# 實分析(證明題)有關llinear vector space

Two norms on a linear vector space are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

### 1 個解答

• Sam
Lv 6
9 年前
最佳解答

Equivalent norms

Two norms |•|_1 and |•|_2 on a vector space V are called equivalent if there exist positive real numbers C and D such that

C|x|_1 ≦ |x|_2≦ D|x|_1. for all x in V.

Two norms on a linear vector space are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

[Proof]

(=>) Let A be a set which is bounded in the first norm |*|_1, and the bound is M.

Then |x|_2 ≦ (1/C)|x|_1 ≦ (1/C)M, for all x in A, i.e. A is bounded in the norm |*|_2 with bound (1/C)M.

Let A be a set which is bounded in the second norm|*|_2, and the bound is M.

Then |x|_1 ≦ D|x|_2 ≦ DM, for all x in A, i.e. A is bounded in the norm |*|_1 with bound DM.

The part (=>) is done.

(<=)

Let US={x: |x|_1=1}, where US is the abbreviation of unit sphere , by the assumption there is M >0, such that

|x|_2 ≦ M, for all x in US.

For any x in V, x=|x|_1*(x/|x|_1), where (x/|x|_1) in US, and

|x|_2=||x|_1*(x/|x|_1)|_2

=|x|_1*|(x/|x|_1)|_2

{since (x/|x|_1) in US}

≦ M|x|_1.

Take D=M, we have

|x|_2≦ D|x|_1.

For the same reason as the above proof, we can easy obtain the other part of the inequality, C|x|_1≦|x|_2, for all x in V, so omit it and leave you as an exercise.

Hence |*|_1 is equivalent to |*|_2.

(<=) is done.

Q.E.D.