Csdrhrt

a)
Consider $(mathbb{R}^n, |.|_infty)$ and $(L_b(mathbb{R}^n, mathbb{R}),|.|)$ which is the space of all linear and bounded functions from $mathbb{R}^n to mathbb{R}$ associated with the operator norm.

$(L_b(mathbb{R}^n, mathbb{R}),|.|)$ is isomorphic to $mathbb{R}^n$ if we interpret $A in (L_b(mathbb{R}^n, mathbb{R}),|.|)$ as a matrix $(a_1,dots a_n)$

I want to show that $|.|$ is equivalent to $|.|_1$ then.

The operator norm is defined as

$$|Ax|=supBig{frac{|Ax|_Y}{|x|_X}:x in Xsetminus{0}Big}=sup{|Ax|_Y:xin X, |x|_X=1}$$

Let $A=(a_1,dots,a_n)$ and $x=(x_1,dots,x_n)^T$. Then $Ax=sum_{i=1}^{n}a_ix_i$.

And $|sum_{i=1}^{n}a_ix_i|lesum_{i=1}^{n}|a_ix_i|lemax_{i=1,dots,n}|x_i|sum_{i=1}^{n}|a_i|=sum_{i=1}^{n}|a_i|=|A|_1$ because $max_{i=1,dots,n}|x_i|=1$. Therefore $|A|le|A|_1$.

For the other direction $"|A|ge|A|_1"$:

Choose $x$ so that $x_i=sgn(a_i), i=1,dots,n$. Then $|x|_infty=1$ and $|sum_{i=1}^{n}a_ix_i|=sum_{i=1}^{n}|a_ix_i|=sum_{i=1}^{n}|a_i|$.
Therefore $|A|ge|A|_1$.

Is that correct so far?

b)

To which norm is the operator norm on $L_b(mathbb{R}^n,mathbb{R})$ equivalent if we associate $mathbb{R}^n$ with $|.|_2$?

$|sum_{i=1}^{n}a_ix_i|lesum_{i=1}^{n}|a_ix_i|=sum_{i=1}^{n}sqrt{(a_ix_i)^2}$ but I don't know how to continue

All norms are equivalent in finite dimensional spaces.