2
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.]

$\endgroup$
  • $\begingroup$ Sorry for necroing this old answer. I was wondering if you could prove this alternate definition of the operator norm. Thanks! $\endgroup$ – rubikscube09 Nov 24 '17 at 0:31
  • $\begingroup$ 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. $\endgroup$ – Sean Haight Feb 18 at 22:19
  • 1
    $\begingroup$ @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 $\| T_n (x) - T(x) \| = \| T_n (x) - \lim_{m \to \infty} T_m (x) \| = \lim_{m \to \infty} \| T_n (x) - T_m (x) \| \leq \epsilon < 2 \epsilon$. $\endgroup$ – Kenny Wong Feb 18 at 22:24
  • $\begingroup$ By operator do you also mean continuous? If so, how would you prove continuity of the limit? $\endgroup$ – An old man in the sea. Jul 14 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.