For an exercise, I need to prove that a functional is a bounded linear functional. The exercise is concretely posed as
Let $C([- \pi, \pi])$ be the Banach space of continuous functions on the interval $[- \pi, \pi]$ equipped with the usual norm $\left\| f \right\|_{\infty} = \max_{t \in [-\pi , \pi]} |f(t)|$. Let $g \in L^1([- \pi, \pi])$ be arbitrary and define therewith the functional $\varphi_g : C([- \pi, \pi]) \rightarrow \mathcal{C}$ by
$\varphi_g(f) = \frac{1}{2\pi} \cdot \int_{- \pi}^{\pi} f(t) g(t) dt \text{\quad } f \in C([-\pi, \pi])$
- Show that $\varphi_g$ belongs to the dual space of $C([-\pi, \pi])$, i.e. show that $\varphi_g$ is a bounded, linear functional on $C([-\pi, \pi])$
- What is the norm $\left\| \varphi_g \right\|$ of this functional?
Remark : Recall that $L^1([-\pi, \pi])$ is the Banach space of functions on $[-\pi, \pi]$ for which $\left\| f \right\|_1 = \frac{1}{2\pi} \cdot \int_{- \pi}^{\pi} |f(t)| dt < \infty$.
My thoughts so far :
- I would argue that it is easy to see that $\varphi_g$ is linear since the integral itself is linear w.r.t. the argument. Hence, I would skip any explicit proof here. Further, to show boundedness, we have $\left\| \varphi_g(f) \right\|_{1} = \left\| \varphi_g \cdot f \right\|_{1} \leq \left\| \varphi_g \right\|_{1} \cdot \left\| f \right\|_{1}$. With $\left\| \varphi_g \right\|_{1} < \infty$ as per the statement in the Remark and with $\left\| f \right\|_{1} < \infty$ as per definition of the norm $\left\| . \right\|_{1}$, we consequently have that $\left\| \varphi_g \right\|_{1} < \infty$. This proves the statement.
- ?
I would like to know whether the explanation in 1. is sufficient for the proof. Further, I am currently stuck with the norm in 2.
Any help is very much appreciated! :-)