# Proving that the space of bounded linear operators is complete

Let $\mathfrak{X}$ be a normed space and let $\mathfrak{Y}$ be a complete normed space. Prove that $\mathfrak{L}(\mathfrak{X},\mathfrak{Y})$ is complete.

As far as I'm aware, proving a space is complete requires proving that every Cauchy sequence converges but I'm unsure on how to do this.

If $T_n$ is a Cauchy sequence in $\mathcal L (X,B)$, then for any chosen $x \in X$, $T_n(x)$ is a Cauchy sequence in $B$. Since $B$ is complete, this sequence converges.
Define a new linear operator $T : X \to B$ mapping $x \mapsto \lim_{n \to \infty} T_n(x)$.
Now prove that $T_n \to T$ in the operator norm. Let $\epsilon > 0$. The Cauchy property tells you that there exists an $N$ such that $$m,n > N \implies \sup_{||x || \leq 1} ||T_n(x) - T_m (x) || < \epsilon.$$ Take the limit $m \to \infty$.
[If your notation $\mathcal L (X, B)$ refers to bounded operators, then you also need to prove that $T$ is bounded. I'll leave you to do that if required.]
• Could you possibly be more explicit when you say "Take the limit as $m \to \infty$"? I agree that $\lim_{m \to \infty}\sup_{||x|| \leq 1}||T_n(x) - T_m(x)|| \leq \epsilon$ if it exists, but I'm not entirely sure why this gives us $\sup_{||x|| \leq 1} ||T_n(x) - T(x)|| = \sup_{||x|| \leq 1} \lim_{m \to \infty}||T_n(x) - T_m(x)|| \leq \epsilon$? It feels like we need to commute the limit past the supremum, and I'm not entirely sure how to do that. – Sean Haight Feb 18 at 22:19
• @SeanHaight We know that, for all $x \in \overline{B(0, 1)}$, and for all $m , n > N$, we have $\| T_n (x) - T_m (x) \| < \epsilon$. Hence, for all $x \in \overline{B(0, 1)}$ and for all $n > N$, we have $\| T_n (x) - T(x) \| = \| T_n (x) - \lim_{m \to \infty} T_m (x) \| = \lim_{m \to \infty} \| T_n (x) - T_m (x) \| \leq \epsilon < 2 \epsilon$. – Kenny Wong Feb 18 at 22:24