A linear operator T: X → Y between Banach spaces has the following property: whenever xₙ → x in X and T(xₙ) → y in Y, it follows that y = T(x). What does the closed graph theorem allow you to conclude?
AT is injective (one-to-one) but not necessarily surjective
BT is continuous (equivalently, bounded) — the closed graph condition is equivalent to continuity for linear operators between Banach spaces
CT has a bounded inverse, but T itself may fail to be continuous
DNothing conclusive — a closed graph implies continuity only for surjective operators
The closed graph theorem states precisely that for a linear operator between Banach spaces, having a closed graph (the sequential condition described) is equivalent to being continuous, i.e., bounded. There is no additional requirement on T beyond linearity — injectivity, surjectivity, and bijectivity are irrelevant to the theorem. The value of the theorem is that this sequential condition is often far easier to verify directly from T's definition than finding an explicit bound C with ‖T(x)‖ ≤ C‖x‖.
Question 2 Multiple Choice
Why does the closed graph theorem fail for linear operators between incomplete normed spaces — that is, normed spaces that are not Banach spaces?
ABecause the norm topology on an incomplete space cannot detect convergence of Cauchy sequences
BBecause completeness is required both to run the open mapping theorem in the proof and to ensure Cauchy sequences arising in the argument actually converge
CBecause linear operators on incomplete spaces are never bounded, making the conclusion vacuously false
DBecause incomplete spaces have no well-defined product topology, so the graph cannot be defined
Completeness is not a technicality — it is load-bearing in the proof. The standard proof of the closed graph theorem invokes the open mapping theorem (a bijective bounded linear operator between Banach spaces has a bounded inverse), which itself requires both spaces to be Banach (complete). In an incomplete space, a linear operator can have a closed graph while failing to be bounded: you can construct sequences where the closed graph hypothesis holds formally but the missing limit points (absent because the space is incomplete) allow boundedness to fail. The theorem is genuinely false without completeness.
Question 3 True / False
For a linear operator T: X → Y between Banach spaces, T having a closed graph is equivalent to T being continuous.
TTrue
FFalse
Answer: True
This is the content of the closed graph theorem. For general maps between metric spaces, these properties are distinct: a function can have a closed graph while failing to be continuous (consider f: ℝ → ℝ with f(x) = 1/x for x ≠ 0 and f(0) = 0 — the graph is closed but f is discontinuous at 0). But for *linear operators between Banach spaces*, the two conditions collapse to the same thing. The linearity and completeness work together to give the equivalence. This is precisely what makes the theorem powerful: it trades a topological property (closed graph) for an analytic one (boundedness).
Question 4 True / False
For any function between metric spaces, having a closed graph is equivalent to being continuous — the closed graph theorem is just the specialization of this general fact to Banach spaces.
TTrue
FFalse
Answer: False
This is false, and the failure of the general case is what makes the theorem non-trivial. For arbitrary functions between metric spaces, closed graph and continuity are genuinely different properties. A function can be discontinuous while still having a closed graph: if f(x) = 1/x for x ≠ 0 and f(0) = 0, the graph {(x, f(x))} is closed in ℝ² (it contains all its limit points), but f is discontinuous at 0. The equivalence is special to *linear operators* on *Banach (complete) spaces* — linearity constrains the behavior of the operator globally, and completeness ensures that Cauchy sequences in the argument converge.
Question 5 Short Answer
What practical advantage does the closed graph theorem provide when proving that a linear operator is continuous, and why is the completeness of both spaces required?
Think about your answer, then reveal below.
Model answer: The practical advantage is a shift in the proof burden: instead of finding an explicit constant C such that ‖T(x)‖ ≤ C‖x‖ for all x (which can be difficult or impossible to derive directly), you instead verify the sequential condition — that whenever xₙ → x and T(xₙ) → y, it follows that y = T(x). For many operators defined by formulas (differential or integral operators), this condition is much easier to check from the operator's definition than finding a direct bound. Completeness is required because the proof uses the open mapping theorem, which requires both spaces to be Banach, and because without completeness the Cauchy sequences that appear in the argument may fail to have limits in the spaces, breaking the chain of reasoning.
This is the theorem's mathematical elegance: it converts a quantitative question (find the bound) into a qualitative/topological question (is the graph closed?). Many important operators in analysis — differential operators, integral operators, unbounded operators on function spaces — are most naturally verified continuous via the closed graph theorem because their definitions make the sequential condition easy to check, even when a direct estimate of the norm would be technically demanding.