$T : X \to Y$ be a surjective linear continuous operator($X, Y$ : Banach Space). Then does $$ T(\overline{B(0,1)}) = \overline{T(B(0,1))} $$ hold? Here $B(0,1)$ means an open ball in $X$, and Bar means its closure.
Is it true that $ T(\overline{B(0,1)}) = \overline{T(B(0,1))} $ for continuous operator $T$?
1
$\begingroup$
functional-analysis
-
0Care to define what $X$ and $Y$ stand for? Are they Banach spaces? – 2012-07-20
-
0@OlivierBégassat Sorry, assume that X,Y be Banach, and $B(0,1):= \{x \in X | \| x \| < 1 \} $ – 2012-07-20
-
0@OlivierBégassat I think that for continuous map, $T( \overline{A}) \subset \overline{T(A)}$, but how about the other side? – 2012-07-20
-
0@OlivierBégassat One more property added, T is surjective. – 2012-07-20
-
0I don't think this holds. Consider the Banach space $c_0$ of all real (or complex) sequences that tend to $0$ equipped with the supremum norm $\|\cdot\|_{\infty}$. Consider the linear functional $T:c_0\rightarrow \mathbb R, (a_n)\mapsto \sum_{n\geq 0} \frac{a_n}{2^n}$, then we obvioulsy have $\|T\|=2$, and yet $T(\overline{B(0,1)})=(-2,+2)$ which is not closed. – 2012-07-20