Questions: Uniform Boundedness Principle

Question 1

A family of bounded linear operators {Tᵢ: X → Y} between Banach spaces satisfies sup_i ‖Tᵢ(x)‖ < ∞ for every fixed x ∈ X. What does the uniform boundedness principle guarantee?

• A. Nothing without additional assumptions — pointwise bounds say nothing about operator norms
• B. There exists a constant C such that ‖Tᵢ‖ ≤ C for all i
• C. The operators converge pointwise to a single bounded operator
• D. The norms ‖Tᵢ‖ are bounded, but only on a dense subset of X

Answer: (B)

The uniform boundedness principle (Banach-Steinhaus theorem) says exactly this: pointwise boundedness of a family of operators on a Banach space implies uniform boundedness — a single constant C bounds all operator norms simultaneously. This is surprising because the hypothesis is weak (each x individually imposes a bound that may depend on x) while the conclusion is strong (one C works for all i and all unit vectors). The proof uses the Baire category theorem, which requires completeness of X.

Question 2

A sequence of bounded linear operators Tₙ: X → Y (where X is a Banach space) has unbounded operator norms: sup_n ‖Tₙ‖ = ∞. What must follow by the contrapositive of the uniform boundedness principle?

• A. Y must not be a Banach space
• B. The operators Tₙ are not actually bounded as claimed
• C. There exists some x ∈ X for which sup_n ‖Tₙ(x)‖ = ∞
• D. The Baire category theorem does not apply to this sequence

Answer: (C)

The UBP states: pointwise bounded on Banach space X ⟹ uniformly bounded. Contrapositive: NOT uniformly bounded ⟹ NOT pointwise bounded. If the norms are unbounded and X is a Banach space, there must exist some x ∈ X where the outputs blow up. In fact, by the Baire category argument, the set of 'bad' x where pointwise boundedness fails is a dense Gδ set — in a category-theoretic sense it is the 'typical' point of X.

Question 3

The uniform boundedness principle holds on Banach spaces but fails on incomplete normed spaces.

• T. True
• F. False

Answer: True

Completeness is essential. The proof proceeds by showing that if the norms were unbounded, one could express X as a countable union of closed nowhere-dense sets — a contradiction with the Baire category theorem, which requires X to be a complete metric space. On an incomplete normed space, the Baire category theorem does not apply, and you can explicitly construct pointwise bounded families with unbounded operator norms.

Question 4

If a family of bounded linear operators {Tᵢ} is uniformly bounded in norm, then for every fixed x the values ‖Tᵢ(x)‖ are automatically bounded.

• T. True
• F. False

Answer: True

This direction is trivial and does not require completeness: ‖Tᵢ(x)‖ ≤ ‖Tᵢ‖ · ‖x‖ ≤ C · ‖x‖ for any fixed x. The deep content of the uniform boundedness principle is the converse — that pointwise boundedness (weak hypothesis) implies uniform boundedness (strong conclusion). The easy direction is: uniform ⟹ pointwise. The hard direction is: pointwise ⟹ uniform (which requires completeness).

Question 5

Why is completeness of the domain space X necessary for the uniform boundedness principle? What goes wrong without it?

Model answer: The proof uses the Baire category theorem: a complete metric space cannot be written as a countable union of closed nowhere-dense sets. If operator norms were unbounded, one can construct closed sets covering X whose nowhere-denseness contradicts Baire — forcing uniform boundedness. On an incomplete space the Baire theorem fails, so the argument breaks down. Explicit counterexample: the space of finitely-supported sequences with sup norm (incomplete). Define Tₙ(x) = n·xₙ·e₁. For any fixed finitely-supported x, xₙ = 0 for large n, so pointwise boundedness holds — but ‖Tₙ‖ = n → ∞.

Completeness is not a technicality but the structural property the theorem requires. The 'big three' theorems of Banach space theory (uniform boundedness, open mapping, closed graph) all depend on the Baire category theorem and all require completeness for exactly this reason.