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Let $U$ and $V$ be normed linear spaces over $\mathbb{R}$, and $L : U \mapsto V$ a linear function. Prove that if $L(S)$ is bounded, where $S$ is the unit sphere of $U$, then $L$ is Lipschitz.

There are several things that don't make sense to me in this question:

Why should $S$ be "the" unit sphere of $U$? Does that mean $U$ is an interval in $\mathbb{R}$ with length $2$? Also why does the question use the term "sphere" when it's only $1-D$?

Also by linear function, do they mean functions in the form $f(x)=ax+b$? (I guess $b=0$ or otherwise it'd be called affine.) If so, then $|f(x)-f(y)|=|a||x-y|$, which is Lipschitz. Why do we even need the "sphere" condition? Thank you.

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A linear function is a function $L:U\rightarrow V$ such that $L(ax+by)=aL(x)+bL(y)$. Here $a,b\in \mathbb{R}$ and $x,y$ are vectors in $U$. The unit sphere of $U$ is the set $S=\{x\in U : \| x \| = 1 \}$. So you want to prove that $\|L(x)-L(y)\|\leq k \|x - y\|$ knowing that there exist $a,b,\in \mathbb{R}$ such that $a\leq L(x) \leq b$ for every $x\in S$. –  Alexander Gruber Jan 13 '13 at 21:33
    
@user57162: Do you know what "normed linear space" means? Do you know examples of normed linear spaces that are not $\mathbb R$? In some way accompanying this problem, do you have a reference where you can look up definitions and other related information? –  Jonas Meyer Jan 14 '13 at 0:10
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