Norm equivalence in a Banach space [closed]

Solution 1:

Here is a full solution: since $\Vert x\Vert_1\le \Vert x\Vert_2, \forall x\in X$ the identity map $I:(X,\Vert \cdot\Vert_2)\to (X,\Vert\cdot\Vert_1)$ is continous and surjective (both should be clear). Since both $(X,\Vert \cdot\Vert_2), (X,\Vert\cdot\Vert_1)$ are Banach spaces, we know by the open mapping theorem that $I$ is open, hence a homeomorphism (i.e. $I^{-1}$ is continous), this means there is some $c>0$ s.t. $\Vert I^{-1}x\Vert_2=\Vert x\Vert_2 \le c\Vert x\Vert_1$, which establishes $1/c\cdot \Vert x\Vert_2 \le \Vert x\Vert_1 \le \Vert x\Vert_2, \forall x\in X$, showing the equivalence of the norms.

Solution 2:

Another way to prove it is using the Closed Graph Theorem.

If $X=(E,\|\cdot\|_1)$ and $Y=(E,\|\cdot\|_2)$, then $F: X\to Y$, with $F(x)=x$ is closed, and hence continuous and thus, there exists a $c>0$, such that $$ \|x\|_2=\|F(x)\|_2\le c\|x\|_1. $$