Dual norm of a functional defined in $c_0$

Question

I'd like a check in the following exercise

In $(c_0, \Vert \cdot \Vert_{\infty}) $ consider for every element $u=(u_1,u_2,u_3,\ldots)$ in $c_0$ the functional $T(u)=\sum_{n=1}^{\infty}\frac{1}{2^n}u_n$.

Check that $T$ is a continuous linear functional and compute $\Vert T \Vert$ (the dual norm). Can one find some $u \in c_0$ s.t $\Vert u \Vert_{\infty}=1$ and $T(u)=\Vert T \Vert$ ?

---

The first part is clear and I'v checked it also here. My problem is on how to formalize the last part: I strongly believe that there's no $u \in c_0$ with that property, but I don't know how to move properly.

From the boundedness one can see that $\Vert T \Vert \leq 1$, and choosing the "cut-off" sequence $u=(\underbrace{1,...,1}_{N \text{times}},0,0,\ldots)$ I have that $\sum_{n=1}^{\infty} \frac{u_{n}}{2^n}=1-2^{-N}$, and then the supremum over $N$ is 1, and than follows that $\Vert T \Vert =1$.

But how can I say that there is no one $u$ with that property? I would say that the only way for which $T(u)$ is equal to $1 (=\Vert T \Vert)$ is the cut-off sequence is made from all entries equal to $1$, which is a contradiction, since $u$ must belong to $c_0$. Is this enough?

Answer 1

While you don't yet have a "proof" the idea is all there:

Suppose for contradiction that $T(u)=1$. Take any $\epsilon<1$. By definition of $c_0$ there is some $N$ such that $|u_n| < \epsilon$ for all $n\geq N$.

Then $|T(u)| \leq \left(\sum\limits_{n=1}^{N} + \sum\limits_{n=N+1}^{\infty}\right)| \frac{1}{2^n}u_n| \leq (1-2^{-N}) + \epsilon2^{-N}<1$, a contradiction.

Answer 2

Since $ \Vert u\Vert_\infty = 1,$ we must have $|u_n|\leq 1$ for all $n$. Evidently, we cannot have $u_n = 1$ for all $n$, since in that case $u \not\in c_0$. Hence $|u_{n_0}| < 1$ for some $n_0$. Thus $$ \left| \sum_{n=1}^\infty \frac{u_n}{2^n} \right| = \sum_{n=1}^\infty \frac{|u_n|}{2^n} = \sum_{\substack{n\geq 1\\n\not=n_0}} \frac{|u_n|}{2^n}+\frac{|u_{n_0}|}{2^{n_0}} < \sum_{\substack{n\geq 1\\n\not=n_0}} \frac{1}{2^n} + \frac{1}{2^{n_0}} = \sum_{n=1}^\infty \frac{1}{2^n}=1, $$ for any $u \in c_0$, so in particular equality cannot hold since the inequality is strict.