Two norms $||\cdot||_1$ and $||\cdot||_2$ on a vector space are equivalent if there exist constants $c_1$, $c_2$ such that
$$c_1||x||_1 \le ||x||_2 \le c_2||x||_1$$
for all $x\in X$ (I have proved that this is a equivalence relation ) .
So I know that $||f||_1 \le ||f||_2 \le ||f||_u$ so since I don't want that this norms are equivalent I want to do this by contradiction.
So I assumed that there were constants such that :
$$c_{1} (\int{|f(x)|}))<(\int{|f(x)|^{2}})^{1/2})<c_{2} (\int{|f(x)|}))$$
But this tell us that $$(\int{|f(x)|^{2}}))<c_{2}^{2}(\int{|f(x)|}^{2})$$
The thing is that how can I use something else than Jensen's inequality to prove this? And what can I do to prove that the uniform norm is not equivalent to the other two? , because there I have to prove that no lower bound can be accepted.
Thanks a lot for your help un advance 🙂 .
Best Answer
If $f_n(x) = (n+1)x^n,$ then $\|f_n\|_1 = 1, \|f_n\|_2 =(n+1)/\sqrt {2n+1}, \|f_n\|_u = n+1.$