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Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.

The volume ratio of a $d$-dimensional normed vector space is $\mathrm{vr}(E)=(\mathrm{vol}(B_E)/\mathrm{vol}(\mathcal{E}){)}^{1/d}$, where $\mathcal{E}$ is its John ellipsoid. It is unchanged by invertible linear changes of coordinates. In John position, polar integration gives $\mathrm{vr}(E{)}^d={\int }_{S^{d-1}}\Vert u{\Vert }_E^{-d}d\sigma (u)$, with normalized spherical measure. Consequently $\sigma \{\Vert u{\Vert }_E\le t\}\le (\mathrm{vr}(E)t{)}^d$.

If the relative weak topology on the unit ball of a normed vector space is first countable at zero, choose a countable collection of finite families of continuous linear functionals defining a local base. For any $f\in E^{\prime }$ and $ϵ>0$, one family $F$ has common kernel $M$ on which $\Vert f{|}_M\Vert \le ϵ$. Extend that restriction to $h\in E^{\prime }$ with $\Vert h\Vert \le ϵ$ using the Hahn-Banach theorem. Then $f-h$ vanishes on $M$, so it belongs to $\mathrm{span}F$ by finite-dimensional factorization. Consequently the countable union of these families has norm-dense linear span in $E^{\prime }$. The converse is the metrization of the weak topology on a bounded set construction from a norm-dense sequence in $E^{\prime }$. In particular, separability of $E$ alone does not suffice: ${\ell }^1$ has nonseparable dual ${\ell }^{\infty }$.

The weak limit of a sequence in a normed vector space is an element at which every continuous linear functional has the limit of its values along the sequence. Thus the definition is precisely weak convergence to that element.

A subset $D$ of a normed vector space is weakly bounded when ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. It is norm bounded: its canonical embedding into the bidual is a pointwise bounded family of functionals on the complete continuous dual space, so the Uniform boundedness principle applies. The converse is immediate. A weakly compact set is weakly bounded because each scalar evaluation has compact image.

Every weakly compact set in a normed vector space is norm bounded, even if the space is incomplete. Its canonical embedding into the bidual is pointwise bounded on the continuous dual space; the Uniform boundedness principle on that Banach dual gives the norm bound.

Every weakly convergent sequence in a normed vector space is a bounded sequence. Apply the Uniform boundedness principle on the Banach space $X^{\ast }$ to the evaluation maps $J{x}_n:f\mapsto f(x_n)$. They are pointwise bounded, while the canonical embedding into the bidual gives $\Vert J{x}_n\Vert =\Vert x_n\Vert$.