Does the Banach-Steinhaus theorem hold for non-complete normed spaces?

Question

The Banach-Steinhaus theorem (Uniform Boundedness Principle) states that for a family of bounded linear operators $\{T_\alpha\}_{\alpha \in A}$ from a Banach space $X$ to a normed space $Y$ , pointwise boundedness implies uniform boundedness. Specifically, if

\sup_{\alpha \in A} \|T_\alpha(x)\| < \infty \quad \forall x \in X

then

\sup_{\alpha \in A} \|T_\alpha\|_{\mathcal{L}(X,Y)} < \infty.

Does this result still hold if $X$ is only assumed to be a normed space rather than a Banach space? If not, what is a counterexample and which specific failure of completeness allows the theorem to break?

Kenji Nakamura · Accepted Answer

No, the Banach-Steinhaus theorem does not hold for non-complete normed spaces. Here is a concrete counterexample. Let X = C c [ 0 , 1 ], the space of continuous functions on [ 0 , 1 ] with compact support (i.e., vanishing outside some compact subset), equipped with the L 2 norm ∥ f ∥ L 2 = ( ∫ 0 1 ∣ f ( x ) ∣ 2 d x ) 1 / 2 . This space is not complete — its completion is L 2 [ 0 , 1 ]. Define a family of linear functionals { T n } n ∈ N ⊂ X ∗ by T n ( f ) = n ∫ 0 1 / n f ( x ) d x . For each fixed f ∈ X, dominated convergence gives T n ( f ) → f ( 0 ) (defined since f is continuous near 0), hence sup ⁡ n ∣ T n ( f ) ∣ < ∞. However, ∥ T n ∥ = sup ⁡ ∥ f ∥ 2 ≤ 1 ∣ T n ( f ) ∣ = n → ∞ , so sup ⁡ n ∥ T n ∥ is infinite. Therefore, pointwise boundedness does not imply uniform boundedness in the non-complete setting. The completeness of X is essential: the proof uses the Baire category theorem, which requires the space to be a complete metric space.

Maria Schmidt · Answer

To add to the counterexample above, here is another one using sequence spaces. Let X = c 00, the space of sequences with only finitely many non-zero entries, equipped with the ℓ ∞ norm ∥ ( x n ) ∥ ∞ = sup ⁡ n ∈ N ∣ x n ∣ . Define T k : X → R by T k ( x ) = k ⋅ x k . For any fixed x ∈ c 00, since only finitely many entries are non-zero, T k ( x ) = 0 for all sufficiently large k, hence sup ⁡ k ∣ T k ( x ) ∣ < ∞. But ∥ T k ∥ = k → ∞ . The completion of c 00 under ∥ ⋅ ∥ ∞ is c 0 (sequences converging to 0). In c 0, the functionals T k are not even well-defined as bounded linear functionals — they're unbounded. This illustrates that the Baire category argument in the Banach-Steinhaus proof critically uses the completeness of the domain to apply the uniform boundedness principle.

James O'Neill · Answer

From a categorical perspective, the Uniform Boundedness Principle is a statement about the barreledness of locally convex spaces. A locally convex space X is called barreled if every absorbing, balanced, convex, closed set in X is a neighborhood of 0. The Banach-Steinhaus theorem holds whenever X is barreled and Y is locally convex. Banach spaces are barreled (by the Baire category theorem), but non-complete normed spaces are never barreled. In fact, a normed space is barreled iff it is a Banach space. So the question reduces to: the theorem fails for non-complete normed spaces precisely because they lack the barreled property, which itself is equivalent to completeness in the normed category.

Does the Banach-Steinhaus theorem hold for non-complete normed spaces?

3 Answers

Does the Banach-Steinhaus theorem hold for non-complete normed spaces?

3 Answers