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Assume $A$ is a map from real normed space $E$ to $F$ such that $A(x+y) = Ax + Ay \ \forall \ x, y \in E$ and $A$ is bounded on unit ball $B(0,1) \subset E$. Prove $A$ is a linear continuous operator.

Could someone please help me with this?

The only problem is I can't see how to prove $A(\alpha x) = \alpha Ax$ for $\alpha \in \mathbb{R}$.

Any hint?

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closed as off-topic by Nosrati, Delta-u, Key Flex, user10354138, Parcly Taxel Oct 16 '18 at 23:08

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For linearity, one can in effect assume that $E$ is one-dimensional. Fix $x\in E$ and define $\phi:\Bbb R\to F$ by $\phi(t)=A(tx)$. It suffices to prove $\phi$ is linear. It is additive, and bounded in a neighbourhood of zero. Then $\phi(t)=t\phi(1)$ for $t\in\Bbb Q$. Boundedness near zero implies (uniform) continuity of $\phi$, so taking limits gives $\phi(t)=t\phi(1)$ for $t\in\Bbb R$.

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  • $\begingroup$ Sr but how can you conclude that "ϕ(t)=tϕ(1) for t∈Q"? $\endgroup$ – Nguyễn Rèo Oct 16 '18 at 10:02
  • $\begingroup$ @NguyễnRèo additivity $\endgroup$ – Lord Shark the Unknown Oct 16 '18 at 13:59

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