Posted 6 years agoUpdated 4 years agoMath Analysis / Functional Analysisa few seconds read (About 55 words)NormNormLp 范数 Lp 范数 ‖x‖p=(|x1|p+|x2|p+⋯+|xN|p)1/p x∈Rn Lp Dual Norm 对偶范数 ‖x‖∗=supzxTz|‖z‖≤1 x∈RN |x|:x的某个范数