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Here is Theorem 2.10-2 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig:

If $Y$ is a Banach space, then $B(X, Y)$ is a Banach space.

Here $X$ and $Y$ are normed spaces, either both real or both complex, and $B(X, Y)$ denotes the normed space of all the bounded linear operators $T \colon X \to Y$, with the norm defined by $$ \Vert T \Vert = \sup \left\{ \ \frac{ \left\Vert T(x) \right\Vert_Y}{ \Vert x \Vert_X} \ \colon \ x \in X, x \neq \mathbf{0}_X \ \right\} = \sup \left\{ \ \left\Vert T(x) \right\Vert \ \colon \ x \in X, \ \Vert x \Vert_X = 1 \ \right\}.$$

Here is Kreyszig's proof:

We consider an arbitrary Cauchy sequence $\left( T_n \right)$ in $B(X, Y)$ and show that $\left( T_n \right)$ converges to an operator $T \in B(X, Y)$.

Since $\left( T_n \right)$ is Cauchy, for every $\varepsilon > 0$ there is an $N$ such that $$ \left\Vert T_m - T_n \right\Vert < \varepsilon \qquad \qquad \qquad \qquad (m, n > N). $$ For all $x \in X$ and $m, n > N$ we thus obtain [cf. (3) in Sec. 2.7] $$ \left\Vert T_n x - T_m x \right\Vert = \left\Vert \left( T_n - T_m \right)x \right\Vert \leq \left\Vert T_n - T_m \right\Vert \left\Vert x \right\Vert. \tag{2} $$ Now for any fixed $x$ and given $\tilde{\varepsilon}$ we may choose $\varepsilon = \varepsilon_x$ so that $\varepsilon_x \Vert x \Vert < \tilde{\varepsilon}$. Then from (2) we have $\left\Vert T_n x - T_m x \right\Vert < \tilde{\varepsilon}$ and see that $\left( T_n x \right)$ is Cauchy in $Y$. Since $Y$ is complete, $\left( T_n x \right)$ converges, say, $T_n x \longrightarrow y$. Clearly, the limit $y \in Y$ depends on the choice of $x \in X$. This defines an operator $T \colon X \longrightarrow Y$, where $y = T x$. The operator $T$ is linear since $$ \lim T_n \left( \alpha x + \beta z \right) = \lim \left( \alpha T_n x + \beta T_n z \right) = \alpha \lim T_n x + \beta \lim T_n z. $$

We prove that $T$ is bounded and $T_n \longrightarrow T$, that is, $\left\Vert T_n - T \right\Vert \longrightarrow 0$.

Since (2) holds for every $m > N$ and $T_m x \longrightarrow T x$, we may let $m \longrightarrow \infty$. Using the continuity of the norm, we then obtain from (2) for every $n > N$ and all $x \in X$ $$ \left\Vert T_n x - T x \right\Vert = \left\Vert T_n x - \lim_{m \longrightarrow \infty} T_m x \right\Vert = \lim_{m \longrightarrow \infty} \left\Vert T_n x - T_m x \right\Vert \leq \varepsilon \Vert x \Vert. \tag{3} $$ This shows that $T_n - T$ with $n > N$ is a bounded linear operator. Since $T_n$ is bounded, $T = T_n - \left( T_n - T \right)$ is bounded, that is, $T \in B(X, Y)$. Furthermore, if in (3) we take the supremum over all $x$ of norm $1$, we obtain $$ \left\Vert T_n - T \right\Vert \leq \varepsilon \qquad (n > N). $$ Hence $\left\Vert T_n - T \right\Vert \longrightarrow 0$.

Although I understand this proof, I would like to modify the presentation as follows:

Let $\left( T_n \right)_{n \in \mathbb{N}}$ be a Cauchy sequence in $\mathrm{B}(X, Y)$. Let $x$ be an arbitrary point of $X$. We show that $\left( T_n(x) \right)_{n \in \mathbb{N}}$ is a Cauchy sequence in $Y$.

For each $n \in \mathbb{N}$, as $T_n \in \mathrm{B}(X,Y)$, so $T_n \colon X \to Y$, which implies that $T_n(x) \in Y$. Thus $\left( T_n(x) \right)_{n \in \mathbb{N}}$ is indeed a sequence in $Y$.

Let $\varepsilon$ be any positive real number. Then, as $\left( T_n \right)_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathrm{B}(X, Y)$, so, corresponding to the positive real number $\frac{\varepsilon}{1+ \Vert x \Vert_X}$, we can find a natural number $N_x$ such that $$\left\Vert T_m - T_n \right\Vert_{\mathrm{B}(X, Y)} < \frac{\varepsilon}{1+ \Vert x \Vert_X} \ \mbox{ for all } m, n \in \left\{ N_x + 1, N_x + 2, N_x + 3, \ldots \right\}. \tag{1} $$ So, for all $m, n \in \left\{ N_x + 1, N_x + 2, N_x + 3, \ldots \right\}$, we obtain $$ \begin{align} \left\Vert T_m (x) - T_n (x) \right\Vert_Y &= \left\Vert \ \left( T_m - T_n \right) (x) \ \right\Vert_Y \\ &\leq \left\Vert T_m - T_n \right\Vert_{\mathrm{B}(X,Y)} \Vert x \Vert_X \\ & \ \ \ \mbox{[ using a property of bounded linear operators; } \\ & \qquad \mbox{ note that as $T_m, T_n \in \mathrm{B}(X, Y)$, so $T_m - T_n \in \mathrm{B}(X,Y)$ also ]} \\ &\leq \frac{\varepsilon}{1+ \Vert x \Vert_X} \Vert x \Vert_X \qquad \mbox{[ by (1) above ]} \\ &< \varepsilon. \qquad \mbox{[ because $0 < \frac{\Vert x \Vert_X}{1+ \Vert x \Vert_X} < 1$, and $\varepsilon > 0$ ]} \end{align} $$ Thus, for every real number $\varepsilon > 0$, we can find a natural number $N_x$ such that $$\left\Vert T_m(x) - T_n(x) \right\Vert_Y < \varepsilon \ \mbox{ for any natural numbers } m \mbox{ and } n \mbox{ such that } m > N_x \mbox{ and } n > N_x.$$ Therefore, the sequence $\left( T_n (x) \right)_{n \in \mathbb{N}}$ is a Cauchy sequence in $Y$, and since $Y$ is a Banach space (i.e. $Y$ is complete with respect to the metric induced by the norm on $Y$), the sequence $\left( T_n (x) \right)_{n \in \mathbb{N}}$ converges in $Y$. Let's put $$T(x) \colon= \lim_{n \to \infty} T_n(x) \ \mbox{ in } Y. \tag{2} $$ As $x \in X$ was arbitrary, so we have a mapping $T \colon X \to Y$.

We first show that this mapping $T$ is linear. The domain of $T$ is $X$, which being a normed space is of course a vector space. Let $u$ and $v$ be some arbitrary points of $X$, and let $\alpha$ and $\beta$ be some arbitrary scalars. Then we note that $$ \begin{align} T \left( \alpha u + \beta v \right) &= \lim_{n \to \infty} T_n \left( \alpha u + \beta v \right) \ \mbox{ in } Y \qquad \mbox{[ using (2) above ]} \\ &= \lim_{n \to \infty} \left( \alpha T_n(u) + \beta T_n (v) \right) \ \mbox{ in } Y \qquad \mbox{[ since each $T_n$ is linear ]} \\ &= \alpha \lim_{n \to \infty} T_n(u) \ \mbox{ in } Y + \beta \lim_{n \to \infty} T_n(v) \ \mbox{ in } Y \\ & \qquad \mbox{[ using the results on limits of sums and scalar multiples of} \\ & \qquad \qquad \mbox{ convergent sequences in normed sapces ]} \\ &= \alpha T(u) + \beta T(v). \qquad \mbox{[ using (2) again ]} \end{align} $$ Hence $T$ is linear.

We now show that $T$ is bounded. For this we first show that $T_n - T$ is bounded for some $n \in \mathbb{N}$.

As $\left( T_n \right)_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathrm{B}(X,Y)$, so we can find a natural number $N$ such that $$ \left\Vert T_m - T_n \right\Vert_{\mathrm{B}(X,Y)} < \frac{\varepsilon}{4} \ \mbox{ for any natural numbers } m, n > N. \tag{3} $$ Therefore for all $n \in \mathbb{N}$ such that $n > N$ and for all $x \in X$, we have $$ \begin{align} \left\Vert T_m (x) - T_n(x) \right\Vert_Y &= \left\Vert \left( T_m - T_n \right)(x) \right\Vert_Y \\ &\leq \left\Vert T_m - T_n \right\Vert_{\mathrm{B}(X,Y)} \Vert x \Vert_X \\ & \ \ \ \mbox{[ using a property of bounded linear operators; } \\ & \qquad \mbox{ note that as $T_m, T_n \in \mathrm{B}(X,Y)$, so $T_m - T_n \in \mathrm{B}(X,Y)$ also ]} \\ &\leq \frac{\varepsilon}{4} \Vert x \Vert_X. \qquad \mbox{[ using (3) above ]} \end{align} $$ Thus we have shown that for any natural numbers $m, n > N$ and for all $x \in X$, we have $$ \left\Vert T_m (x) - T_n(x) \right\Vert_Y \leq \frac{\varepsilon}{4} \Vert x \Vert_X. \tag{4} $$ Therefore, for any natural number $n > N$ and for all $x \in X$, we obtain $$ \begin{align} \left\Vert T(x) - T_n(x) \right\Vert_Y &= \left\Vert \lim_{m \to \infty} T_m(x) - T_n(x) \right\Vert_Y \\ & \mbox{[ using (2); in (4), we let $m \to \infty$, keeping $n>N$ fixed ]} \\ &= \left\Vert \lim_{m \to \infty} T_m(x) - \lim_{m \to \infty} T_n(x) \right\Vert_Y \\ & \ \ \ \mbox{[ note that $T_n(x)$ remains unchanged as $m \to \infty$ ]} \\ &= \left\Vert \lim_{m \to \infty} \left( T_m(x) - T_n(x) \right) \right\Vert_Y \\ &= \lim_{m \to \infty} \left\Vert T_m(x) - T_n (x) \right\Vert_Y \\ & \ \ \ \mbox{[ if a sequence $\left(u_n\right)_{n\in\mathbb{N}}$ converges to $u$ in a normed space $V$, } \\ & \qquad \mbox{ then the sequence $\left( \left\Vert u_n \right\Vert_V \right)_{n\in\mathbb{N}}$ of real numbers converges to $\Vert u \Vert_V$ ]} \\ &\leq \frac{\varepsilon}{4} \Vert x \Vert_X. \ \ \mbox{[ using (4) ]} \end{align} $$ Thus we have shown that for any natural number $n > N$ and for all $x \in X$, $$ \left\Vert T(x) - T_n(x) \right\Vert_Y \leq \frac{\varepsilon}{4} \Vert x \Vert_X. \tag{5} $$ For each $n \in \mathbb{N}$, the mapping $T_n \colon X \to Y$ is a bounded linear operator; so there exists a positive real number $c_n$ such that, for all $x \in X$, the following holds:
$$ \left\Vert T_n(x) \right\Vert_Y \leq c_n \Vert x \Vert_X. \tag{6} $$ Thus from (5) and (6), we can conclude that, for all $x \in X$, $$ \begin{align} \left\Vert T(x) \right\Vert_Y &= \left\Vert T(x) - T_{N+1}(x) + T_{N+1}(x) \right\Vert_Y \\ &\leq \left\Vert T(x) - T_{N+1}(x) \right\Vert_Y + \left\Vert T_{N+1}(x) \right\Vert_Y \\ &\leq \frac{\varepsilon}{4} \Vert x \Vert_X + c_{N+1} \Vert x \Vert_X \ \mbox{[ using (5) and (6) ]} \\ &= \left( \frac{\varepsilon}{4} + c_{N+1} \right) \Vert x \Vert_X, \end{align} $$ which shows that the linear mapping $T \colon X \to Y$ is also bounded; that is, $T \in \mathrm{B}(X,Y)$.

Finally, from (5), we also see that, for any natural number $n > N$ and for all $x \in X$ such that $x \neq \mathbf{0}_X$, $$ \frac{\left\Vert \left( T_n - T \right)(x) \right\Vert_Y }{\Vert x \Vert_X } = \frac{\left\Vert T_n(x) - T (x) \right\Vert_Y }{\Vert x \Vert_X } \leq \frac{\varepsilon}{4}, $$ which implies that, for any natural number $n > N$, the real number $\varepsilon/4$ is an upper bound for the set $$ \left\{ \ \frac{\left\Vert \left( T_n - T \right)(x) \right\Vert_Y }{\Vert x \Vert_X } \ \colon \ x \in X, x \neq \mathbf{0} \ \right\} \subset \mathbb{R}, $$ and so $$ \left\Vert T_n - T \right\Vert_{\mathrm{B}(X,Y)} = \sup \left\{ \ \frac{\left\Vert \left( T_n - T \right)(x) \right\Vert_Y }{\Vert x \Vert_X } \ \colon \ x \in X, x \neq \mathbf{0} \ \right\} \leq \frac{\varepsilon}{4}.$$ Thus, for every real number $\varepsilon > 0$, we can find a natural number $N$ such that $$ \left\Vert T_n - T \right\Vert_{\mathrm{B}(X,Y)} < \varepsilon \ \mbox{ for any natural number } n > N. $$ Therefore our original Cauchy sequence $\left( T_n \right)_{n \in \mathbb{N} }$ converges to $T$ in $\mathrm{B}(X,Y)$.

But $\left( T_n \right)_{n \in \mathbb{N} }$ was an arbitrary Cauchy sequence in $\mathrm{B}(X,Y)$. Thus every Cauchy sequence in $\mathrm{B}(X,Y)$ is convergent. Hence $\mathrm{B}(X,Y)$ is a Banach space.

Is this proof good enough? If so, then is it any better than the proof by Kreyszig?

Is my presentation of this proof any simpler than that of Kreyszig, especially at the beginner level?

If not, then where have I erred? What is lacking in my proof (apart from brevity of course)?

The Kreyszig's text is available at the following link.

http://www-personal.acfr.usyd.edu.au/spns/cdm/resources/Kreyszig%20-%20Introductory%20Functional%20Analysis%20with%20Applications.pdf

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    $\begingroup$ I haven't read your proof but I can certainly answer the second question you posed. I don't see what is not "simple" about Kreyszig's proof -- it does the obvious thing at each step. If anything, your proof (at a cursory glance) makes the statement look like it requires some technical epsilon-pushing when in reality it is definition pushing. $\endgroup$ – user217285 May 31 '17 at 6:20
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    $\begingroup$ For starters, your conclusion that $T$ is bounded is unwarranted because $1+c_{1+M_x}$ should be independent of $x$. $\endgroup$ – user384138 May 31 '17 at 6:41
  • $\begingroup$ @Nitin I've just editted my post to correct an error. So can you please take time going through my proof in detail and comparing it with Kreyszig's? I think the latter has a problem, similar to the one pointed out by Open Ball. in my original proof. $\endgroup$ – Saaqib Mahmood May 31 '17 at 19:11
  • $\begingroup$ @OpenBall can you please take a look at my proof now? I've editted my original post so as to correct the error that you pointed out. But please carefully go through Kreyszig's proof. I'm he too has made a similar mistake when he introduces $\tilde({\varepsilon}$ and $\varepsilon_x$. Can you please have a look? $\endgroup$ – Saaqib Mahmood May 31 '17 at 19:14
  • $\begingroup$ The proof is correct now, but I don't thinks it's better than Kreyszig's proof. That proof seemed simple enough to understand. Why did you want to modify Kreyszig's proof? In essence they are the same, yours just covers every small detail which shouldn't be necessary at this level. $\endgroup$ – Demophilus Jun 5 '17 at 12:41

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