12
$\begingroup$

I am interested in some geometrical aspects of spaces $L(E)$, of bounded operators on a given Banach space $E$. I am unable to estimate if my problem deserves to be asked at MO, but let me try.

Is there an infinite-dimensional Banach space (non-separable preferably) $E$ such that for some non-zero

$T\in L(E)$

the set

$$\{S\in L(E)\colon \|S-T\|=\|S+T\|\}$$

contains an open ball? In fact, I am more interested in the negation:

Is there a Banach space such that for none non-zero $T\in L(E)$ this can happen?

I cannot (dis)prove it even if $E$ is a Hilbert space.

$\endgroup$
2
  • $\begingroup$ Of course, you mean $T\ne0$. $\endgroup$ Oct 8, 2011 at 12:45
  • $\begingroup$ Yes, you're right. Corrected. $\endgroup$ Oct 8, 2011 at 13:50

2 Answers 2

9
$\begingroup$

In what follows I show that such an operator exists if $E$ can be written (isometrically) as the $\ell_\infty$-direct sum of two (nonzero) subspaces (I have not tried the Hilbert space case, but I started writing my answer before the edits were made to the question.)

Let $E = X\oplus_\infty Y$, where $X$ and $Y$ are nonzero (infinite dimensional, if you like). Each $V\in L(E)$ satisfies $\Vert V \Vert = \max ( \Vert P_X V \Vert, \Vert P_Y V\Vert )$, where $P_X$ and $P_Y$ denote the projections onto the complemented subspaces $X$ and $Y$.

Let $T= P_X$ and $S=3P_Y$, so that $\Vert T-S\Vert =3=\Vert T+S\Vert $. To construct the desired example, we show that if $\Vert R-S\Vert <1$, then $\Vert T-R\Vert = \Vert T+R\Vert $. So take such $R$ and note that then $\Vert P_YR \Vert >2$ and $\Vert P_XR\Vert<1$. It follows that $$ \Vert T-R\Vert = \max (\Vert P_X(T-R) \Vert ,\Vert P_Y(T-R)\Vert ) = \max (\Vert T-P_XR \Vert ,\Vert P_YR\Vert ) = \Vert P_Y R\Vert $$ (since $\Vert T-P_XR \Vert \leq \Vert T\Vert + \Vert P_XR \Vert$<2 and $\Vert P_Y R\Vert >2$).

Similarly, we conclude that $\Vert T+R\Vert = \Vert P_Y R\Vert$, hence $\Vert T+R\Vert = \Vert T-R\Vert $.

Edit: Note that since each $U\in L(X\oplus_1 Y)$ satisfies $\Vert U\Vert = \max (\Vert UP_X\Vert , \Vert UP_Y\Vert )$, a similar construction gives an example of such a ball for spaces isometrically isomorphic to $X\oplus_1 Y$ for nonzero $X$ and $Y$.

$\endgroup$
1
  • $\begingroup$ Perhaps if we could prove that this set is a vector space for some Banach space $E$, then we would be done (as it is always proper subspace). $\endgroup$ Oct 8, 2011 at 19:24
0
$\begingroup$

For a given $\varepsilon>0$ the set never contains the operator $\varepsilon T$ unless $T=0$.

$\endgroup$
1
  • $\begingroup$ Right, but I am not assuming that the ball must be centered at $T$. $\endgroup$ Oct 8, 2011 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.