The proof constructs the limit functional pointwise, shows it is bounded, and then upgrades the pointwise convergence to operator-norm convergence. The completeness of the scalar field $\mathbb{F}$ is the only ingredient beyond the definition of the operator norm. **Step 1: Pointwise convergence.** Let $(T_n)_{n=1}^\infty \subset X^*$ be a [Cauchy sequence](/page/Cauchy%20Sequence) in the operator norm. For each fixed $v \in X$, the [sequence](/page/Sequence) $(T_n(v))_{n=1}^\infty \subset \mathbb{F}$ satisfies \begin{align*} |T_n(v) - T_m(v)| = |(T_n - T_m)(v)| \le \|T_n - T_m\|_{X^*} \|v\|_X, \end{align*} so it is Cauchy in $\mathbb{F}$. Since $\mathbb{F}$ is complete, the limit exists. Define $T: X \to \mathbb{F}$ by $T(v) := \lim_{n \to \infty} T_n(v)$. **Step 2: Linearity.** For any $v, w \in X$ and $\lambda, \mu \in \mathbb{F}$: \begin{align*} T(\lambda v + \mu w) = \lim_{n \to \infty} T_n(\lambda v + \mu w) = \lim_{n \to \infty} [\lambda T_n(v) + \mu T_n(w)] = \lambda T(v) + \mu T(w), \end{align*} using the linearity of each $T_n$ and the linearity of [limits](/page/Limit) in $\mathbb{F}$. **Step 3: Boundedness of $T$.** Since $(T_n)_n$ is Cauchy, it is bounded: $M := \sup_n \|T_n\|_{X^*} < \infty$. For any $v \in X$ with $\|v\|_X \le 1$: \begin{align*} |T(v)| = \lim_{n \to \infty} |T_n(v)| \le \sup_n \|T_n\|_{X^*} = M. \end{align*} Therefore $\|T\|_{X^*} \le M < \infty$, so $T \in X^*$. **Step 4: Convergence in the operator norm.** Let $\varepsilon > 0$. Since $(T_n)_n$ is Cauchy, there exists $N$ such that $\|T_n - T_m\|_{X^*} < \varepsilon/2$ for all $n, m \ge N$. Fix any $v \in X$ with $\|v\|_X \le 1$ and any $n \ge N$. Choose $m \ge N$ large enough that $|T_m(v) - T(v)| < \varepsilon/2$. Then: \begin{align*} |(T_n - T)(v)| \le |T_n(v) - T_m(v)| + |T_m(v) - T(v)| < \varepsilon/2 + \varepsilon/2 = \varepsilon. \end{align*} The choice of $N$ was independent of $v$, so $\|T_n - T\|_{X^*} = \sup_{\|v\| \le 1} |(T_n - T)(v)| \le \varepsilon$ for all $n \ge N$. Hence $T_n \to T$ in the operator norm.