Let $X$ have basis $e_1,\ldots,e_d$, and write $x=\sum_ja_je_j$. On the compact coordinate sphere $\sum_j|a_j|=1$, the continuous positive function
$$
(a_1,\ldots,a_d)\longmapsto
\left\lVert\sum_ja_je_j\right\rVert
$$
has a positive minimum $m$. Homogeneity and the <triangle inequality> therefore give constants $m,M>0$ such that
$$
m\sum_j|a_j|
\leq\left\lVert\sum_ja_je_j\right\rVert
\leq M\sum_j|a_j|.
$$
In particular, each coordinate functional $\phi_j(x)=a_j$ belongs to the <continuous dual space> $X^*$.

Every member of $X^*$ is norm-continuous, so the <weak topology> is coarser than the <norm topology>. Conversely, given $x_0\in X$ and $\varepsilon>0$, the basic weak neighbourhood
$$
U=\left\{x:|\phi_j(x-x_0)|<\frac{\varepsilon}{Md}
\text{ for }1\leq j\leq d\right\}
$$
satisfies $U\subseteq\{x:\lVert x-x_0\rVert<\varepsilon\}$. 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 22H / a / Solution

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.

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