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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).

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    $\begingroup$ 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 $\ell^1$. $\endgroup$ – Jochen Dec 10 '14 at 7:27
  • $\begingroup$ 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. $\endgroup$ – Tyler Hilton Dec 10 '14 at 22:17
  • $\begingroup$ But $c$ must be INDEPENDENT of $n$ which is impossible. $\endgroup$ – Jochen Dec 11 '14 at 7:49
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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 $\|\cdot\|_1$.

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  • $\begingroup$ 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. $\endgroup$ – Tyler Hilton Dec 15 '14 at 19:20
  • $\begingroup$ 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? $\endgroup$ – Tyler Hilton Dec 15 '14 at 19:45
  • $\begingroup$ @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. $\endgroup$ – 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?

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  • $\begingroup$ I've been thinking about it. Could you give me a hint? $\endgroup$ – Tyler Hilton Dec 15 '14 at 19:18
  • $\begingroup$ @TylerHilton: Hint: $\frac{1}{n} + \dots + \frac{1}{n}$ ($n$ times) is 1. $\endgroup$ – Nate Eldredge Dec 15 '14 at 19:53
  • $\begingroup$ Why is sufficient? $\endgroup$ – eraldcoil Dec 19 '18 at 17:40
  • $\begingroup$ @eraldcoil: If the $\ell^1$ unit ball $B$ were open in $\ell^\infty$ norm, then it would be an $\ell^\infty$-open neighborhood of 0. So a sequence $f_n$ which converges to $0$ in $\ell^\infty$ norm would have to have all but finitely many terms contained in $B$. But the sequence we construct has no terms in $B$. $\endgroup$ – Nate Eldredge Dec 19 '18 at 18:12

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