Suppose that $X,Y$ are Banach space and $T:X\rightarrow Y$ is linear, continuous and overjective then $\overline{T(B(0,1))} + \overline{T(B(0,1))} \subseteq 2\overline{T(B(0,1))}$
Guys a question, I am reading the theorem of the open function and in a part of that demonstration I am asked to prove some things, among them this, I still can't do it, someone could give me some help. Thank you
We have $$B(0,1) + B(0,1) \subseteq 2B(0,1)$$ (in fact we have equality here, both sides being equal to $B(0,2)$). Applying the linear map $T$ we get $$T(B(0,1)) + T(B(0,1)) =T(B(0,1)+B(0,1))\subseteq T(2B(0,1)) = 2T(B(0,1)).$$ Applying the closure we get $$\overline{T(B(0,1)) + T(B(0,1))} \subseteq \overline{2T(B(0,1))}.$$ Now, for any two sets $A,B \subseteq Y$ by continuity of addition $+ : Y \times Y \to Y$ we have
$$\overline{A}+\overline{B} = +(\overline{A}\times \overline{B}) = + (\overline{A\times B}) \subseteq \overline{+(A\times B)} = \overline{A+B}.$$ Also, since $y \mapsto 2y$ is a homeomorphism $Y \to Y$, we have $\overline{2A} = 2\overline{A}$.
Therefore $$\overline{T(B(0,1))} + \overline{T(B(0,1))} \subseteq \overline{T(B(0,1)) + T(B(0,1))} \subseteq \overline{2T(B(0,1))} = 2\overline{T(B(0,1))}.$$