Spectral Theory in Hilbert Space

A. FUNCTIONAL ANALYSIS 119
other words, we can solve T x = y for any y ∈ B2, so the inverse of
T is defined everywhere, and the inverse is bounded by 2 , so it is
C
continuous. The proof is complete.
In these notes we do not actually use Banach’s theorem, but the
following simple corollary (which is actually equivalent to Banach’s
theorem). Recall that a linear map T : B1 → B2 is called closed if the
graph {(u, T u) | u ∈ D(T )} is a closed subset of B1 ⊕B2. Equivalently,
if uj → u in B1 and T uj → v in B2 implies that u ∈ D(T ) and T u = v.
Corollary A.4 (Closed graph theorem). Suppose T is a closed
linear operator T : B1 → B2, defined on all of B1. Then T is bounded.
Proof. The graph {(u, T u) | u ∈ B1} is by assumption a Banach
space with norm (u, T u) = u1 + T u2, where ·j denotes the
norm of Bj. The map (u, T u) ↦→ u is linear, defined in this Banach
space, with range equal to B1, and it has norm ≤ 1. It is obviously
injective, so by Banach’s theorem the inverse is bounded, i.e., there is
a constant so that (u, T u) ≤ Cu1. Hence also T u2 ≤ Cu1, so
that T is bounded.
In Chapter 3 we used the Banach-Steinhaus theorem, Theorem 3.10.
Since no extra effort is involved, we prove the following slightly more
general theorem.
Theorem A.5 (Banach-Steinhaus; uniform boundedness principle).
Suppose B is a Banach space, L a normed linear space, and M
a subset of the set L(B, L) of all bounded, linear maps from B into L.
Suppose M is pointwise bounded, i.e., for each x ∈ B there exists a
constant Cx such that T xB ≤ Cx for every T ∈ M. Then M is uniformly
bounded, i.e., there is a constant C such that T xB ≤ CxL
for all x ∈ B and all T ∈ M.
Proof. Put Fn = {x ∈ B | T xB ≤ n for all T ∈ M}. Then
Fn is closed, as the intersection of the closed sets which are inverse
images of the closed interval [0, n] under a continuous function B1 ∋
x ↦→ T xL ∈ R. The assumption means that ∪ ∞ n=1Fn = B. By
Baire’s theorem at least one Fn must have an interior point. Since Fn
is convex (if x, y ∈ Fn and 0 ≤ t ≤ 1, then tT x + (1 − t)T yL ≤
tT xL + (1 − t)T yL ≤ n) and symmetric with respect to the origin
it follows, like in the proof of Banach’s theorem, that 0 is an interior
point in Fn. Thus, for some r > 0 we have T xL ≤ n for all T ∈ M, if
xB ≤ r. By homogeneity follows that T xL ≤ n
r xB for all T ∈ M
and x ∈ B.