All norms on a finite dimensional space are equivalent

The theorem as stated above holds in general¹, but is currently beyond me. The case for $ℂ 𝑛$ and $ℝ 𝑛$ is simpler.

Complex vector space

Any two norms $‖ \cdot ‖ 𝑎$ and $‖ \cdot ‖ 𝑏$ on a finite-dimensional complex vector space $𝑉$ are equivalent. #m/thm/anal/vec

Proof

Let $(𝑒 𝑖) 𝑛 𝑖 = 1$ be a vector basis of $𝑉$ , so that any $𝑥 \in 𝑉$ may be uniquely written as $𝑥 = \sum 𝑛 𝑖 = 1 𝑥 𝑖 𝑒 𝑖$ . Let $‖ \cdot ‖ 1$ denote the 1-norm $‖ 𝑥 ‖ 1 = \sum 𝑛 𝑖 = 1 𝑥 𝑖$ . Now we will show that any norm $‖ \cdot ‖ 𝑎$ is continuous under $‖ \cdot ‖ 1$ , i.e. for all $𝜖 > 0$ there exists $𝛿 > 0$ such that

‖ 𝑥 - 𝑦 ‖ 1 < 𝛿 ⟹ | ‖ 𝑥 ‖ 𝑎 - ‖ 𝑦 ‖ 𝑎 | < 𝜖

By the Reverse triangle inequality, it is sufficient to show that

‖ 𝑥 - 𝑦 ‖ 1 < 𝛿 ⟹ ‖ 𝑥 - 𝑦 ‖ 𝑎 < 𝜖

Letting $𝑥 = \sum 𝑛 𝑖 = 1 𝑥 𝑖 𝑒 𝑖$ and $𝑦 = \sum 𝑛 𝑖 = 1 𝑦 𝑖 𝑒 𝑖$ , it follows from the triangle inequality of $‖ \cdot ‖ 𝑎$ that

‖ 𝑥 - 𝑦 ‖ 𝑎 \leq 𝑛 \sum 𝑖 = 1 | 𝑥 𝑖 - 𝑦 𝑖 | ‖ 𝑒 𝑖 ‖ 𝑎 \leq ‖ 𝑥 - 𝑦 ‖ 1 𝑛 m a x 𝑖 = 1 ‖ 𝑒 𝑖 ‖ 𝑎

‖ 𝑥 - 𝑦 ‖ 1 < 𝜖 m a x 𝑛 𝑖 = 1 ‖ 𝑒 𝑖 ‖ 𝑎 ⟹ ‖ 𝑥 - 𝑦 ‖ 𝑎 < 𝜖

Hence $‖ \cdot ‖ 𝑎$ is 1-continuous. Since the $‖ \cdot ‖ 1$ unit sphere is compact, by the Extreme Value Theorem $‖ \cdot ‖ 𝑎$ has an absolute minimum $𝑎$ and maximum $𝑏$ on this domain. Hence

𝑎 < ‖ 𝑣 ‖ 𝑎 < 𝑏

for all $𝑣 \in 𝑉$ with $‖ 𝑣 ‖ 1 = 1$ . Therefore all norms on $𝑉$ are equivalent to $‖ \cdot ‖ 1$ , and by transitivity equivalent.

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See these lecture notes. ↩

All norms on a finite dimensional space are equivalent

Complex vector space

Footnotes