# linear image of closed convex set

Let $T:X\rightarrow Y$ be a continuous operator between Banch spaces. Denote by $T':Y'\rightarrow X'$ the norm adjoint map of $T$. Assume that $T'$ is injective on some subset $M$ of the unit sphere $U'$ of $Y'$. Under which conditions on $T'$ is the image $T'(C)$ of any closed convex set $C\subseteq M$, closed in the weak*-topology of $X'$?

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Does "the unit sphere" mean $\{y'\colon \|y'\|=1\}$ or $\{y'\colon \|y'\|\le 1\}$ (the unit ball)? It's a bit strange to consider convex subsets of the sphere. –  user31373 Jun 11 '12 at 18:56
@Norbert weak* closedness is not the same as norm-closedness even for convex sets (e.g., a norm-closed subspace of $\ell_1=c_0'$ may fail to be weak*-closed). –  user31373 Jun 11 '12 at 19:08
@Norbert The set $A=\lbrace x\in\ell_1\colon \sum_{n=1}^\infty \frac{n}{n+1}x_n=0\rbrace$ is norm-closed and convex. The sequence $a_n=\frac{n}{2n+1}e_1-\frac{n+1}{2n+1}e_n$ has weak* limit $\frac{1}{2}e_1$, which is outside of $A$. –  user31373 Jun 11 '12 at 19:27
@Norbert Yes, for the weak topology it's true. –  user31373 Jun 11 '12 at 20:06
@Andy: Sure! The same is true for $L^1$ spaces. I think all I was saying is that for a wide class of Banach spaces, it's a slightly weird question. It might help if the original question could pin down what Banach spaces they are interested in... –  Matthew Daws Jun 12 '12 at 19:58