I would like to ask a counterexample for the classical theorem in functional analysis: the open mapping theorem in the case that $Y$ is Banach, but $X$ is not Banach to show that the completeness of X is crucial.

In details, find a continuous linear mapping $T:X \to Y$ such that $T(X)=Y$ and $Y$ is Banach but $T$ is not open.

If we can construct this, we could get an interesting example: there exists a bijective linear (contiuous) mapping between two normed space $X$ and $Y$, and only one of them is Banach. The counterexamples for the case when $Y$ is not Banach is simple, but I didn't come up if I need $X$ is not Banach and $Y$ is Banach. Thanks!

  • $\begingroup$ I voted to close, since this question does not seem to be a question on a research level. It is almost perfectly suited for Math Stack Exchange (I think), since the basic tools to find the required example (like a Hamel basis, the existence of unbonded linear functionals etc.) are sufficient. In some sense it is an exercise everybody should solve, but not get solved within one hour on MO. I am sure that a crowd of interested students would have been able to answer this question quickly. $\endgroup$ – Andreas Thom Nov 4 '10 at 7:16
  • $\begingroup$ I wonder if there is an example without using AC. $\endgroup$ – Pietro Majer Nov 4 '10 at 9:01
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    $\begingroup$ True, Andreas, but I learned something thinking about this standard problem not in a course environment: Notice that my solution shows that the absolutely closed convex hull of any Hamel basis for a Banach space contains a neighborhood of zero. Did you know this? I certainly did not. $\endgroup$ – Bill Johnson Nov 4 '10 at 13:19
  • $\begingroup$ Yes, that's very nice, I didn't notice before either. In fact, it's a neighborhood of zero because it's a barrel in a Banach space. $\endgroup$ – Pietro Majer Nov 4 '10 at 21:37
  • $\begingroup$ Oh, yes, Pietro; that is simpler. $\endgroup$ – Bill Johnson Nov 6 '10 at 2:49

Minh, this is not a new answer since you have already a satisfying number of them, but it's a riddle for you, since you are interested in this topic (too long to be posted as a comment).

"Theorem". All Banach norms on a real vector space $X$ are equivalent.

"Proof". (sketch). Let $\|\cdot\|_1$ and $\|\cdot\|_2$ two Banach norms on $X$. Consider $\|\cdot\|_3:=\|\cdot\|_1+ \|\cdot\|_2. $ Prove that it is actually a norm. Prove that a sequence converge to $x\in X$ w.r.to $\|\cdot\|_3$ if and only if it converges to $x$ both w.r.to $\|\cdot\|_1$ and w.r.to $\|\cdot\|_2$. Prove that a sequence is Cauchy wrto $\|\cdot\|_3$ if and only if it is Cauchy both w.r.to $\|\cdot\|_1$ and w.r.to $\|\cdot\|_2$. Deduce that $\|\cdot\|_3$ is complete. Apply the OMT tho the identity from $(X, \|\cdot\|_3)$ to $(X, \|\cdot\|_1)$ and from $(X, \|\cdot\|_3)$ to $(X, \|\cdot\|_2)$, and deduce that the three norms are equivalent.

However, if all Banach norms on $X$ are equivalent, all linear forms are continuous, and in infinite dimension there are non-continuous linear forms.

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    $\begingroup$ Took me a bit, but I see it now. [rot13] Gurer znl or n frdhrapr gung pbairetrf gb bar irpgbe va gur svefg abez, naq gb nabgure va gur frpbaq abez. Fhpu n frdhrapr vf Pnhpul va gur guveq abez ohg qbrf abg pbairetr. $\endgroup$ – Nate Eldredge Nov 3 '10 at 23:24
  • $\begingroup$ Thanks. Abgr gung abez 3 vf Onanpu vss gurer rkvfgf n Unhfqbess gbcbybtl ba K juvpu vf jrnxre guna obgu gur gbcbybtl bs abez1 naq gur gbcbybtl bs abez 2. $\endgroup$ – Pietro Majer Nov 4 '10 at 8:51
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    $\begingroup$ Note that the existence of an actual counterexample to the "theorem" requires the Axiom of Choice. Eric Schechter in his analysis book talks about negations of choice where the quoted "theorem" actually becomes true. $\endgroup$ – arsmath Feb 14 '11 at 12:49
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    $\begingroup$ (Kamran, the explanation is written above, cyphered in rot 13 not to spoil the riddle for other people. You can read it using e.g. rot13.com/index.php) $\endgroup$ – Pietro Majer Feb 10 '12 at 22:02
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    $\begingroup$ Nevertheless, this "proof" actually proves something if suitably corrected. You need an extra condition, but it is one that would usually be true in applications. $\endgroup$ – Michael Renardy Oct 12 '13 at 0:12

Take a normalized Hamel basis $y_a$, $a\in A$, for $Y$ a separable infinite dimensional Banach space, so that $A$ has cardinality the continuum. Write $A$ as a disjoint union of $B$ and a sequence $a_n$. Define $T$ from the linear span $X$ of the unit vector basis $e_a$, $a\in A$, of $\ell_1(A)$ to $Y$ by setting $T e_{a_n} = n^{-1} y_{a_n}$ and $T e_a = x_a$ for $a\in B$. Then $T$ has norm one and is surjective and injective, but is not open.


Choose an infinite dimensional Banach space $\left( Y,\|\cdot\| \right)$ with a non-continuous linear form $\phi$. Take $X$ to be the same vector space $Y$, with the norm $\|x\|_ \phi:=\|x\|+|\langle \phi,x\rangle|.$ Take $T$ the identity. Then $T$ is continuous, as $ \|x\| \leq \|x\|_ \phi $. However, $ \|\cdot\|_ \phi $ can't be complete, or it would be equivalent to $ \|\cdot\|$ by the open mapping theorem, and $\phi$ would be continuous w.r.to $\|\cdot\|$.

$$* $$

Rmk: existence of a non-continuous linear form on any infinite dimensional normed space. On any infinite dimensional normed space $Y$ you can define a non-continuous linear form as the projection map on the quotient of $Y$ on a non-closed linear subspace $N$ of codimension 1 (hence dense), identifying the quotient $Y/N$ with $\mathbb{R}$. Such a linear subspace can be defined starting from a Hamel basis. Indeed, let $\{u_\lambda\}_{\lambda\in\Lambda}$ be a Hamel basis for $Y,$ and pick $\lambda_0\in\Lambda$. We may assume $\inf_{\lambda\in\Lambda}\|u_\lambda\|=0$ because $\Lambda$ is infinite and one can re-normalize the basis. Define $N:=\mathrm{span}\left( \{u_\lambda-u_{\lambda_0}\, : \,\lambda\in\Lambda\}\right).$ Then $N$ is of codimension 1 and $\bar N=Y.$ Of course, the existence of a Hamel basis requires the Zorn lemma.


Problem 5.31 of Folland's Real Analysis provides such an example.

Let $Y$ be a Banach space, $Z$ any normed space, and let $S : Y \to Z$ be an unbounded, everywhere defined linear operator (you need the axiom of choice to construct such a thing). Let $X \subset Y \times Z$ be the graph of $S$: $X = \{(y,Sy) : y \in Y\}$; by the closed graph theorem $X$ is not complete. Let $T : X \to Y$ be the map $T(y,Sy)=y$. $T$ is bijective and bounded but its inverse cannot be bounded, so it is not open.

  • $\begingroup$ This is actually the same as Pietro Majer's example: take $Z = \mathbb{R}$, $S = \phi$, and my $T$ is effectively an isomorphism from my $X$ (with the graph norm) to his $X$. $\endgroup$ – Nate Eldredge Apr 25 '11 at 2:37

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