[proofplan] We prove closedness by the [sequential characterization of closed sets in normed spaces](/theorems/9985). Take an arbitrary sequence in $Y$ that converges in the ambient space $V$; ambient convergence makes it Cauchy, and the restricted norm makes the same sequence Cauchy in $Y$. Completeness of $Y$ supplies a limit in $Y$, and [uniqueness of limits](/theorems/625) in the ambient norm identifies that limit with the original ambient limit. [/proofplan]

[step:Reduce closedness of $Y$ to limits of convergent sequences from $Y$] Since $(V,\|\cdot\|_V)$ is a [Banach space](/page/Banach%20Space), it is a [normed space](/page/Normed%20Space). By [citetheorem:9985], it is enough to prove the following sequential closure property: whenever $(y_n)_{n\in\mathbb{N}}$ is a sequence in $Y$ and $y_n\to v$ in $V$ for some $v\in V$, then $v\in Y$. [/step]

[step:Use completeness of $Y$ to obtain a limit inside the subspace]Let $(y_n)_{n\in\mathbb{N}}$ be a sequence in $Y$, and let $v\in V$ satisfy $y_n\to v$ in $V$. We first show that $(y_n)_{n\in\mathbb{N}}$ is Cauchy in $(Y,\|\cdot\|_Y)$. Let $\varepsilon>0$. Since $y_n\to v$ in $V$, there exists $N\in\mathbb{N}$ such that for every $n\geq N$, \begin{align*} \|y_n-v\|_V<\frac{\varepsilon}{2}. \end{align*} Hence, for all $m,n\geq N$, the triangle inequality in $V$ gives \begin{align*} \|y_n-y_m\|_Y=\|y_n-y_m\|_V\leq \|y_n-v\|_V+\|v-y_m\|_V<\varepsilon. \end{align*} Thus $(y_n)_{n\in\mathbb{N}}$ is Cauchy in $Y$. Since $(Y,\|\cdot\|_Y)$ is Banach, there exists $y\in Y$ such that $y_n\to y$ in $Y$.[/step]

[guided]Let $(y_n)_{n\in\mathbb{N}}$ be a sequence in $Y$, and suppose that there is some $v\in V$ such that $y_n\to v$ in the ambient norm $\|\cdot\|_V$. The aim is to show that this ambient limit $v$ actually belongs to $Y$. The first point is that convergence in $V$ forces the sequence to be Cauchy in $V$. We verify this directly because the same estimate is what transfers the argument to $Y$. Let $\varepsilon>0$. Since $y_n\to v$ in $V$, there exists $N\in\mathbb{N}$ such that for every $n\geq N$, \begin{align*} \|y_n-v\|_V<\frac{\varepsilon}{2}. \end{align*} Therefore, whenever $m,n\geq N$, the triangle inequality in the normed space $V$ gives \begin{align*} \|y_n-y_m\|_V\leq \|y_n-v\|_V+\|v-y_m\|_V<\varepsilon. \end{align*} Now we use the fact that the norm on $Y$ is the restricted norm from $V$. Since $y_n,y_m\in Y$, the difference $y_n-y_m$ belongs to $Y$ because $Y$ is a linear subspace, and the restricted norm satisfies \begin{align*} \|y_n-y_m\|_Y=\|y_n-y_m\|_V. \end{align*} Thus the same estimate proves that $(y_n)_{n\in\mathbb{N}}$ is Cauchy in $(Y,\|\cdot\|_Y)$. The hypothesis says precisely that this normed space is complete, so there exists an element $y\in Y$ such that $y_n\to y$ in the norm $\|\cdot\|_Y$.[/guided]

[step:Identify the subspace limit with the ambient limit] The convergence $y_n\to y$ in $Y$ means that for every $\varepsilon>0$ there exists $N_Y\in\mathbb{N}$ such that, for all $n\geq N_Y$, \begin{align*} \|y_n-y\|_Y<\varepsilon. \end{align*} Since $\|y_n-y\|_Y=\|y_n-y\|_V$, the same sequence also converges to $y$ in $V$. We now prove uniqueness of the ambient limit. Let $\varepsilon>0$. Since $y_n\to v$ in $V$ and $y_n\to y$ in $V$, there exists $N_V\in\mathbb{N}$ such that for all $n\geq N_V$, \begin{align*} \|v-y_n\|_V<\frac{\varepsilon}{2} \end{align*} and \begin{align*} \|y_n-y\|_V<\frac{\varepsilon}{2}. \end{align*} For such $n$, the triangle inequality gives \begin{align*} \|v-y\|_V\leq \|v-y_n\|_V+\|y_n-y\|_V<\varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, $\|v-y\|_V=0$, so $v=y$. Because $y\in Y$, it follows that $v\in Y$. [/step]

[step:Conclude that $Y$ is closed in $V$] We have shown that every sequence in $Y$ that converges in $V$ has its limit in $Y$. By the sequential characterization of closed sets in normed spaces, [citetheorem:9985], the subset $Y$ is closed in $V$ with respect to the norm topology induced by $\|\cdot\|_V$. [/step]