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Let there be $f$ such that $$f: C_0 \mapsto \mathbb{R}$$ With $x=(x_n)_{n \in \mathbb{N}} $, and $$f(x)= \sum_{n=1}^{\infty}\frac{x^n}{2^n}$$ 1) Show that $f \in (C_0)'$ (dual space).

$\\$ 2) Show that there is no $x\in C_0$ such that $\mid\mid f \mid\mid = \mid f(x)\mid$

I have no problem with the first one, I get that $$\mid\mid f \mid\mid = 2$$ Which is the answer I should be getting.

For 2) I should suppose there exists an $x$ such that $ \mid\mid f \mid\mid = \mid f(x)\mid $ and $x$ converges to $0$. $$\mid f(x) \mid = \mid \sum_{n=1}^{\infty}\frac{1}{2^n} \mid$$ $$ \sum_{n=1}^{\infty}\frac{x^n}{2^n}=\sum_{n=1}^{\infty}\frac{1}{2^n} = 2 $$ I'm guessing I should now prove that $x=(x_n)_{n \in \mathbb{N}}=1$ for all n (a succesion of 1s), which means it converges to one (and not 0), and we have an absurd. How do I go on about doing that?

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    Your question is unclear and missing definitions. One moment $f$ is an operator which sends a continuous function (?) to a real number, and then $f$ is just a normal function. – 2017-02-01
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    Sorry, $f$ is a function not an operator. $C_0$ is the set of Cauchy sequences convergent to 0. – 2017-02-01
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    Do you mean $f(x)=\sum_{n\geq 1}\frac{x_n}{2^n}$? What structure does $C_0$ have? Do you consider it simply as a vector space? Normed space? – 2017-02-01
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    I copied the statement straight from the exercise sheet. Mind telling me how your definition changes things? – 2017-02-01
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    Well, how would you define $x^n$ where $x$ is a cauchy sequence? You could define this, but the result $f(x)$ has to be an element of $\mathbb{R}$ and not a sequence. I'd say that's a huge difference, because this does not make sense. – 2017-02-01
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    The norm of $f$ is $1$, not $2$. If $x$ is in $c_0$ (not $C_0$) and has norm $1$, then every coordinate of $x$ is at most $1$ in absolute value and some (most) coordinate is less than $1/2$ in absolute value. Use the triangle inequality to show $|f(x)|<1$. – 2017-02-01

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