Prove the the uniform, $L^1$ and $L^2$ norms are not equivalent. 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 :) .
 A: If $f_n(x) = (n+1)x^n,$ then $\|f_n\|_1 = 1, \|f_n\|_2 =(n+1)/\sqrt {2n+1}, \|f_n\|_u = n+1.$
A: Jensen's inequality?  Oh geeze, I'm a bad analyst, so I can never remember something like that.  Here's a trick you might try.  Suppose that there is a constant $c > 0$ so that $||f||_{L^1(\mathbb{R}^n)} \le c ||f||_{L^2(\mathbb{R}^n)}$.
Let's make the change of variables $x \to \epsilon x $ for some $\epsilon > 0$.  That is, let's let $y = \epsilon x$.
Notice then that $\int |f(x)| dx = \int |f(\epsilon^{-1}y)| \epsilon^{-n} dy = \epsilon^{-n} ||f_{\epsilon^{-1}}||_{L^1}$
So, $||f||_{L^1} = \epsilon^{-n} ||f_{\epsilon^{-1}}||_{L^1} \le \epsilon^{-n}c ||f_{\epsilon^{-1}}||_{L^2} $
A similar computation will show that $||f_{\epsilon^{-1}}||_{L^2}|| = \epsilon^{\frac{n}{2}} ||f||_{L^2} $
Plugging that it in, we get that $||f||_{L^1} \le  \epsilon^{-\frac{n}{2}} c ||f||_{L^2}$
That holds for any $\epsilon > 0$.  Can you finish the contradiction?
Edit:
To elaborate a bit (skip this if you don't want to see the solution).  Arguing as in the other direction, if a constant $c > 0$ existed so that $||f||_{L^2} \le c ||f||_{L^1}$, we'd have for all $\epsilon > 0$ that
$||f||_{L^2} = \epsilon^{-\frac{n}{2}} ||f_{\epsilon^{-1}}||_{L^2} \le c \epsilon^{-\frac{n}{2}} ||f_{\epsilon^{-1}}||_{L^1} = c\epsilon^{\frac{n}{2}}||f||_{L^1} $
Now let $\epsilon \to 0$ and deduce a contradiction from an $f$ which isn't $0$ ae.
The point of these arguments is that $L^p$ norms scale under the scaling operator $\tau_{\epsilon}$ defined by $\tau_{\epsilon}f(x) = f(\epsilon x)$.  In particular, we should get that $|| \tau_{\epsilon}f||_{L^p} = \epsilon^{-\frac{n}{p}} ||f||_{L^p}$.  Were the norms equivalent, this operator would have to scale both norms proportionally in order to maintain the equivalence.  But that doesn't happen.
