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Let $X$ be a normed space and let $\varphi :X\rightarrow \mathbb{R}$ be linear and not continuous. Show that there exists such sequence $\left\{ x_n\right\}\subset X$, such that $\Vert x_n\Vert\rightarrow 0$ and $|\varphi(x_n)|\rightarrow\infty$, when $n\rightarrow\infty$.

I know that in linear operators between normed space, the boundess is the same as the countinuity. I did such excercises for concrete functionals, then the solution was just finding the counterexample (example of the sequence which is not bounded in the nbhd of $0$).

But what in general case? Maybe this excercise, in other words, is showing what I mentioned above, that not continuous functionals are not bounded?

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    $\begingroup$ If your functional is not bounded then for any $n$ there must be a vector $x_n$ so that $|\phi(x_n)|≥ n^2\,\|x_n\|$ (do you see why if this were not the case the functional must be bounded?). Do you see how this helps you? $\endgroup$ – s.harp Sep 27 '17 at 19:44
  • $\begingroup$ the inequality looks like, hmm, the result of using contraposition on the definition of boundness? $\endgroup$ – Yelon Sep 27 '17 at 19:54
  • $\begingroup$ Yes, the contraposition of the statement of the inequality is that there must exist an $n$ so that $|\phi(x)|≤n^2\|x\|$ for all $x$, implying boundedness. $\endgroup$ – s.harp Sep 27 '17 at 20:00
  • $\begingroup$ yes, but the main problem I have is probably that this excercise sounds for me, like repeating definisions. From unboundness we got your inequality (using contraposition of the other definition) and so on.. u know what I mean? just using equivalence definitions :) $\endgroup$ – Yelon Sep 27 '17 at 20:06
  • $\begingroup$ Often these kinds of exercises follow from unpacking definitions in the right way. I'm not entirely sure what you mean. $\endgroup$ – s.harp Sep 27 '17 at 20:18

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