/sci/ - Science & Math

>>11486057
Theorem 4: Any real or complex vector space admits a norm.
Proof: We choose a basis [math]B[/math] of V. For [math]v \in V[/math], we can write [math]v = \Sigma {i=1}^n ki b_i[/math]. This is well defined, because the linear combination exists, is unique, and is finite. Set [math]f(v) = \Sigma_{i=1}^n |k_i|[/math]. [math]f(v)=0[/math] if and only if [math]v=0[/math] follows trivially. Similarly, [math]f( \lambda v) = \Sigma_{i=1}^n | \lambda k_i| = \lambda \Sigma_{i=1}^n |k_i|[/math]. Finally, we have [math]u = \Sigma_{i=1}^n u_i b_i[/math] and [math]w = \Sigma_{i=1}^n w_i b_i[/math], where we’ve hidden the process of reindexing, etc. Then [math]f( u + w) = \Sigma_{i=1}^n |u_i+w_i| \leq \Sigma_{i=1}^n |u_i| + \Sigma_{i=1}^n |w_i| \leq f(u) + f(w)[/math].