TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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If (Y) is Banach then (\mathcal B(X,Y)) is Banach

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field. If $Y$ is Banach, then $B (X, Y)$ is Banach for the operator norm.

Facts & Assumptions

Given: A Banach space $Y$ and an operator-norm Cauchy sequence $(T_{n})$ in $B (X, Y)$ .

[L1]

A Banach space is complete for its norm metric (Banach space).

[L2]

For a bounded operator, the operator norm is the unit-ball supremum and satisfies $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ for every $x$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L3]

$B (X, Y)$ is the vector space of bounded linear operators (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

Proof

technique · direct

1.1

Fix $x \in X$ . Since $(T_{n})$ is Cauchy in operator norm, [L2] gives $∥ T_{n} x - T_{m} x ∥ \leq ∥ T_{n} - T_{m} ∥ ∥ x ∥$ , so $(T_{n} x)$ is a Cauchy sequence in $Y$ . Because $Y$ is Banach, there is $T x \in Y$ with $T_{n} x \to T x$ .

L1L2choose

2.1

Step 1.1 defines a map $T : X \to Y$ . If $x, y \in X$ , then $T_{n} (x + y) = T_{n} x + T_{n} y$ for every $n$ , so passing to the limit gives $T (x + y) = T x + T y$ . The same argument with $λ x$ gives $T (λ x) = λ T x$ . Thus $T$ is linear.

step 1.1L3algebra

2.2

Choose $N$ such that $∥ T_{n} - T_{m} ∥ \leq 1$ for all $m, n \geq N$ . Fix $n \geq N$ and $x \in X$ . For every $m \geq N$ , $∥ T_{m} x ∥ \leq ∥ T_{N} x ∥ + ∥ (T_{m} - T_{N}) x ∥ \leq (∥ T_{N} ∥ + 1) ∥ x ∥$ by [L2]. Letting $m \to \infty$ in step 1.1 gives $∥ T x ∥ \leq (∥ T_{N} ∥ + 1) ∥ x ∥$ , so $T$ is bounded and hence $T \in B (X, Y)$ .

step 1.1L2L3

2.3

Let $ε > 0$ . Since $(T_{n})$ is operator-norm Cauchy, choose $N$ so that $∥ T_{n} - T_{m} ∥ < ε$ for all $m, n \geq N$ . Fix $n \geq N$ and $x$ with $∥ x ∥ \leq 1$ . Step 1.1 gives $T_{m} x \to T x$ , so $∥ (T_{n} - T) x ∥ = \lim_{m} ∥ (T_{n} - T_{m}) x ∥ \leq ε$ by [L2]. Taking the supremum over the unit ball yields $∥ T_{n} - T ∥ \leq ε$ .

step 1.1L2

3.1

Step 2.3 shows $T_{n} \to T$ in operator norm, with $T \in B (X, Y)$ by step 2.2. Therefore every operator-norm Cauchy sequence converges in $B (X, Y)$ , so $B (X, Y)$ is Banach by [L1].

step 2.2step 2.3L1∎

Depends on

Used by

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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)