[proofplan] We must show that every Cauchy sequence $(T_n)_{n=1}^\infty$ in $\mathcal{L}(X, Y)$ converges to a bounded linear operator $T \in \mathcal{L}(X, Y)$. The strategy has three stages: (1) For each fixed $x \in X$, use the operator-norm Cauchy condition to show that $(T_n x)_{n=1}^\infty$ is Cauchy in $Y$, then define $Tx$ as its limit using the completeness of $Y$. (2) Verify that $T$ is linear and bounded. (3) Show $\|T_n - T\|_{\mathcal{L}(X,Y)} \to 0$, which requires upgrading the pointwise convergence to uniform convergence over the unit ball. [/proofplan]

[step:Show that $(T_n x)_{n=1}^\infty$ is Cauchy in $Y$ for each fixed $x \in X$]Let $(T_n)_{n=1}^\infty$ be a Cauchy sequence in $(\mathcal{L}(X, Y), \|\cdot\|_{\mathcal{L}(X,Y)})$. Fix $x \in X$. For all $m, n \in \mathbb{N}$, \begin{align*} \|T_m x - T_n x\|_Y = \|(T_m - T_n)x\|_Y \le \|T_m - T_n\|_{\mathcal{L}(X,Y)} \cdot \|x\|_X. \end{align*} Given $\varepsilon > 0$, the Cauchy property of $(T_n)$ in $\mathcal{L}(X, Y)$ provides $N \in \mathbb{N}$ with $\|T_m - T_n\|_{\mathcal{L}(X,Y)} < \varepsilon / (\|x\|_X + 1)$ for all $m, n \ge N$. Then \begin{align*} \|T_m x - T_n x\|_Y < \frac{\varepsilon \|x\|_X}{\|x\|_X + 1} < \varepsilon \quad \text{for all } m, n \ge N. \end{align*} Hence $(T_n x)_{n=1}^\infty$ is a Cauchy sequence in $Y$.[/step]

[guided]The first task is to define the candidate limit operator. We cannot directly define $T$ as an abstract object — we must construct it pointwise. Fix any $x \in X$ and examine the sequence of images $(T_n x)_{n=1}^\infty$ in $Y$. Since $(T_n)$ is Cauchy in the operator norm, for every $\eta > 0$ there exists $N \in \mathbb{N}$ with $\|T_m - T_n\|_{\mathcal{L}(X,Y)} < \eta$ for all $m, n \ge N$. The key observation is that the operator norm controls the pointwise behaviour: \begin{align*} \|T_m x - T_n x\|_Y = \|(T_m - T_n)x\|_Y \le \|T_m - T_n\|_{\mathcal{L}(X,Y)} \cdot \|x\|_X. \end{align*} This follows directly from the definition of the operator norm: $\|Sx\|_Y \le \|S\|_{\mathcal{L}(X,Y)} \|x\|_X$ for any $S \in \mathcal{L}(X, Y)$, applied with $S = T_m - T_n$ (which is in $\mathcal{L}(X, Y)$ by the vector space structure). For the given $\varepsilon > 0$, choose $\eta = \varepsilon / (\|x\|_X + 1)$ (the $+1$ handles the case $x = 0$, though that case is vacuous since $T_n(0) = 0$ for all $n$). Then for $m, n \ge N$, \begin{align*} \|T_m x - T_n x\|_Y \le \|T_m - T_n\|_{\mathcal{L}(X,Y)} \cdot \|x\|_X < \frac{\varepsilon \|x\|_X}{\|x\|_X + 1} < \varepsilon. \end{align*} So $(T_n x)_{n=1}^\infty$ is Cauchy in $Y$. This is where the completeness of the codomain will be consumed: $Y$ is a Banach space, so every Cauchy sequence in $Y$ converges.[/guided]

[step:Define $T$ as the pointwise limit and verify linearity]Since $Y$ is a [Banach space](/page/Banach%20Space), every Cauchy sequence in $Y$ converges. For each $x \in X$, define \begin{align*} T: X &\to Y \\ x &\mapsto \lim_{n \to \infty} T_n x. \end{align*} **Linearity.** Let $x_1, x_2 \in X$ and $\lambda$ be a scalar. Then \begin{align*} T(\lambda x_1 + x_2) &= \lim_{n \to \infty} T_n(\lambda x_1 + x_2) \\ &= \lim_{n \to \infty} (\lambda T_n x_1 + T_n x_2) \\ &= \lambda \lim_{n \to \infty} T_n x_1 + \lim_{n \to \infty} T_n x_2 \\ &= \lambda T x_1 + T x_2, \end{align*} where the second equality uses the linearity of each $T_n$, and the third uses the linearity of the limit in the normed space $Y$ (the limits exist by the previous step). Hence $T$ is linear.[/step]

[guided]The completeness of $Y$ is invoked precisely here: for each $x \in X$, the sequence $(T_n x)_{n=1}^\infty$ is Cauchy in $Y$ (by the previous step), and since $Y$ is complete, this sequence converges to a unique limit in $Y$. We call this limit $Tx$. Note that this step would fail if $Y$ were merely a normed space rather than a Banach space. For example, if $Y = c_{00}$ (the space of eventually-zero real sequences with the sup norm), then $Y$ is not complete, and a Cauchy sequence in $\mathcal{L}(X, Y)$ could have pointwise limits that escape $Y$. The linearity of $T$ follows from the linearity of each $T_n$ and the continuity of the algebraic operations in $Y$. In detail, the norm limit preserves linear combinations: if $y_n \to y$ and $z_n \to z$ in $Y$, then $\lambda y_n + z_n \to \lambda y + z$. Applying this with $y_n = T_n x_1$ and $z_n = T_n x_2$ gives the conclusion.[/guided]

[step:Show $T$ is bounded using the uniform boundedness of Cauchy sequences]Every Cauchy sequence in a normed space is bounded: there exists $M > 0$ with $\|T_n\|_{\mathcal{L}(X,Y)} \le M$ for all $n \in \mathbb{N}$. (To see this, choose $N$ with $\|T_m - T_n\|_{\mathcal{L}(X,Y)} < 1$ for all $m, n \ge N$. Then $\|T_n\|_{\mathcal{L}(X,Y)} \le \|T_N\|_{\mathcal{L}(X,Y)} + 1$ for $n \ge N$, and we set $M = \max\{\|T_1\|, \ldots, \|T_{N-1}\|, \|T_N\| + 1\}$.) For any $x \in X$ with $\|x\|_X \le 1$, \begin{align*} \|Tx\|_Y = \left\|\lim_{n \to \infty} T_n x\right\|_Y = \lim_{n \to \infty} \|T_n x\|_Y \le \sup_{n \in \mathbb{N}} \|T_n x\|_Y \le \sup_{n \in \mathbb{N}} \|T_n\|_{\mathcal{L}(X,Y)} \le M, \end{align*} where the second equality uses the continuity of the norm $\|\cdot\|_Y: Y \to \mathbb{R}$. Taking the supremum over all $x \in X$ with $\|x\|_X \le 1$ gives $\|T\|_{\mathcal{L}(X,Y)} \le M < \infty$. Hence $T \in \mathcal{L}(X, Y)$.[/step]

[guided]Before proving operator-norm convergence, we must verify that $T$ is bounded — otherwise $T \notin \mathcal{L}(X, Y)$ and the statement of operator-norm convergence is meaningless. **Uniform bound on $(T_n)$.** Every Cauchy sequence in a metric space is bounded. We verify this explicitly for completeness. Choose $N \in \mathbb{N}$ with $\|T_m - T_n\|_{\mathcal{L}(X,Y)} < 1$ for all $m, n \ge N$ (using the Cauchy condition with $\varepsilon = 1$). Then for $n \ge N$, the triangle inequality gives $\|T_n\|_{\mathcal{L}(X,Y)} \le \|T_n - T_N\|_{\mathcal{L}(X,Y)} + \|T_N\|_{\mathcal{L}(X,Y)} < 1 + \|T_N\|_{\mathcal{L}(X,Y)}$. Setting $M = \max\{\|T_1\|_{\mathcal{L}(X,Y)}, \ldots, \|T_{N-1}\|_{\mathcal{L}(X,Y)}, \|T_N\|_{\mathcal{L}(X,Y)} + 1\}$ gives $\|T_n\|_{\mathcal{L}(X,Y)} \le M$ for all $n \in \mathbb{N}$. **$T$ is bounded.** The norm $\|\cdot\|_Y: Y \to \mathbb{R}$ is a continuous function (by the reverse triangle inequality $|\|y_1\|_Y - \|y_2\|_Y| \le \|y_1 - y_2\|_Y$). Therefore, if $T_n x \to Tx$ in $Y$, then $\|T_n x\|_Y \to \|Tx\|_Y$. For $\|x\|_X \le 1$, each term satisfies $\|T_n x\|_Y \le \|T_n\|_{\mathcal{L}(X,Y)} \le M$. Since the limit of a sequence bounded by $M$ is at most $M$, we conclude $\|Tx\|_Y \le M$. Taking the supremum: $\|T\|_{\mathcal{L}(X,Y)} = \sup_{\|x\|_X \le 1} \|Tx\|_Y \le M$.[/guided]

[step:Upgrade pointwise convergence to operator-norm convergence $\|T_n - T\|_{\mathcal{L}(X,Y)} \to 0$]Let $\varepsilon > 0$. By the Cauchy property of $(T_n)$, choose $N \in \mathbb{N}$ with $\|T_m - T_n\|_{\mathcal{L}(X,Y)} < \varepsilon$ for all $m, n \ge N$. Fix $n \ge N$ and any $x \in X$ with $\|x\|_X \le 1$. For every $m \ge N$, \begin{align*} \|(T_n - T_m)x\|_Y \le \|T_n - T_m\|_{\mathcal{L}(X,Y)} \cdot \|x\|_X \le \|T_n - T_m\|_{\mathcal{L}(X,Y)} < \varepsilon. \end{align*} Taking $m \to \infty$, the continuity of the norm gives \begin{align*} \|(T_n - T)x\|_Y = \lim_{m \to \infty} \|(T_n - T_m)x\|_Y \le \varepsilon. \end{align*} Since this bound holds for every $x \in X$ with $\|x\|_X \le 1$, taking the supremum yields \begin{align*} \|T_n - T\|_{\mathcal{L}(X,Y)} = \sup_{\substack{x \in X \\ \|x\|_X \le 1}} \|(T_n - T)x\|_Y \le \varepsilon \quad \text{for all } n \ge N. \end{align*} Since $\varepsilon > 0$ was arbitrary, $\|T_n - T\|_{\mathcal{L}(X,Y)} \to 0$ as $n \to \infty$. This shows that $(T_n)$ converges to $T$ in $\mathcal{L}(X, Y)$, completing the proof that $\mathcal{L}(X, Y)$ is a Banach space.[/step]

[guided]This is the most delicate step. We have pointwise convergence $T_n x \to Tx$ for each fixed $x$, but we need the stronger statement $\|T_n - T\|_{\mathcal{L}(X,Y)} \to 0$, which requires uniform convergence over the unit ball $\{x \in X : \|x\|_X \le 1\}$. The key technique is the **Cauchy-to-limit transfer**: we use the Cauchy estimate $\|T_m - T_n\| < \varepsilon$ (valid for $m, n \ge N$), hold $n$ fixed, and let $m \to \infty$. The pointwise convergence $T_m x \to Tx$ allows us to replace $T_m$ by $T$ in the limit, but we must be careful — the convergence $T_m x \to Tx$ is in the $Y$-norm for each fixed $x$, and we need the resulting bound to be uniform in $x$. Here is the detailed argument. Fix $\varepsilon > 0$ and choose $N$ from the Cauchy condition. Fix any $n \ge N$ and any $x$ with $\|x\|_X \le 1$. For every $m \ge N$, we have the pointwise estimate: \begin{align*} \|(T_n - T_m)x\|_Y \le \|T_n - T_m\|_{\mathcal{L}(X,Y)} < \varepsilon. \end{align*} Now let $m \to \infty$. Since $T_m x \to Tx$ in $Y$, and the map $y \mapsto \|T_n x - y\|_Y$ is continuous in $y$ (by the triangle inequality), we obtain \begin{align*} \|(T_n - T)x\|_Y = \|T_n x - Tx\|_Y = \lim_{m \to \infty} \|T_n x - T_m x\|_Y \le \varepsilon. \end{align*} The inequality $\le \varepsilon$ (rather than $< \varepsilon$) arises because limits preserve non-strict inequalities. This bound holds for every $x$ with $\|x\|_X \le 1$ — this uniformity comes not from the pointwise convergence but from the operator-norm Cauchy estimate, which was already uniform in $x$. Taking the supremum: \begin{align*} \|T_n - T\|_{\mathcal{L}(X,Y)} = \sup_{\|x\|_X \le 1} \|(T_n - T)x\|_Y \le \varepsilon \quad \text{for all } n \ge N. \end{align*} Since $\varepsilon > 0$ was arbitrary, $T_n \to T$ in $\mathcal{L}(X, Y)$. The Cauchy sequence $(T_n)$ converges in $\mathcal{L}(X, Y)$, so $\mathcal{L}(X, Y)$ is complete.[/guided]