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Suppose $X$ is Banach and $N$ its closed subspace, then $X/N$ is in turn Banach. Now suppose we have a convergent sequence $\bar{x_n}\to \bar{x_0}$ in $X/N$, is it then possible to construct a sequence $x_n\in X$, with $\|x_n\|=\|\bar x_n\|, n=0\cup\Bbb N^+$ and $x_n\to x$?

By the first requirement, $x_n$ are actually already uniquely determined, i.e., they are the unique minimisers of $\|x\|,x\in \bar{x_n}$. So we must show that $x_n$ so defined must converge. Indeed, geometrical intuition tells me that $\bar{x_n}$ should be a collection of "parallel" hyperplanes in $X$, and that $\|x_n-x_0\|$ should be exactly the distance between the corresponding hyperplanes. But rather disappointingly I couldn't prove it. So, is my initial conjecture really true? If so is there any easy way to prove it?


As commented by Daniel, for a generic Banach space $X$, such a unique determination of $x_n$ may not exist. That closed convex sets always admit a unique modulus minimiser is only a result of Hilbert spaces (or more generally, reflexive spaces as in Daniel's comment).

Anyway, now let's relax the first requirement a little. We now drop the requirement that $\|\bar x_n\|$ be equal to $\|x_n\|$, rather, we only require that $\|x_n\|\le c\|\bar x_n\|$ (including $x_0$) for some $c<\infty$ and $x_n\to x_0$ in $X$. So is there a positive answer to this weaker problem?

I feel I should've included some context here. It all went from the following:

$X,N$ as above, if $Y$ is Banach, then if $A$ is a bounded linear operator from $X$ to $Y$, then for each convergent sequence $y_n\to y_0$, there exists a convergent sequence $x_n\to x_0$ such that $Ax_n=y_n$ (including $x_0$) and that $\|x_n\|\le M\|y_n\|$ for some $M<\infty$.

My approach was get everything into the quotient space $X/N$, then the resulting quotient operator $\bar A$ is bijective, whence, if we can recover $x_n\in\bar x_n$, convergent and dominated by $\|\bar x_n\|$, we readily prove the result with the Inverse Operator Theorem.


Okay after thinking for a while I finally got the answer to my modified question. It isn't hard actually, just an application of the inf definition of the quotient norm: for each $\bar x\ne 0$ we have $$\|\bar x\|:=\inf_{x\in\bar x}\|x\|>0,$$ and thus we are able to find $x'\in\bar x$ such that $\|x'\|\le \|\bar x\|+\|\bar x\|$ (taking $\epsilon=\|\bar x\|>0$ here). Then for our question, fixing any $x_0\in\bar x_0$, we just replace $\bar x$ with each $\bar x_n-\bar x_0=\overline{x_n-x_0}$ and we're almost done (noteworthy that in this step the well definedness of the quotient norm, as mentioned in Omnomnomnom's answer, is crucial.)

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  • $\begingroup$ @user251257 that's the sad news of course. And my task is to find $y_n$ that do. $\endgroup$ – Vim Nov 21 '16 at 18:33
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    $\begingroup$ Problem 1, there is not always an $x_n$ with $\lVert x_n\rVert_X = \lVert \overline{x}_n\rVert_{X/N}$. Problem 2, if there is, such a minimiser isn't necessarily unique. $\endgroup$ – Daniel Fischer Nov 21 '16 at 19:07
  • $\begingroup$ @DanielFischer but $X$ is a Banach space, and $\bar x_n$ is a closed, convex set. $\endgroup$ – Omnomnomnom Nov 21 '16 at 19:08
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    $\begingroup$ If this construction fails to work, it would have to be in a space for which $$ X \ncong N \oplus X/N $$ that is, $N$ can't be complemented. $\endgroup$ – Omnomnomnom Nov 21 '16 at 19:12
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    $\begingroup$ @Omnomnomnom If $X$ is reflexive, a minimiser always exists, and if $X$ is uniformly convex (strictly convex suffices), it is unique. But if $X$ isn't reflexive, there needn't exist one. Take $X = C([0,1])$, and $N = \Bigl\{ f \in X : \int_0^{1/2} f(t)\,dt = \int_{1/2}^1 f(t)\,dt\Bigr\}$. Then $\lVert \overline{f}\rVert_{X/N} < \lVert f\rVert_X$ for all $f \neq 0$. $\endgroup$ – Daniel Fischer Nov 21 '16 at 19:14
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Edit: This answer doesn't reach the desired conclusion, but hopefully it will be a helpful step.


The key, one way or the other, is to remember that the addition/subtraction of the hyperplanes in $X/N$ is well-defined.

Consider the sequence $\bar y_n = (\bar x_n - \bar x_0)$. Then we can construct a sequence $y_n$ of minimizers in $\bar y_n$, and clearly $\|y_n\| \to 0$ just as $\|\bar y_n \| \to 0$.

Now, select the minimizer $x \in \bar x_0$, and define $x_n = x + y_n$. Clearly, $x_n \to x$, and we have $x_n \in \bar x_n$. It remains to be shown, however, that $\|x_n\| = \bar x_n$.

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    $\begingroup$ I don't think it's ok to select any $x_n$ since they must not exceed the lengths of $\bar x_n$, by the first requirement. $\endgroup$ – Vim Nov 21 '16 at 18:45
  • $\begingroup$ Ah, fair point, I missed that part $\endgroup$ – Omnomnomnom Nov 21 '16 at 18:47
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    $\begingroup$ Do we always have a projection onto a closed subspace? Shouldn't we switch to a minimizing sequence? $\endgroup$ – user251257 Nov 21 '16 at 18:48
  • $\begingroup$ @user251257 I'm not sure what either of your statements mean. Yes, the projection onto a closed subspace is always continuous, but I'm not sure this is what you're asking. What's a "minimizing sequence" in this context? $\endgroup$ – Omnomnomnom Nov 21 '16 at 18:59
  • $\begingroup$ @Omnomnomnom the projection is continuous if and if $N$ is complemented. That I am pretty sure. I am not sure if the projection always exists. But, for $n$ we can select a $y_n \in \bar y_n$ with $\|y_n\| \le \|\bar y_n\| + 1/n$. $\endgroup$ – user251257 Nov 21 '16 at 19:04

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