[proofplan] The restricted norm inherits the norm axioms from the ambient norm because $Y$ is a linear subspace of $V$. To prove completeness, take an arbitrary [Cauchy sequence](/page/Cauchy%20Sequence) in $Y$ and view it as a Cauchy sequence in $V$, where it converges because $V$ is Banach. Closedness of $Y$ in the norm topology then forces the ambient limit to lie in $Y$, and the equality of the two norms on $Y$ gives convergence in the restricted norm. [/proofplan]

[step:Verify that the restricted norm makes $Y$ a normed vector space] Since $Y$ is a linear subspace of $V$ over $\mathbb{F}$, the vector addition and scalar multiplication on $Y$ are the restrictions of those on $V$. Define $\|\cdot\|_Y:Y\to[0,\infty)$ by $y\mapsto \|y\|_V$. For $y\in Y$, positivity and definiteness follow from the corresponding properties of $\|\cdot\|_V$: $\|y\|_Y\geq 0$, and $\|y\|_Y=0 \iff \|y\|_V=0 \iff y=0$. For $\lambda\in\mathbb{F}$ and $y\in Y$, homogeneity follows from homogeneity in $V$: $\|\lambda y\|_Y=\|\lambda y\|_V=|\lambda|\,\|y\|_V=|\lambda|\,\|y\|_Y$. For $y,z\in Y$, the triangle inequality follows from the triangle inequality in $V$: $\|y+z\|_Y=\|y+z\|_V\leq \|y\|_V+\|z\|_V=\|y\|_Y+\|z\|_Y$. Thus $(Y,\|\cdot\|_Y)$ is a [normed vector space](/page/Normed%20Vector%20Space) over $\mathbb{F}$. [/step]

[step:View a Cauchy sequence in $Y$ as a Cauchy sequence in $V$]Let $(y_n)_{n=1}^{\infty}$ be an arbitrary Cauchy sequence in $(Y,\|\cdot\|_Y)$. By the definition of Cauchy sequence, for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that, for all $m,n\geq N$, $\|y_n-y_m\|_Y<\varepsilon$. Since $\|\cdot\|_Y$ is the restriction of $\|\cdot\|_V$ to $Y$, and since $y_n-y_m\in Y$, this is exactly $\|y_n-y_m\|_V<\varepsilon$. Therefore $(y_n)_{n=1}^{\infty}$ is a Cauchy sequence in $(V,\|\cdot\|_V)$.[/step]

[guided]We start with an arbitrary Cauchy sequence in the subspace because completeness is a statement about every Cauchy sequence having a limit inside the space under consideration. Let $(y_n)_{n=1}^{\infty}$ be a Cauchy sequence in $(Y,\|\cdot\|_Y)$. This means that for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that, whenever $m,n\geq N$, $\|y_n-y_m\|_Y<\varepsilon$. The point of using the restricted norm is that distances between points of $Y$ are unchanged when we regard those same points as elements of $V$. Since $Y$ is a linear subspace, $y_n-y_m\in Y$ for all $m,n\in\mathbb{N}$. By the definition of the restricted norm, $\|y_n-y_m\|_Y=\|y_n-y_m\|_V$. Substituting this equality into the Cauchy estimate gives, for every $\varepsilon>0$, an index $N\in\mathbb{N}$ such that for all $m,n\geq N$, $\|y_n-y_m\|_V<\varepsilon$. This is precisely the statement that $(y_n)_{n=1}^{\infty}$ is a Cauchy sequence in the ambient [normed space](/page/Normed%20Space) $(V,\|\cdot\|_V)$.[/guided]

[step:Use completeness of $V$ to obtain an ambient limit] Since $(V,\|\cdot\|_V)$ is Banach, every Cauchy sequence in $V$ converges in $V$. Therefore there exists $v\in V$ such that $\lim_{n\to\infty}\|y_n-v\|_V=0$. Equivalently, $y_n\to v$ in the norm topology of $V$. [/step]

[step:Use closedness of $Y$ to keep the limit inside the subspace] The sequence $(y_n)_{n=1}^{\infty}$ lies in $Y$, and it converges to $v$ in $V$. Since $Y$ is closed in the norm topology of $V$, the [sequential characterization of closed sets in normed spaces](/theorems/9985) implies that $v\in Y$. [/step]

[step:Conclude convergence in the restricted norm] Because $v\in Y$ and $y_n\in Y$ for every $n\in\mathbb{N}$, we have $y_n-v\in Y$ for every $n\in\mathbb{N}$. By the definition of the restricted norm, $\|y_n-v\|_Y=\|y_n-v\|_V$. Since $\|y_n-v\|_V\to 0$, it follows that $\|y_n-v\|_Y\to 0$. Thus every Cauchy sequence in $(Y,\|\cdot\|_Y)$ converges to an element of $Y$. Hence $(Y,\|\cdot\|_Y)$ is complete, and therefore it is a [Banach space](/page/Banach%20Space) over $\mathbb{F}$. [/step]