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Unbounded Operators 6 | Closed Graph Theorem thumbnail

Unbounded Operators 6 | Closed Graph Theorem

5 min read

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TL;DR

For a linear operator T: DT → Y between Banach spaces, T is bounded if and only if its graph is closed, provided DT is a Banach space (e.g., DT is closed in X).

Briefing

Closed operators on Banach spaces turn out to be automatically bounded once their domains are “large enough.” More precisely, for a linear operator T between Banach spaces X and Y, if the domain DT is a closed subset of X (equivalently, DT is itself a Banach space), then T is closed exactly when it is bounded—so continuity and closedness become the same condition in this setting. That equivalence matters because it explains why genuinely unbounded operators must have “small” domains: a closed unbounded operator cannot be defined on all of a Banach space.

The result is framed as the Closed Graph Theorem for Banach spaces. Start with a linear map T: DT → Y, where DT is assumed to be a Banach space (the transcript notes this can be arranged by restricting to DT). The operator is called closed when its graph GT = {(x, Tx) : x ∈ DT} is closed in the product space X × Y. Under the Banach-space domain assumption, closedness of the graph forces boundedness of T, and conversely boundedness forces closedness.

One direction is straightforward: if T is bounded (hence continuous), then limits pass through T. Concretely, if xn → x in X and Txn → y in Y, continuity gives Tx = y. Since x lies in the domain, the graph contains the limit point (x, y), so the graph is closed.

The harder direction uses the graph itself as a Banach space. If T is closed, then GT is a closed subspace of the Banach space X × Y, so GT is Banach. Two natural projection maps are introduced: PX: GT → X taking (x, Tx) to x, and PY: GT → Y taking (x, Tx) to Tx. The projection PX is linear and bounded, and it is also bijective because every x in X (with DT = X after restriction) appears in exactly one pair in the graph. The bounded inverse theorem (linked to the open mapping theorem) then implies that PX^{-1}: X → GT is bounded.

With PX^{-1} in hand, T can be written as a composition of bounded maps: PY ∘ PX^{-1}. Since PX^{-1} sends x to (x, Tx) and PY extracts the second component, the composition returns Tx. A composition of continuous (equivalently bounded) linear maps is continuous, so T is bounded. This completes the equivalence: for Banach-space domains, “closed graph” and “bounded operator” are the same.

An immediate takeaway for unbounded operators follows: if a linear operator is closed but unbounded, its domain cannot be a Banach space. In particular, a closed unbounded operator cannot have DT = X when X is Banach. This is why analysts spend so much effort specifying domains carefully when working with unbounded operators.

Cornell Notes

For linear operators between Banach spaces, having a closed graph is equivalent to being bounded—provided the operator’s domain is itself a Banach space (e.g., DT is closed in X). If T is bounded, continuity ensures that whenever xn → x and Txn → y, one must have Tx = y, so the graph is closed. If T is closed, its graph GT is a closed subspace of X × Y and hence Banach; projection PX: GT → X becomes a bijective bounded map, so PX^{-1} is bounded by the bounded inverse theorem. Then T can be recovered as PY ∘ PX^{-1}, making T bounded and therefore continuous. This equivalence explains why closed unbounded operators must have non-Banach (proper) domains.

Why does boundedness imply closedness of a linear operator’s graph?

Boundedness for linear maps is equivalent to continuity. If xn → x in X and Txn → y in Y, continuity lets limits pass through T: T(xn) → T(x). Since Txn already converges to y, it follows that y = T(x). Therefore the limit point (x, y) lies in the graph GT, making GT closed.

What role does the assumption “DT is a Banach space” play in the closed graph theorem?

The proof of “closed graph ⇒ bounded” relies on GT being Banach. When DT is a Banach space, the graph GT = {(x, Tx)} sits as a closed subspace of X × Y, so GT is complete. That completeness is what allows the bounded inverse theorem to apply to the bijective bounded projection PX: GT → X, guaranteeing PX^{-1} is bounded.

How do the projection maps PX and PY help turn closedness into boundedness?

Once GT is Banach, define PX(x, Tx) = x and PY(x, Tx) = Tx. PX is linear and bounded, and it is bijective because each x corresponds to exactly one pair (x, Tx) in the graph (after restricting so DT = X). The bounded inverse theorem then gives that PX^{-1}: X → GT is bounded. Finally, T is obtained by PY ∘ PX^{-1}, so T is bounded as a composition of bounded maps.

Why is PX bijective in this setup?

Bijectivity comes from how the graph encodes the operator. For each x in the (restricted) domain, there is exactly one value Tx, so exactly one pair (x, Tx) lies in GT. That means PX sends each pair in GT to its unique first component x, and every x in X is hit by some pair in GT, making PX both injective and surjective.

What does the theorem imply about closed unbounded operators?

If T is closed and unbounded, its domain DT cannot be a Banach space. In particular, if X is Banach and DT = X, then the theorem would force T to be bounded—contradicting unboundedness. So closed unbounded operators must have “smaller” domains that are not complete under the inherited norm.

Review Questions

  1. In the proof that closed graph implies boundedness, where exactly is completeness used?
  2. How does the identity T = PY ∘ PX^{-1} follow from the definitions of PX and PY?
  3. Why would a closed unbounded operator be impossible if its domain were all of a Banach space?

Key Points

  1. 1

    For a linear operator T: DT → Y between Banach spaces, T is bounded if and only if its graph is closed, provided DT is a Banach space (e.g., DT is closed in X).

  2. 2

    Boundedness implies closedness because continuity forces Tx = y whenever xn → x and Txn → y.

  3. 3

    If T is closed, its graph GT is a closed subspace of X × Y and therefore Banach, enabling functional-analytic tools.

  4. 4

    The projection PX: GT → X is a bijective bounded linear map, so PX^{-1} is bounded by the bounded inverse theorem.

  5. 5

    The operator T can be reconstructed as T = PY ∘ PX^{-1}, making T continuous and hence bounded.

  6. 6

    A closed unbounded operator cannot have a Banach-space domain; in particular, it cannot be defined on all of a Banach space.

Highlights

Closed graph and boundedness become equivalent for linear operators once the domain is a Banach space.
The proof turns the graph GT into a Banach space and then uses projection maps plus the bounded inverse theorem.
T is obtained as PY ∘ PX^{-1}, converting closedness of GT into continuity of T.
A closed unbounded operator must have a non-Banach domain; otherwise boundedness would follow.
Continuity gives the easy direction: limits of Txn determine Tx directly.

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