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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.

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Of course, you mean $T\ne0$. – Denis Serre Oct 8 '11 at 12:45
Yes, you're right. Corrected. – Sellapan Nathan Oct 8 '11 at 13:50

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 $Y$.

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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). – Sellapan Nathan Oct 8 '11 at 19:24

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

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Right, but I am not assuming that the ball must be centered at $T$. – Sellapan Nathan Oct 8 '11 at 14:02

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