[proofplan] We verify that a bounded [linear operator](/page/Linear%20Map) $T: D \to Y$ on a [dense](/page/Dense%20Subset) subspace of a [normed space](/page/Normed%20Vector%20Space) is uniformly [continuous](/page/Continuity), then apply the [Continuous Extension from Dense Subsets](/theorems/964) to obtain a unique continuous extension $\bar{T}: X \to Y$. The completeness of $Y$ (a [Banach space](/page/Banach%20Space)) supplies the hypothesis that the target is a complete [metric space](/page/Metric%20Space). It remains to verify that the extension $\bar{T}$ is linear and that $\|\bar{T}\| = \|T\|$. [/proofplan] [step:Verify that $T$ is uniformly continuous on $D$] The operator $T: D \to Y$ is bounded, so there exists a constant $M \geq 0$ with $\|Tx\|_Y \leq M\|x\|_X$ for all $x \in D$. For any $x_1, x_2 \in D$, linearity gives \begin{align*} \|Tx_1 - Tx_2\|_Y = \|T(x_1 - x_2)\|_Y \leq M\|x_1 - x_2\|_X. \end{align*} Hence $T$ is Lipschitz [continuous](/page/Continuity) with Lipschitz constant $M$. In particular, $T$ is uniformly continuous on $D$: for every $\varepsilon > 0$, setting $\delta = \varepsilon / (M + 1)$ ensures $\|Tx_1 - Tx_2\|_Y < \varepsilon$ whenever $\|x_1 - x_2\|_X < \delta$. [guided] We need to verify uniform continuity of $T: D \to Y$ because this is a hypothesis of the [Continuous Extension from Dense Subsets](/theorems/964). Boundedness of $T$ means the operator norm $\|T\| = \sup_{x \in D, \, x \neq 0} \|Tx\|_Y / \|x\|_X$ is finite; let $M = \|T\|$. For any $x_1, x_2 \in D$, the difference $x_1 - x_2$ lies in $D$ since $D$ is a subspace. By linearity of $T$ and the operator norm bound, \begin{align*} \|Tx_1 - Tx_2\|_Y = \|T(x_1 - x_2)\|_Y \leq M\|x_1 - x_2\|_X. \end{align*} This is a Lipschitz condition, which is strictly stronger than uniform continuity. Given $\varepsilon > 0$, choose $\delta = \varepsilon / (M + 1)$ (we add $1$ to handle the case $M = 0$). Then $\|x_1 - x_2\|_X < \delta$ implies $\|Tx_1 - Tx_2\|_Y \leq M\delta = M\varepsilon/(M+1) < \varepsilon$. Why does linearity matter here? A continuous (but nonlinear) map on a [dense subset](/page/Dense%20Subset) need not be uniformly continuous — consider $f: \mathbb{Q} \to \mathbb{R}$, $q \mapsto q^2$, which is continuous but not uniformly continuous on all of $\mathbb{Q}$. The linearity of $T$ converts the pointwise bound $\|Tx\|_Y \leq M\|x\|_X$ into the *difference* bound $\|Tx_1 - Tx_2\|_Y \leq M\|x_1 - x_2\|_X$, which is the Lipschitz condition. [/guided] [/step] [step:Apply the Continuous Extension theorem to obtain $\bar{T}: X \to Y$] We verify the hypotheses of the [Continuous Extension from Dense Subsets](/theorems/964): 1. $(X, \|\cdot\|_X)$ is a [metric space](/page/Metric%20Space) (with the metric induced by the norm). 2. $D \subset X$ is a [dense](/page/Dense%20Subset) subspace by hypothesis. 3. $(Y, \|\cdot\|_Y)$ is a complete metric space, since $Y$ is a [Banach space](/page/Banach%20Space). 4. $T: D \to Y$ is uniformly [continuous](/page/Continuity), as established in the previous step. All four hypotheses are satisfied. By the [Continuous Extension from Dense Subsets](/theorems/964), there exists a unique uniformly continuous map $\bar{T}: X \to Y$ satisfying $\bar{T}|_D = T$. [guided] The [Continuous Extension from Dense Subsets](/theorems/964) requires four conditions: (i) $(X, d_X)$ is a metric space, (ii) $D \subset X$ is dense, (iii) $(Y, d_Y)$ is a complete metric space, and (iv) $f: D \to Y$ is uniformly continuous. We verify each. For (i): $X$ is a [normed vector space](/page/Normed%20Vector%20Space), so $d_X(x_1, x_2) = \|x_1 - x_2\|_X$ defines a metric. For (ii): $D \subset X$ is dense by the hypothesis of the BLT Theorem. For (iii): $Y$ is a Banach space by hypothesis, meaning $(Y, \|\cdot\|_Y)$ is a complete normed vector space. The induced metric $d_Y(y_1, y_2) = \|y_1 - y_2\|_Y$ is therefore complete. For (iv): $T: D \to Y$ is uniformly continuous, as shown in the previous step. The theorem therefore yields a unique uniformly continuous map $\bar{T}: X \to Y$ with $\bar{T}(x) = T(x)$ for all $x \in D$. Note that completeness of $Y$ is essential here. The extension is constructed by taking limits: for $x \in X \setminus D$, one chooses a sequence $(x_k)_{k=1}^\infty$ in $D$ with $x_k \to x$, and the uniform continuity of $T$ ensures $(Tx_k)_{k=1}^\infty$ is Cauchy in $Y$. Without completeness of $Y$, this [Cauchy sequence](/page/Cauchy%20Sequence) need not converge, and the extension cannot be defined. [/guided] [/step] [step:Verify that $\bar{T}$ is linear] We show $\bar{T}(\lambda x + \mu z) = \lambda \bar{T}(x) + \mu \bar{T}(z)$ for all $x, z \in X$ and all scalars $\lambda, \mu$. Fix $x, z \in X$ and scalars $\lambda, \mu$. Since $D$ is [dense](/page/Dense%20Subset) in $X$, there exist sequences $(x_k)_{k=1}^\infty$ and $(z_k)_{k=1}^\infty$ in $D$ with $x_k \to x$ and $z_k \to z$ in $X$. Then $\lambda x_k + \mu z_k \in D$ (since $D$ is a subspace) and $\lambda x_k + \mu z_k \to \lambda x + \mu z$ in $X$. Since $\bar{T}$ is [continuous](/page/Continuity) (being uniformly continuous) and $T$ is linear on $D$, \begin{align*} \bar{T}(\lambda x + \mu z) &= \lim_{k \to \infty} \bar{T}(\lambda x_k + \mu z_k) = \lim_{k \to \infty} T(\lambda x_k + \mu z_k) \\ &= \lim_{k \to \infty} \bigl(\lambda T(x_k) + \mu T(z_k)\bigr) = \lambda \lim_{k \to \infty} T(x_k) + \mu \lim_{k \to \infty} T(z_k) \\ &= \lambda \bar{T}(x) + \mu \bar{T}(z). \end{align*} The second equality uses $\bar{T}|_D = T$. The fourth equality uses continuity of scalar multiplication and vector addition in $Y$. Therefore $\bar{T}$ is linear. [guided] The Continuous Extension theorem guarantees that $\bar{T}$ is uniformly continuous, but it says nothing about algebraic structure — linearity must be verified separately. The strategy is to approximate arbitrary elements of $X$ by elements of $D$ and use the continuity of $\bar{T}$ together with the linearity of $T$ on $D$. Fix $x, z \in X$ and scalars $\lambda, \mu$. Since $D$ is dense in $X$, choose sequences $(x_k)_{k=1}^\infty$ and $(z_k)_{k=1}^\infty$ in $D$ with $x_k \to x$ and $z_k \to z$. Since $D$ is a *subspace* (not merely a subset), the linear combination $\lambda x_k + \mu z_k$ lies in $D$ for every $k$. Moreover, by continuity of the vector space operations in $X$, \begin{align*} \lambda x_k + \mu z_k \to \lambda x + \mu z \quad \text{in } X. \end{align*} Now we compute $\bar{T}(\lambda x + \mu z)$ by using the sequential characterisation of continuity. Since $\bar{T}: X \to Y$ is continuous and $\lambda x_k + \mu z_k \to \lambda x + \mu z$, \begin{align*} \bar{T}(\lambda x + \mu z) = \lim_{k \to \infty} \bar{T}(\lambda x_k + \mu z_k). \end{align*} Since $\lambda x_k + \mu z_k \in D$ and $\bar{T}|_D = T$, this equals $\lim_{k \to \infty} T(\lambda x_k + \mu z_k)$. Applying the linearity of $T$ on $D$, \begin{align*} \lim_{k \to \infty} T(\lambda x_k + \mu z_k) = \lim_{k \to \infty} \bigl(\lambda T(x_k) + \mu T(z_k)\bigr). \end{align*} Vector addition and scalar multiplication are continuous operations in the [normed space](/page/Normed%20Vector%20Space) $Y$, so the limit distributes: \begin{align*} \lim_{k \to \infty} \bigl(\lambda T(x_k) + \mu T(z_k)\bigr) = \lambda \lim_{k \to \infty} T(x_k) + \mu \lim_{k \to \infty} T(z_k) = \lambda \bar{T}(x) + \mu \bar{T}(z). \end{align*} The last equality again uses continuity of $\bar{T}$ and the fact that $\bar{T}|_D = T$. Combining, $\bar{T}(\lambda x + \mu z) = \lambda \bar{T}(x) + \mu \bar{T}(z)$, so $\bar{T}$ is linear. This argument illustrates a general principle: algebraic identities that hold on a dense subset and involve continuous operations extend automatically to the closure. [/guided] [/step] [step:Verify that $\|\bar{T}\| = \|T\|$] Since $\bar{T}|_D = T$, we have $\|T\| \leq \|\bar{T}\|$ (the [supremum](/page/Supremum%20and%20Infimum) over a larger set is at least as large). For the reverse inequality, fix $x \in X$ with $x \neq 0$. Choose a sequence $(x_k)_{k=1}^\infty$ in $D$ with $x_k \to x$. By [continuity](/page/Continuity) of $\bar{T}$ and of the norm, \begin{align*} \|\bar{T}(x)\|_Y = \lim_{k \to \infty} \|T(x_k)\|_Y \leq \lim_{k \to \infty} \|T\| \cdot \|x_k\|_X = \|T\| \cdot \|x\|_X. \end{align*} Therefore $\|\bar{T}\| = \sup_{x \neq 0} \|\bar{T}(x)\|_Y / \|x\|_X \leq \|T\|$. Combining both inequalities, $\|\bar{T}\| = \|T\|$. [guided] The operator norm of $\bar{T}$ is defined as $\|\bar{T}\| = \sup_{x \in X, \, x \neq 0} \|\bar{T}(x)\|_Y / \|x\|_X$, while $\|T\| = \sup_{x \in D, \, x \neq 0} \|T(x)\|_Y / \|x\|_X$. We need both directions of the inequality. **Lower bound $\|T\| \leq \|\bar{T}\|$.** Since $D \subset X$ and $\bar{T}|_D = T$, the supremum defining $\|\bar{T}\|$ is taken over a set that includes all nonzero elements of $D$. Therefore \begin{align*} \|\bar{T}\| = \sup_{x \in X, \, x \neq 0} \frac{\|\bar{T}(x)\|_Y}{\|x\|_X} \geq \sup_{x \in D, \, x \neq 0} \frac{\|T(x)\|_Y}{\|x\|_X} = \|T\|. \end{align*} **Upper bound $\|\bar{T}\| \leq \|T\|$.** Fix any $x \in X$ with $x \neq 0$. Since $D$ is [dense](/page/Dense%20Subset) in $X$, there exists a sequence $(x_k)_{k=1}^\infty$ in $D$ with $x_k \to x$. Since $\bar{T}$ is continuous ($\bar{T}$ is Lipschitz with constant $\|T\|$ on $D$, and the extension preserves the Lipschitz constant), $\bar{T}(x_k) \to \bar{T}(x)$ in $Y$. By continuity of the norm in $Y$, \begin{align*} \|\bar{T}(x)\|_Y = \lim_{k \to \infty} \|\bar{T}(x_k)\|_Y = \lim_{k \to \infty} \|T(x_k)\|_Y. \end{align*} Since $\|T(x_k)\|_Y \leq \|T\| \cdot \|x_k\|_X$ for each $k$ (by the definition of the operator norm of $T$ on $D$), passing to the limit gives \begin{align*} \|\bar{T}(x)\|_Y \leq \|T\| \cdot \lim_{k \to \infty} \|x_k\|_X = \|T\| \cdot \|x\|_X. \end{align*} Since this holds for every $x \in X$, we conclude $\|\bar{T}\| \leq \|T\|$. Combining, $\|\bar{T}\| = \|T\|$. The extension preserves the operator norm exactly — no information is lost or gained by extending from the dense subspace to the full space. [/guided] [/step]