# Prove $A$ is a linear continuous operator, if $A(x+y) = Ax + Ay$. [closed]

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.

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

Any hint?

## closed as off-topic by Nosrati, Delta-u, Key Flex, user10354138, Parcly TaxelOct 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 $$t\in\Bbb R$$.