TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field, and let $T : X \to Y$ be linear. Then the following are equivalent:

$T$ is bounded.
$T$ is continuous at $0$ .
$T$ is continuous on $X$ .
$T$ is Lipschitz.

Facts & Assumptions

Given: Normed spaces $X$ and $Y$ , a linear map $T : X \to Y$ , a real $ε > 0$ , and a vector $x_{0} \in X$ .

[L1]

A bounded linear operator has a constant $C \geq 0$ with $∥ T x ∥ \leq C ∥ x ∥$ for every $x \in X$ (A bounded linear operator between normed spaces).

[L2]

A map is Lipschitz when one constant controls all distances, and every Lipschitz map between metric spaces is continuous (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent).

[L3]

Continuity at a point in a metric space is the $ε$ - $δ$ condition of Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form.

Proof

technique · direct

1.1

Assume $T$ is bounded, with constant $C$ from [L1]. Then $∥ T x - T y ∥ = ∥ T (x - y) ∥ \leq C ∥ x - y ∥$ for all $x, y \in X$ , so $T$ is Lipschitz.

L1L2

1.2

If $T$ is continuous on $X$ , then in particular it is continuous at $0$ , so $(3) \Rightarrow (2)$ .

1.3

Assume $T$ is continuous at $0$ . Applying [L3] with $ε = 1$ gives $δ > 0$ such that $∥ u ∥ < δ$ implies $∥ T u ∥ < 1$ .

L3choose

2.1

Step 1.1 proves $(1) \Rightarrow (4)$ , and [L2] gives $(4) \Rightarrow (3)$ .

step 1.1L2

2.2

Let $x \in X$ with $x \neq 0$ and put $u : = \frac{δ}{2 ∥ x ∥} x$ . Then $∥ u ∥ = δ / 2 < δ$ , so $∥ T u ∥ < 1$ by step 1.3. By linearity, $T u = \frac{δ}{2 ∥ x ∥} T x$ , hence $∥ T x ∥ < \frac{2}{δ} ∥ x ∥$ . The same inequality is trivial at $x = 0$ , so $T$ is bounded.

step 1.3L1algebra

3.1

Thus $(2) \Rightarrow (1)$ . Combining steps 2.1, 1.2, and 2.2 gives all four equivalences.

step 2.1step 1.2step 2.2∎

Depends on

Used by

A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm Theorem

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)