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On a finite-dimensional vector space equipped with a norm, the weak topology and norm topology coincide. The weak topology is no finer because every element of the dual is norm-continuous. Conversely, finitely many coordinate functionals in a basis control the norm, so every sufficiently small basic weak neighbourhood lies in a prescribed norm ball.

Let $X$ have basis $e_1,\dots ,e_d$, and write $x={\sum }_j{a}_j{e}_j$. On the compact coordinate sphere ${\sum }_j|a_j|=1$, the continuous positive functionhas a positive minimum $m$. Homogeneity and the triangle inequality therefore give constants $m,M>0$ such thatIn particular, each coordinate functional ${\varphi }_j(x)=a_j$ belongs to the continuous dual space $X^{\ast }$.

Every member of $X^{\ast }$ is norm-continuous, so the weak topology is coarser than the norm topology. Conversely, given $x_0\in X$ and $\epsilon >0$, the basic weak neighbourhoodsatisfies $U\subseteq \{x:\Vert x-x_0\Vert <\epsilon \}$. Thus every norm-open set is weakly open, proving that the two topologies coincide. This is the finite-dimensional weak and norm topologies coincide theorem.

Suppose $a^k\rightharpoonup a$ in ${\ell }^1$ but $a^k$ does not converge to $a$ in the norm topology. There are $\epsilon >0$ and a subsequence $(a^{m_k})$ such thatSet $x^k=a^{m_k}-a$. Thenwhich proves the first assertion.

We now use a gliding hump argument. Put $b_0=0$. Having chosen $b_{n-1}$ and $k_{n-1}$, coordinatewise convergence $x_i^k\to 0$ lets us choose $k_n>k_{n-1}$ so thatFor this fixed element of ${\ell }^1$, choose $b_n>b_{n-1}$ so far out thatDefine one sequence $y$ byThe blocks partition the positive integers and $|y_i|\le 1$, so $y\in {\ell }^{\infty }$. On the $n$th assigned block, the signs agree; outside it, use $|y_i|\le 1$. The duality of l1 and l infinity givesBut $x^{k_n}\rightharpoonup 0$ requires $⟨x^{k_n},y⟩\to 0$ for this fixed $y\in ({\ell }^1{)}^{\ast }={\ell }^{\infty }$, a contradiction. Therefore every weakly convergent sequence in ${\ell }^1$ converges in norm. This is the Schur property of l1.

A Banach space has the Schur property when every weakly convergent sequence converges with respect to the norm topology.