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Let $X$ be the space of all sequences of complex numbers $(x_1,x_2,\ldots)$ such that $\lim_{n}{x_{n}}$ exists, and consider in this space the norm $\|x\|=\sup_{n}{|x_n|}$.

I need to find a linear continuous functional in $X$, that doesn't attain its norm in the closed unit ball.

I thought a functional like, $f:X\to\mathbb{K}$, defined as $$f(x)=\sum_{n\geq1}{\frac{x_n}{n!}}.$$

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Ok, in your case norm is easily attainable if you take $x = (1,1,1,\ldots)$.

Proper answer: take $$f(x) = \sum_{n\ge 1} \frac{e^{i\cdot n}}{n!}\cdot x_n,$$ this way to get $x$ which maximizes $|f(x)|$ you need to choose $x_i = e^{- i\cdot n}$, but a sequence with such elements doesn't converge.

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    $\begingroup$ If the scalar field is $\mathbb R$ then we can use $(-1)^{n}$ in place of $e^{in}$. To see that the norm is not attained we can argue as follows: suppose $||\{x_n\}|| =1$ and $f(x)=\sum \frac 1 {n!}$. Then $|\sum \frac 1 {n!}| =\sum \frac {(-1)^{n}} {n!} x_n \leq \sum \frac 1 {n!}$ since $|x_n| \leq 1$ for all n. Thus equality holds throughout and this forces $x_n$ to be $(-1)^{n}$ for each n. But $\{(-1)^{n}\}$ is not in c. $\endgroup$ – Kavi Rama Murthy Feb 16 '18 at 8:00
  • $\begingroup$ @KaviRamaMurthy oh yeah, that's a good one. I just took the first non-convergent sequence that came to my mind :) $\endgroup$ – mike239x Feb 16 '18 at 8:19

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