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Let $T$ be a surjective, continuous linear operator between two Banach spaces $E$ and $F$. Assume that it is $B_F(y_0,4c)\subset \overline{T(B_E(0,1))}$, where $c>0$, $y_0 \in F$ ($B$ is for "ball", the line means "closure"). Using Minkowski addition, we get

$$ B(0,4c) \subset \overline{T(B_E(0,1))} + \overline{T(B_E(0,1))}. $$

Is $B(0,4c)$ included in the single $\overline{T(B_E(0,1))}$ too?

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You can remove the constant $4$, which is useless (I guess it comes from a proof of the open mapping theorem). At least, we have the more accurate inclusion $B(0,c)\subset y_0+\overline{T(B_E(0,1))}$. – Davide Giraudo Jul 13 '12 at 19:04
The proof goes on by proving that $B(0,4c) \subset \overline{T(B_E(0,1))} + \overline{T(B_E(0,1))}$ implies $B(0,4c) \subset 2\overline{T(B_E(0,1))}$ (applying convexity) and so $B(0,2c) \subset \overline{T(B_E(0,1))}$. But I was not interested in the final result (contained in your message), I just want to know if the inclusion holds if we consider just one $\overline{T(B_E(0,1))}$. – Flast9 Jul 14 '12 at 11:24
up vote 2 down vote accepted

Yes. More generally, the following is true: if $K$ is a convex centrally symmetric subset of a normed space and $B(y,r) \subset K$, then $B(0,r) \subset K$.

Indeed, by symmetry we have $B(-y,r) \subset K$. If $z\in B(0,r)$ then both $z+y$ and $z-y$ belong to $K$. By convexity $z\in K$.

I guess the reason people usually don't bother with this argument is that the Minkowski sum approach is quicker.

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