Uncompletion of a Banach Space

Question

Given any Banach space there is a way to define a norm such that is no longer complete.

I know you can reach the result by using a Hamel base $H$ for this given space and doing this:

If $$ \forall x \in X, x=\sum_{i=1}^{n_x}c_ih_{x_i} ,c_i \in \mathbb{K},h_{i_x} \in H$$ then define $$ \|x\|=\sum_{i=1}^{n_x}|c_i|$$

Then with this norm the space is no longer complete I think or at least that is what I believe, but I haven't been able to find a Cauchy secuence that does not converge.

Answer 1

I don't see a way forward with a Hamel basis (maybe it's the lack of coffee). Anyway, I think a more natural way is to identify $X$ with a dense subset of $X\oplus \mathbb K$ (where $\mathbb K$ is your field of scalars, real or complex). The product can be equipped with any product norm, e.g., $\|(x,t)\| =\|x\|+|t|$. Then the norm of $X\oplus \mathbb K$, restricted to this dense subset, makes it a non-complete normed space.

And a natural way to get such dense subset is to take an unbounded linear functional $\varphi$ and consider its graph $(x,\varphi(x))$. The graph is dense in $X\oplus \mathbb K$, and is in a natural bijection with $X$.

The rest is spoilered.

Let the new norm on $X$ be $\|x\|+|\varphi(x)|$. On the kernel of $\varphi$ this is the same as before. Take a sequence $(x_k)$ in $\ker \varphi$ that converges in the original norm to some $x\notin \ker \varphi$. Then $(x_k)$ is Cauchy in the new norm, but does not converge in it.