Krdytkyu

Let $$f:overline{B(0,1)}rightarrowmathbb{C}$$ be continuous and holomorphic on $B(0,1)$. Consider the function $$zmapsto F(z):=f(z)overline{f(overline{z})}.$$ Show

(i) $int_{gamma}F(z)dz=0$ for the semicircle around $0$ and radius $1$ in the upper half-plane

(ii) $int_{-1}^{1}|f(x)|^2dxleqsqrt{int_{0}^{pi}|f(e^{iTheta})|^2dTheta}sqrt{int_{0}^{pi}|f(e^{-iTheta})|^2dTheta}$

Note: Cauchy-Schwarz inequality for $L^2([0,pi])$

(iii) $2int_{-1}^{1}|f(x)|^2dxleqint_{0}^{2pi}|f(e^{iTheta})|^2dTheta$

Can anyone help?

I am assuming you mean the $textit{closed}$ curve consisting of the segment $-1$ to $1$ followed by the upper unit semicircle. Hints:

$1). F$ is analytic on $B(0,1)$. Once you show this, i) is immediate.

$2). F(x+iy)=|f(x)|^2$ on the segment. Now split up the contour and use i).

$3). $ What happens if you integrate over the contour consisting of the segment and the lower semicircle?

For (i):

Using the differential quotient i showed that $overline{f(overline{z})}$ is holomorphic on $B(0,1)$. Hence $F(z)$ is holomorphic on $B(0,1)$ as a product of two holomorphic functions and thus $F(z)$ is analytic on $B(0,1)$.

When $gamma$ would be a closed curve i would be able to use the Cauchy's integral theorem. But $gamma$ is an arc of a circle and thus not closed.

Do i have an error in reasoning?

Edit:

My idea: I consider first the full circle $psi$ around $0$ and radius $1$. According to Cauchy's integral theorem $int_{psi}F(z)dz=0$. Now $psi$ is composed of $gamma$ and $rho$ in which $rho$ is the semicircle in the lower half-plane. Using the computational rules for curvilinear integrales i can maybe show (i):

$$gamma(t)=e^{it},tin[0,pi]$$
$$rho(t)=e^{it},tin[pi,2pi]$$

$$psi(t)=e^{it},tin[0,2pi]$$

$$Rightarrow0=int_{psi}F(z)dz=int_{gamma}F(z)dz+int_{rho}F(z)dz$$
$$Leftrightarrowint_{gamma}F(z)dz+int_{rho}F(z)dz=0$$

When $$int_{rho}F(z)dz=-int_{-gamma}F(z)dz=int_{gamma}F(z)dz$$ would apply we would have $$int_{gamma}F(z)dz=0$$
Is this right or nonsense?