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Let $(X, \Vert . \Vert)$ be a normed space with $\dim X = \infty$.

Show that if $f: X \rightarrow R$ is linear and not continuous, then

$$f(B_X(x_0, r)) = \mathbb{R} \quad \forall x_0 \in X, \forall r > 0$$ where $B_X(x_0, r)$ is the open ball with center $x_0$, and radius $r$.

If $f$ is linear and not continuous then it is not bounded. In particular, $f(B_x(x_0, r))$ is not bounded (?). Why it is necessarily $\mathbb{R}$?

Another User
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Ricardo
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    Hint: You can make $f(x)$ arbitrarily large for $x \in B_X(x_0, r)$. What can you say about $f(\lambda x)$ for $\lambda \in [-1, 1]$? – Theo Bendit Oct 28 '22 at 13:04
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    Even more simply, $\Bbb{R}$ is a one-dimensional space, and $\operatorname{range} f$ is a subspace of $\Bbb{R}$. What are the possible subspaces? – Theo Bendit Oct 28 '22 at 13:06
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    Here are postings in MSE ask and answer your question: exaple 1, example 2 this one deals with openness of linear functionals, one solution deals with bounded and unbounded cases separately, the later case answers your question. – Mittens Jan 08 '23 at 13:18

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