Inequivalent complete norms

I am looking for an explicit example of two complete inequivalent norms.

Given a Banach space, I know how to explicitly construct an inequivalent complete norm. However, I am looking for an explicit example.

Edit 1: I came across the following here Norm equivalence and Banach spaces

However it does not hold that if two spaces are Banach spaces that their norm would be equivalent. Take for example the space of absolutely summable sequences and take ∑|xn| as one norm, since it's absolutely summable also the sum ∑$\frac{|xn|}{n}$ would be convergent, but now you can consider the sequence δj (that's zero, except it's jth term is 1). You have that ||$\delta_j||_1$=1, but ||$\delta_j||_b$ = 1/j you can't have the equivalence between these norms.

I want to know how the space is complete wrt to both norms.

Edit 2: Daniel Fischer explained in the comments why above example is wrong. So, I'm still looking for an explicit example.

Edit 3: The question I asked here: Complete Inequivalent Norms asks something completely different. I'm not sure why everyone thinks it is a duplicate.

• I'd be surprised if one could give an explicit example. – Daniel Fischer Sep 30 '16 at 13:22
• @DanielFischer Could you please explain why? – Surb Sep 30 '16 at 13:32
• @Surb The existence of inequivalent norms making a vector space (over $\mathbb{R}$, say, and of appropriate dimension) into a Banach space is an important fact. If there was an explicit example of a space and two such norms, it would probably be widely mentioned. I've never seen any mention of an explicit example. – Daniel Fischer Sep 30 '16 at 13:48
• @DanielFischer Thanks, this is exactly the kind of justification I was expecting and I should say: "makes sense". – Surb Sep 30 '16 at 13:51
• The space $\ell^1(\mathbb{N})$ is not complete with the norm $\lVert x\rVert_b = \sum \lvert x_n\rvert/n$. If two norms both make the same space into a Banach space, then either the two norms are equivalent, or the two norms are incomparable. Since $\lVert\,\cdot\,\rVert_b$ is comparable to $\lVert\,\cdot\,\rVert_1$, but not equivalent, the space cannot be complete with that norm. – Daniel Fischer Sep 30 '16 at 14:20