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I have the following problem. Its supposed to be easy but i cant really get to a solution so i would be very thankful if somebody could point me in the right direction.

Show that for every $\lambda\in(l^\infty\mathbb(N))^{*}$ (the dualspace of bounded sequences), there exists unique elements $b\in l^{1}\mathbb(N)$ and $\mu\in(l^\infty\mathbb(N))^{*}$ such that $\lambda = \lambda_b + \mu$, where $\mu\big|_{c_0\mathbb(N)}=0$ and $\lambda_b(a)=\sum_{n=1}^{\infty}b_n a_n$.

Some of my ideas:

  1. Evaluate the standard sequence $e_n$ on both sides to try and determine $b_n$. But since the sequence are just bounded, i have to control the "size" of b to be a nullsequence, but even if i do that, i dont see how it is unique.
  2. Assume i have such a b, then i could restrict myself to the closed subspace of convergent sequences and extend the limes function to the whole of $l^{\infty}$ by Hahn-Banach. This would give my $\mu$ the desired property, but then again i lack uniqueness.

Im very thankful for any help.

NoIdea
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  • Uniqueness can be proved directly: If $\sum b_na_n=0$ whenever $a_n \to 0$ then it is obvious that $b_n=0$ for all $n$. – geetha290krm Jul 31 '23 at 07:48

1 Answers1

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If any such $b$ and $\mu$ with the given properties exist, there is no room to choose what $b$ is. In this very limited sense, there is nothing to "try and determine" about $b$.

In more detail, letting $e_n \in \ell^{\infty}$ denote the sequence with a $1$ in the $n$th entry and zeros everywhere else, assuming such $\lambda_b$ and $\mu$ exist we must have \begin{align*} \lambda(e_n) & = \lambda_b(e_n) + \mu(e_n) & \text{as $\lambda = \lambda_b + \mu$} \\ & = b_n + \mu(e_n) & \text{by the formula for $\lambda_b$} \\ & = b_n & \text{as $e_n \in c_0$ and $\mu|_{c_0} = 0$}. \end{align*} So if such a representation exists, it must be with the sequence $b$ defined by setting $b_n = \lambda(e_n)$ for every $n$.

To see that $b$ thus defined is in $\ell^1$, note that whatever the sequence $b$ is, it is possible to choose for each $n$ a scalar $|\omega_n| = 1$ such that $\omega_n b_n = |b_n|$, and that for any positive integer $N$, the element $\omega_N = \sum_{n=1}^N \omega_n e_n$ then satisfies $\|\omega_N\|_{\infty} = 1$. Using the fact that $\lambda \in (\ell^{\infty})^*$ is a bounded linear functional, it then follows that $$ \|\lambda\| = \|\lambda\| \cdot \|\omega_N\|_{\infty} \geq |\lambda(\omega_N)| = \left|\sum_{n=1}^N \omega_n b_n \right|= \sum_{n=1}^N |b_n|, $$ and since $N$ was arbitrary we deduce $b \in \ell^1$.

With $b \in \ell^1$ in hand, there is no need for Hahn-Banach to define $\mu$ in the representation; just note that $\mu := \lambda - \lambda_b$ vanishes on $c_0$ by construction. (I assume that you know, or are able to prove, that $\lambda_b$ defined by the given formula does indeed give an element of $(\ell^{\infty})^*$ whenever $b$ is in $\ell^1$, so that the difference $\lambda - \lambda_b$ makes sense as an element of $(\ell^{\infty})^*$.) The uniqueness of $\mu$ with the given properties follows from the uniqueness of $b$.

leslie townes
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