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I am trying to show counter examples to the Open Mapping Theorem. In this particular case, I am trying to show that both spaces need to be Banach.

First the OMT:

Let $X, Y$ be Banach spaces. Let $T : X \rightarrow Y$ be a surjective bounded linear operator. $T$ is an open mapping.

Now, for the counterexample, let $X = l^1$ equipped with $||\cdot||_1$. Let $Y = l^1$ equipped with $||\cdot||_\infty$. Claim: Y is not complete. Let $T: X \rightarrow Y$ be the identity operator defined by $Tx = x$. Clearly T is bijection and is bounded (thus continuous). Claim: T is not an open mapping.

Basically, the main question is that $X$ is Banach, $Y$ is not and I want to show that the operator is not an open mapping.

I am also having a hard time proving that $Y$ is not complete (I know its not, I just cant prove it).

Tyler Hilton
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    If $Y$ were complete the OMT would imply that the norms $|\cdot|1$ and $|\cdot|\infty$ are equivalent on $\ell^1$ which is certainly wrong (for $x=(1,.\ldots,1,0,\ldots)$ you have $|x|1=n$ but $|x|\infty=1$). A more direct way to see the incompleteness: The sequence $x_n=(1,1/2,\ldots,1/n,0,\ldots)$ is $|\cdot|_\infty$ Cauchy in $\ell^1$ but it does not have a limit in $\ell^1$. – Jochen Dec 10 '14 at 07:27
  • Actually from your example, I would think the norms are equivalent, no? Norms are equivalent if there exists some $c >0$ such that $||\cdot||1 \leq c||\cdot||\infty$. Take $c = n$ and the norms would become equivalent. Also, is there a quick way to see that the limit does not exist in $\ell^1$ for your second part. – Tyler Hilton Dec 10 '14 at 22:17
  • But $c$ must be INDEPENDENT of $n$ which is impossible. – Jochen Dec 11 '14 at 07:49

2 Answers2

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For a counterexample with incomplete $X$ you can use e.g. a Hamel basis of $\ell^1$ to produce a discontinuous linear functional $f:\ell^1\to\mathbb R$. Then define a norm on $X=\ell^1$ by $\|x\|_f=\|x\|_1+|f(x)|$. The identity $(\ell^1,\|\cdot\|_f)\to (\ell^1,\|\cdot\|_1)$ is continuous and surjective but not open since otherwise $f$ would be continuous with respect to $\|\cdot\|_1$.

Jochen
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  • I am not sure about your last time, I don't understand it. What do you mean by the "identity ..." and how can you say its not open. – Tyler Hilton Dec 15 '14 at 19:20
  • Ahh I think I get it. If $X = \ell^1$ was complete, then the norms would be equivalent. But we defined $||\cdot||_f$ as a sum of continuous norm + a discontinuous linear functional. So does this imply that $||\cdot||_f$ is discontinuous? – Tyler Hilton Dec 15 '14 at 19:45
  • @TylerHilton Both $f$ and $|\mathord\cdot|_f$ are discontinuous on $(\ell^1,|\mathord\cdot|_1)$. They are, however, continuous on $(\ell^1, |\mathord\cdot|_f)$. Remember that continuity has to be determined with respect to the topologies on the domain and the codomain. Assuming the standard topology on $\mathbb R$, a norm is always continuous with respect to the topology on the domain induced by itself. It might not be continuous with respect to a different topology. – epimorphic Apr 27 '15 at 12:35
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To show $T$ is not an open mapping, let $B$ be the open unit ball of $X$. We have to show $T(B)$ is not open in $Y$. That is, the $\ell^1$ unit ball is not open in $\ell^\infty$ norm. It is sufficient to find a sequence $f_n$ with $f_n \to 0$ in $\ell^\infty$ but $\|f\|_1 \ge 1$ for all $n$. (Why?) Can you think of such a sequence?

Nate Eldredge
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