1范数小于等于根号n倍的2范数

1范数小于等于根号n倍的2范数

∥ x ∥ 1 ⩽ n ⋅ ∥ x ∥ 2 \left\|x\right\|_1\leqslant\sqrt{n}\cdot\left\|x\right\|_2 x1n x2
证明:
上式即为:
∑ i = 1 n ∣ a i ∣ ⩽ n ∑ i = 1 n ∣ a i ∣ 2 \sum^n_{i=1}|a_i|\leqslant\sqrt{n\sum^n_{i=1}|a_i|^2} i=1naini=1nai2
令n维向量 α = ( 1 , 1 , ⋯   , 1 ) T β = ( ∣ a 1 ∣ , ∣ a 2 ∣ , ⋯   , ∣ a n ∣ ) T \alpha=(1,1,\cdots,1)^T\quad\beta=(|a_1|,|a_2|,\cdots,|a_n|)^T α=(1,1,,1)Tβ=(a1,a2,,an)T
由内积的柯西不等式:
∣ ( α , β ) ∣ ⩽ ∣ α ∣ ⋅ ∣ β ∣ |(\alpha,\beta)|\leqslant|\alpha|\cdot|\beta| (α,β)αβ
其中:
( α , β ) = ∑ i = 1 n ∣ a i ∣ (\alpha,\beta)=\sum^n_{i=1}|a_i| (α,β)=i=1nai
∣ α ∣ = ( α , α ) = n |\alpha|=\sqrt{(\alpha,\alpha)}=\sqrt{n} α=(α,α) =n
∣ β ∣ = ( β , β ) = ∑ i = 1 n ∣ a i ∣ 2 |\beta|=\sqrt{(\beta,\beta)}=\sqrt{\sum^n_{i=1}|a_i|^2} β=(β,β) =i=1nai2

你可能感兴趣的:(1范数小于等于根号n倍的2范数)