If $(X, \Vert \cdot \Vert_1)$ and $(X, \Vert \cdot \Vert_2)$ are Banach and $\Vert \cdot \Vert_1 \leq C \Vert \cdot \Vert_2 $ then the norms are equivalent by the open mapping theorem. I can't find an example where $(X, \Vert \cdot \Vert_2)$ is not Banach and the norms are not equivalent, can you help me?
I've tried $\ell^1$ and variations of $\mathcal{C}^0$ but I only get positive results when I assume $(X, \Vert \cdot \Vert_1)$ is not Banach.
The solution can be performed without referring to the open mapping theorem.
Let $\varphi$ be an unbounded linear functional on $\ell^2.$ Consider $$\|x\|_1=\left (\sum_{n=1}^\infty |x_n|^2\right )^{1/2},\quad \|x\|_2=\|x\|_1+|\varphi(x)|$$ Then $\|x\|_1\le \|x\|_2.$ Since $\varphi$ is unbounded there is a sequence $y_n$ such that $\|y_n\|=1$ and $|\varphi(y_n)|\to \infty.$ Let $x_n={y_n\over \varphi(y_n)|}.$ Then $\|x_n\|_1\to 0$ and $|\varphi(x_n)|=1.$ WLOG we may assume that the sequence $\varphi(x_n)$ is convergent. The elements $x_n$ constitute a Cauchy sequence with respect to $\|\cdot||_2$ norm. However this sequence is not convergent . Indeed if $\|x_n-x\|_2\to 0,$ then $\|x_n-x\|_1\to 0.$ Thus $x=0.$ However $1=|\varphi(x_n)|\not\to 0.$