With $f(x)=\int_{|t|=3/2} x(t)dt$, $f$ is in the dual space of $X$

Question

Let $\Omega=\{t\in \mathbb{C}:1\leq |t| \leq 2\}$ and let $X$ be the complex Banach space of $C(\Omega)$ of complex-valued continuous functions defined on $\Omega$ with the supremum norm.

Define $f:X\to \mathbb{C}$ such that for $x\in X$, $$ f(x)=\int_{|t|=3/2} x(t)dt$$ Show that $f\in X^*$ (i.e. $f$ is in the dual space of $X$)

First, $f$ is linear. Then I want to show the integral is well-defined and $f$ is bounded.

If $x$ has no singularity inside the circle $|t|=3/2$, then $f(x)=0$. And hence the integral is well-defined.

If $x$ has singularity inside the circle $|t|=3/2$, then I don't know how to show the integral is well-defined and $\lVert f \rVert \leq C$, $C$ is some constant.

Answer 1

It does not really matter if the function has a singularity, its continuous on the annulus and therefore on the circle with radius 3/2. Your assumption that $x(t)$ has an antiderivative (and therefore the contour integral vanishes) is not required in any way, nor true. Take e.g. $|z|$ for a complex arguments. I think you try to apply your knowledge from complex analysis here, but it isnt required here. The function is also not holomorphic, just continuous and doesnt have to have values outside of the annulus.
The integral is well defined as the set $\partial D=\big\{t \in \mathbb{C}| \;|t|=3/2 \big\}$ is compact and the function is continuous on this set. Therefore, the integral is well defined. Furthermore, we get the estimate:
$$ |f(x)|=|\int_{\partial D} x(t)dt| \leq \int_{\partial D} |x(t)|dt \leq |\partial D| \sup_{t \in \partial D}|x(t)| \leq C \sup_{t \in \Omega} |x(t)| $$ As for linearity, it follows from the fact that well defined integrals are indeed linear. It also applies to integrals on $\partial D$.