# About linear bijection between Banach spaces!

It is well-known that, by Banach theorem, every continuous, linear and bijective operator between Banach spaces is a isomorphism. There must be a linear bijective and discontinuous operator between Banach spaces! How can we show/construct such a map?

Thanks for all helpings!

Take any infinite-dimensional Banach space and choose its Hamel basis $\left(e_i\right)_{i\in I}$, say, with $\left\|e_i\right\|=1$. Define $f\colon X \to X$ by

$$f(e_i) = \alpha_i e_i$$

where ${\alpha_i}$ is any unbounded collection of (nonzero) numbers. Clearly $f$ is a bijection, and $f$ is unbounded, hence discontinuous.

Now, the above relies critically on the Axiom of Choice. If I recall correctly, there was a similar question here, and it's been pointed out that in certain models with negation of Axiom of Choice there are no discontinuous linear maps from Banach spaces to normed linear spaces at all. So there can be no "truly constructive" example.

It's probably a bit off-topic, but such natural examples exist for incomplete normed linear spaces. A well-known one is a derivative operator for $C^\infty\left([0,1]\right)$ with supremum norm. It's not injective, but it is on a quotient by subspace of constant functions.

• Los: infinite-dimensional Banach space cannot have countable Hamel base, so you shouldn't denote $e_i$, it should be $e_\alpha$. Mar 2, 2014 at 9:43
• Well, yes, I ment it as $i\in I$, but haven't written it explicitly. Admittedly, it may be confusing. Will edit, thanks. Mar 2, 2014 at 9:44
• @Los: The set $\{\alpha_i: i \in I\}$ how to choose when the index $I$ is given in order to satisfy unbouded? Mar 2, 2014 at 9:55
• For example, choose countable subset of $I$, $\{i_1,\ldots\}=I_0\subset I$ and put $\alpha_{i_n}=n$, and $\alpha_i=1$ for all the $i\not\in I_0$ Mar 2, 2014 at 9:59

This requires some form of the Axiom of Choice.
Take two Banach spaces that are not isomorphic, but both have Hamel bases of the same cardinality (which will be the case e.g. if they are both infinite-dimensional and have cardinality $\bf c$) and define an operator using a one-to-one correspondence between Hamel bases.

We can build a bijective linear function which is not continuous even from a Banach space $E$ in itself (clearly $B$ can't be finite dimensional).

Using $AC$ (we need the Hamel basis) it is easy to build a discontinuous linear functional $$\varphi : B \to \mathbb{R}$$ Let $u\in B$ such that $\varphi(u)=1$. Then define $$S : B \to B$$ $$S(x) = x - 2\varphi(x)u$$ such $S$ is linear and bijective, but clearly not continuous.

linearity

$S(\alpha x +\beta y ) = \alpha x +\beta y - 2\varphi(\alpha x +\beta y)u = \alpha x -2\alpha\varphi(x)u +\beta y - 2\beta\varphi(y)u = \alpha S(x) + \beta S(y)$

injectivity

$S(x)=0 \Leftrightarrow x- 2\varphi(x)u =0 \Leftrightarrow x=2\varphi(x)u$ But $S(2\varphi(x)u) = -2\varphi(x)u = 0 \Leftrightarrow \varphi(x)=0$ but again (using the observation just made) $S(x)= x =0$

surjectivity

For every $y\in B$, $S(y)$ is a preimage of it. In order to prove this result consider this chain of obvious equalities (we are using the definition, the fact that $\varphi(u)=1$ and the linearity of $\varphi$). $$S(S(y))=S(y)-2\varphi(S(y))u=y-2\varphi(y)u-2\varphi(y-2\varphi(y)u)u= y-2\varphi(y)u -2\varphi(y)u+4\varphi(y)u=y$$

$$\times \times \times \times \times \times$$

I came across this construction when searching for non equivalent complete norm over the same set (so two Banach space over the same set whose norms are not equivalent), in fact $S: B \to B$ induce a new complete norm over $B$. Let's denote with $|| \cdot ||_1$ the original norm, then we have $|| x ||_2 := ||S(x)||_1$. Using linearity, injectivity and surjectivity of $S$ it's not hard to prove the completeness of such norm. And final addendum, the two norm can't be equivalent!

Edit 1 added proof of injectivity surjectivy and linearity

Edit 2 added a factor $2$ in the definition of the operator

Edit 3 added extended calculation as required by OP

• @Ped: Can you explain why S is injective and surjective? Mar 2, 2014 at 9:50
• @Ped: Bijection is ok. But the discontinuous have something wrong! Mar 2, 2014 at 10:09
• @Richkent Why? It can't be continuous, try arguing by absurd :) (the sum of two continuous function with value in a vector space is again continuous) Mar 2, 2014 at 10:12
• If $\|S(x)\|\leq K\|x\|$ then $\|\varphi(x)\|\leq \frac{1+K}{2\|u\|}\|x\|,$ this is contradicts to the discontinuty of $\varphi$. Is this right? Mar 2, 2014 at 10:15
• Mar 2, 2014 at 10:21