Criterion for B(X, Y) to be a Banach Space

# Criterion for B(X, Y) to be a Banach Space

Theorem 1: Let $X$ and $Y$ be normed linear spaces. If $Y$ is a Banach space then $\mathcal B(X, Y)$ is a Banach space. |

**Proof:**Let $X$ and $Y$ be normed linear spaces. From the theorem on The Normed Linear Space B(X, Y) page we have that $\mathcal B(X, Y)$ is a normed linear space. All we need to show is that every Cauchy sequence in $B(X, Y)$ converges in $B(X, Y)$.

- Let $(T_n)_{n=1}^{\infty}$ be a Cauchy sequence of linear operators in $\mathcal B(X, Y)$. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| T_m - T_n \| < \epsilon$.

- In particular, for $\displaystyle{\epsilon_x = \frac{\epsilon}{\| x \|} > 0}$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| T_m - T_n \| < \epsilon_x$ and:

\begin{align} \quad \| T_m(x) - T_n(x) \| \leq \| (T_m - T_n)(x) \| \leq \| T_m - T_n \| \| x \| < \epsilon_x \| x \| = \frac{\epsilon}{\| x \|} \cdot \| x \| = \epsilon \end{align}

- So $(T_n(x))_{n=1}^{\infty}$ is a Cauchy sequence in $Y$ for every $x \in X$ with $x \neq 0$. And trivially, $(T_n(0))_{n=1}^{\infty}$ is a Cauchy sequence in $Y$. So $(T_n(x))_{n=1}^{\infty}$ is a Cauchy sequence in $Y$ for every $x \in X$.

- Since $Y$ is a Banach space, each of these Cauchy sequences converge to some element $y_x \in Y$, that is, for each $x \in X$:

\begin{align} \quad \lim_{n \to \infty} T_n(x) = y_x \end{align}

- Define a function $T : X \to Y$ for all $x \in X$ by:

\begin{align} \quad T(x) = y_x \end{align}

- We first show that $T$ is a linear operator from $X$ to $Y$. Let $x_1, x_2 \in X$ and let $\lambda \in \mathbb{C}$. Then:

\begin{align} \quad T(x_1 + x_2) = y_{x_1 + x_2} = \lim_{n \to \infty} T_n(x_1 + x_2) = \lim_{n \to \infty} [T_n(x_1) + T_n(x_2)] = \lim_{n \to \infty} T_n(x_1) + \lim_{n \to \infty} T_n(x_2) = y_{x_1} + y_{x_2} = T(x_1) + T(x_2) \end{align}

(5)
\begin{align} \quad T(\lambda x_1) = y_{\lambda x_1} = \lim_{n \to \infty} T_n(\lambda x_1) = \lim_{n \to \infty} \lambda T_n(x_1) = \lambda \lim_{n \to \infty} T_n(x_1) = \lambda y_{x_1} = \lambda T(x_1) \end{align}

- So indeed, $T$ is a linear operator from $X$ to $Y$. We now show that $T$ is a bounded linear operator. Let $\epsilon > 0$. Since $(T_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathcal B(X, Y)$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| T_m - T_n \| < \epsilon$. So by the continuity of the norm we have that for every $x \in X$:

\begin{align} \quad \| T(x) - T_n(x) \| = \| \lim_{m \to \infty} T_m(x) - T_n(x) \| = \lim_{m \to \infty} \| T_m(x) - T_n(x) \| \leq \lim_{m \to \infty} \| T_m - T_n \| \| x \| < \epsilon \| x \| \end{align}

- This shows that $(T - T_n) \in \mathcal B(X, Y)$. But we have that $T_n \in \mathcal B(X, Y)$. Since $\mathcal B(X, Y)$ is a linear space, it is closed under its operations. So $T = (T - T_n) + T_n \in \mathcal B(X, Y)$, that is, $T$ is a bounded linear operator.

- We lastly show that $(T_n)_{n=1}^{\infty}$ converges to $T$. From above, we have that if $n \geq N$ then:

\begin{align} \quad \| T - T_n \| \| x \| < \epsilon \| x \| \end{align}

- That is, for $n \geq N$ we have that $\| T - T_n \| < \epsilon$. So every Cauchy sequence $(T_n)_{n=1}^{\infty}$ in $\mathcal B(X, Y)$ converges to some $T \in \mathcal B(X, Y)$. So $\mathcal B(X, Y)$ is a Banach space. $\blacksquare$