Let's say we have a (obviously infinite-dimensional) normed space $(V,||\cdot ||)$.
Using a discontinuous linear map $T:V\to\mathbb{K}$ we can construct the norm $||x||'=||x||+|T(x)|$, which is not equivalent to $||\cdot||$. However, $||\cdot||'$ dominates $||\cdot ||$, so that the two norms are comparable.
My question is: can we always find a norm that is not comparable with $||\cdot ||$?
I have tried in classical spaces and this is always the case but I cannot prove it in the general case.
Yes. Choose a Hamel basis $\{e_i\}_{i \in I}$ of $V$ s.t. $\|e_i\| = 1$ for all $i$. Let $I_1, I_2 \subset I$ be two disjoint countably infinite subsets of $I$. Enumerate $I_k = \{i^k_j\}_{j=1}^\infty$ for $k = 1, 2$. We then define a norm $\|\cdot\|’$ on $V$ by,
$$\|\sum_{j=1}^\infty a_je_{i_j^1} + \sum_{j=1}^\infty b_je_{i_j^2} + \sum_{i \in I \setminus (I_1 \cup I_2)} c_ie_i\|’ = \sum_{j=1}^\infty \frac{|a_j|}{j} + \sum_{j=1}^\infty j|b_j| + \sum_{i \in I \setminus (I_1 \cup I_2)} |c_i|$$
where only finitely many $a_j$, $b_j$, and $c_i$ are nonzero. Then $\|\cdot\|’$ is not comparable to $\|\cdot\|$, as $\|e_{i_j^1}\|’ = \frac{1}{j} \to 0$ while $\|e_{i_j^1}\| = 1$ for all $j$, and $\|e_{i_j^2}\|’ = j \to \infty$ while $\|e_{i_j^2}\| = 1$ for all $j$.
Let $(X, \|\cdot\|_{1})$ be an infinite-dimensional normed space. There exists a linear bijection $T\colon X\to X$ with $T^{-1} = T$ such that $T$ is not a bounded linear operator on $(X, \|\cdot\|_{1})$. (See Lemma 1 here.) Define $\|\cdot\|_{2}\colon X\to [0, \infty )$ by $\|x\|_{2} := \|Tx\|_{1}$. Then $\|\cdot\|_{2}$ is a norm on $X$. (See Lemma 2 here. In addition, $\|\cdot\|_{2}$ is a complete norm on $X$ if $\|\cdot\|_{1}$ is a complete norm on $X$.) As there is no $C > 0$ such that $\|Tx\|_{1} \leq C\|x\|_{1}$ for all $x\in X$, it follows from the relation $T^{-1} = T$ that the norms $\|\cdot\|_{1}$ and $\|\cdot\|_{2}$ on $X$ are not comparable.
