I am having trouble solving this task:
Let $X:=C_b([0,\infty))$ be the space of bounded functions $f:[0,\infty)\rightarrow\mathbb{R}$ equipped with the norm $\|f\|=\sup_{x\in[0,\infty)}|f(x)|$.
Show that there exists a continuous linear functional $L:X\rightarrow\mathbb{R}$ such that $$Lf=\lim_{x\rightarrow\infty}f(x)$$
for all $f\in X$ for which the limit on the right hand side exists.
I really have no idea how to do this at all. I am currently studying for an exam and found this task in an old exam of 2013.
Basically i have to find a continuous function $g:[0,\infty)\rightarrow\mathbb{R}$ such that $$\int_0^\infty f(x)g(x)dx=\lim_{x\rightarrow\infty}f(x)$$
This means that $g\in L^1([0,\infty))\cap C^0([0,\infty))$. But i have no idea where to start.
Maybe one could first find a functional on the subspace $Y\subseteq X$ given by $Y=\{f\in X: \|f\|\leq 1\}$ and then extend it with Hahn Banach, since then for any $f\in X$ we could use
$$Lf=\|f\|L\left(\frac{f}{\|f\|}\right)=\|f\|\lim_{x\rightarrow\infty}\frac{f(x)}{\|f\|}=\lim_{x\rightarrow\infty}f(x)$$
any help would be very much appreciated.
Let $Y$ be the subset of $Y$ such that $lim_{x\rightarrow +\infty}f(x)$ exists Just write $L(x)=lim_{x\rightarrow+\infty}f(x)$ on $Y$, it is linear and its bounded. Suppose that $\|f\|\leq 1$, this is equivalent to $Sup_{x\in [0,+\infty)}|f(x)|\leq 1$, you deduce that $|lim_{x\rightarrow +\infty}f(x)|\leq 1$. So $L$ is a linear bounded function and is continuous. You can apply Hahn Banach to extend this to $X$.