Integral on $\ell^{\infty}$

Question

I begin with measure and integral theory. I want to give answer on the following statement:

Suppose $l^{\infty}$ is the Rieszspace of all bounded functions on $\mathbb{N}$. Define $\phi\colon\ell^{\infty}\rightarrow\mathbb{C}$ by: $$\phi(f)=2^{-1}f(1)+2^{-2}f(2)+\cdots.$$ Prove that $\phi$ defines an integral on $l^{\infty}$.

I did it on the following way: That $\phi$ is positiv is clear. Now suppose the sequence $(f_n)$ converges pointwise to $0$ with all $f_n$ positive. Thus the following holds:

$$\forall x\in\mathbb{N}\ \forall\epsilon>0\ \exists N>0\ \forall n>N: f_n(x)<\epsilon$$ But then we have:

$$\sum_{x=1}^\infty{\frac{f_n(x)}{2^x}}<\epsilon\sum_{x=1}^{\infty}{\frac{1}{2^x}}$$

This proves that $(\phi(f_n))$ also tends to zero, true?! Thus $\phi$ defines an integral.

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Is this proof good?

Thank you for help.

Answer 1

As you say, $\phi$ is obviously positive. For continuity, you need for there to exists $M>0$ such that $$ |\phi(f)| < M \quad \forall f \in \ell^{\infty} \text{ such that } \|f\|_{\infty} \leq 1 $$ Take $$ M = \sum_{n=1}^{\infty} \frac{1}{2^n} $$ Then, for any $f\in \ell^{\infty}$ such that $\|f\|_{\infty} \leq 1$, it follows that $$ |f(j)| \leq 1 \quad \forall j\in \mathbb{N} $$ Now just use the inequality you have up there.