1. ## Equivalent Norms

Let ||.||_1 and ||.||_2 be norms on a vector space X

such that X_1 = (X, ||.||_1) and X_2=(X,||.||_2) are complete.

If ||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0,

show that convergence in X_1 implies convergence in X_2.

and conversely show that there are positive numbers

a and b such that for all x element of X,

a||x||_1 <= ||x||_2 <= b ||x||_1.

2. Originally Posted by frater_cp
Let ||.||_1 and ||.||_2 be norms on a vector space X

such that X_1 = (X, ||.||_1) and X_2=(X,||.||_2) are complete.

If ||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0,

show that convergence in X_1 implies convergence in X_2.
If x_1 converges to x in X_1, then x_1- X converges to 0. Therefore...

and conversely show that there are positive numbers

a and b such that for all x element of X,

a||x||_1 <= ||x||_2 <= b ||x||_1.

3. Hi Thank you for looking at this problem.

4. Oh, c'mon! If $\displaystyle x_n$ converges to x in ||.||_1, then $\displaystyle x_n- x$ converges to 0 in ||.||_1. But your condition is "||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0" so $\displaystyle x_n- x$ converges to 0 in ||.||2 from which $\displaystyle x_n$ converges to x in ||.||_2.

5. Originally Posted by HallsofIvy
Oh, c'mon! If $\displaystyle x_n$ converges to x in ||.||_1, then $\displaystyle x_n- x$ converges to 0 in ||.||_2.

You mean it converges to 0 in ||.||_1 and not ||.||_2

6. Thanks. I had just noticed that myself and editted it. (So I could pretend I didn't make that mistake!)

7. First part of the problem is perfectly explained by HallsofIvy.

Second part consists of proving that || ||_1 and || ||_2 are equivalent norms. But the first part implies that the identity

I: (X,|| ||_1)----------(X,|| ||_2)
I(x)=x

is continuous at zero, and hence continuous. Since both spaces are complete, Open Mapping Theorem implies that I is an isomorphism, and we are done.

8. Originally Posted by frater_cp
Let ||.||_1 and ||.||_2 be norms on a vector space X

such that X_1 = (X, ||.||_1) and X_2=(X,||.||_2) are complete.

If ||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0,

show that convergence in X_1 implies convergence in X_2.

and conversely show that there are positive numbers

a and b such that for all x element of X,

a||x||_1 <= ||x||_2 <= b ||x||_1.

Do you mean that ||x_n||_1------>0 means :

$\displaystyle lim_{n\rightarrow\infty}{||x_{n}||_{1}} =0$??