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In a finite-dimensional vector space $X$, if $\phi: X \rightarrow \mathbb{R}$ is linear, then it can be shown to be continuous and bounded.

Bounded in this context means that there exists an $M < \infty$ such that for all $x\in X$, $|\phi(x)| \le M \|x\|$.

Does the converse hold? I.e. if we have a functional $\phi: X \rightarrow \mathbb{R}$ that is bounded and continuous, need it be linear?

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    $\begingroup$ Please do not delete questions once they got an answer. $\endgroup$ – quid Aug 11 '18 at 10:30
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    $\begingroup$ Which (nonlinear) examples did you check? $\endgroup$ – Did Aug 11 '18 at 10:38
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    $\begingroup$ You also shouldn't edit questions to ask something else once they got an answer. It appears you are happy with the answer you received since you've copied it into your new question here. That being the case you should revert to the original version of the question, accept the answer and ask your new question separately. $\endgroup$ – Rhys Steele Aug 11 '18 at 19:02
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No. For instance, consider $\phi(x)=\|x\|$.

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