—uniform boundedness and uniform equi-continuity—the Arzelà–Ascoli theorem applies. That theorem turns these properties into relative compactness (and hence compactness of the closure) of the image of the unit ball. Therefore, the integral operator defined by a continuous kernel k on a compact domain is compact, illustrating how compact operators generalize finite-dimensional behavior in a rigorous way.

Cornell Notes

Compact operators are bounded linear maps whose action on the unit ball produces a relatively compact set: the closure of T(B_X) is compact in Y. In finite-dimensional spaces, boundedness already implies compactness of the unit-ball image closure, so every linear operator is compact. In infinite-dimensional spaces this fails: the identity operator on L^p has a closed unit ball that is not compact, so it is not compact. A standard infinite-dimensional example is an integral operator on C([0,1]) defined by (Tf)(s)=∫_0^1 k(s,t)f(t)dt with continuous k. Continuity and uniform equi-continuity of the image set come from uniform continuity of k on the compact domain, boundedness comes from sup-norm estimates using ||f||∞≤1, and Arzelà–Ascoli then yields compactness.

What is the operational definition of a compact operator in normed spaces?

A bounded linear operator T: X → Y is called compact if the set T(B_X), where B_X is the unit ball in X, has compact closure in Y. Equivalently: take all vectors x with ||x||≤1, apply T, then the resulting set becomes compact once closed in Y.

Why is the identity operator on L^p not compact?

For the identity operator I on L^p, the image of the unit ball is exactly the unit ball itself. The closed unit ball in L^p is closed and bounded but not compact (in infinite dimensions). Since compactness of the closure fails, I cannot be a compact operator.

How does the integral operator (Tf)(s)=∫_0^1 k(s,t)f(t)dt map C([0,1]) into itself?

To show Tf is continuous in s, compare outputs at s1 and s2: |(Tf)(s1)−(Tf)(s2)| = |∫_0^1 (k(s1,t)−k(s2,t)) f(t) dt|. Using ||f||∞ and uniform continuity of k on [0,1]×[0,1], the kernel difference k(s1,t)−k(s2,t) becomes uniformly small for all t when s1 and s2 are close, forcing the integral difference to be small. Hence Tf is continuous.

What does “uniformly equi-continuous” mean for the image set of the unit ball?

Let A = T(B_X). The set A is uniformly equi-continuous if for every ε>0 there exists δ>0 such that for all s1,s2 with |s1−s2|<δ and for every g in A, one has |g(s1)−g(s2)|<ε. In the transcript, the same δ works simultaneously for all functions produced by applying T to unit-norm inputs.

How does boundedness of the integral operator follow from the supremum norm?

With ||f||∞=1, estimate |(Tf)(s)| = |∫_0^1 k(s,t)f(t)dt| ≤ ∫_0^1 |k(s,t)| |f(t)| dt ≤ ∫_0^1 |k(s,t)| dt. The transcript simplifies this by bounding with a constant derived from the supremum norm of k, showing the operator norm is finite and depends only on k, not on f.

Why does Arzelà–Ascoli imply compactness here?