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

Theorem 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. Then $\mathcal B(X, Y)$ is a Banach space if and only if $Y$ is a Banach space.

At this time we can only provide one direction of the proof. The other direction is much harder and will be proven later.

• Proof: $\Rightarrow$ Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. We already know that $\mathcal B(X, Y)$ with the operator norm is a normed space, so all we need to show is that it is complete.

• 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 $\| T_m - T_n \|_{\mathrm{op}} < \epsilon$.

• In particular, for each $x \neq 0$, by letting $\displaystyle{\epsilon_x = \frac{\epsilon}{\| x \|_X} > 0}$ there exists an $N_x \in \mathbb{N}$ such that if $m, n \geq N_x$ then $\| T_m - T_n \|_{\mathrm{op}} < \epsilon_x$ and:

(1)

\begin{align} \quad \| T_m(x) - T_n(x) \|_Y \leq \| (T_m - T_n)(x) \|_Y \leq \| T_m - T_n \|_{\mathrm{op}} \| x \|_X < \epsilon_x \| x \|_X = \frac{\epsilon}{\| x \|_X} \cdot \| x \|_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, $(T_n(0))_{n=1}^{\infty} = (0)$ 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 $x \in X$:

(2)

\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:

(3)

\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:

(4)

\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 $\| T_m - T_n \|_{\mathrm{op}} < \epsilon$. So by the continuity of the norm we have that for every $x \in X$:

(6)

\begin{align} \quad \| T(x) - T_n(x) \|_Y = \| \lim_{m \to \infty} T_m(x) - T_n(x) \|_Y = \lim_{m \to \infty} \| T_m(x) - T_n(x) \|_Y \leq \lim_{m \to \infty} \| T_m - T_n \|_{\mathrm{op}} \| x \_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:

(7)

\begin{align} \quad \| T - T_n \|_{\mathrm{op}} \| x \|_X < \epsilon \| x \|_X \end{align}

• That is, for $n \geq N$ we have that $\| T - T_n \|_{\mathrm{op}} < \epsilon$. 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. $\blacksquare$