DE
May 15, 2026
Does the Banach-Steinhaus theorem hold for non-complete normed spaces?
The Banach-Steinhaus theorem (Uniform Boundedness Principle) states that for a family of bounded linear operators from a Banach space to a normed space , pointwise boundedness implies uniform boundedness. Specifically, if
then
Does this result still hold if 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?
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3 Answers
PK
Prof. Kenji NakamuraMay 15, 2026 AcceptedNo, 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
$$\|f\|_{L^2} = \left(\int_0^1 |f(x)|^2 \, dx\right)^{1/2}.$$
This space is not complete — its completion is $L^2[0,1]$. Define a family of linear functionals $\{T_n\}_{n\in\mathbb{N}} \subset X^*$ by
$$T_n(f) = n \int_0^{1/n} f(x) \, dx.$$
For each fixed $f \in X$, dominated convergence gives $T_n(f) \to f(0)$ (defined since $f$ is continuous near $0$), hence $\sup_n |T_n(f)| < \infty$. However,
$$\|T_n\| = \sup_{\|f\|_2 \leq 1} |T_n(f)| = \sqrt{n} \to \infty,$$
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.
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MS
Maria SchmidtMay 15, 2026To 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 $\ell^\infty$ norm
$$\|(x_n)\|_\infty = \sup_{n \in \mathbb{N}} |x_n|.$$
Define $T_k: X \to \mathbb{R}$ by
$$T_k(x) = k \cdot x_k.$$
For any fixed $x \in c_{00}$, since only finitely many entries are non-zero, $T_k(x) = 0$ for all sufficiently large $k$, hence $\sup_k |T_k(x)| < \infty$. But
$$\|T_k\| = k \to \infty.$$
The completion of $c_{00}$ under $\|\cdot\|_\infty$ 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.
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JO
James O'NeillMay 15, 2026From 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.
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