Prove inequalities with Cauchy's integral theorem

Question

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

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

(ii) $\int_{-1}^{1}|f(x)|^2dx\leq\sqrt{\int_{0}^{\pi}|f(e^{i\Theta})|^2d\Theta}\sqrt{\int_{0}^{\pi}|f(e^{-i\Theta})|^2d\Theta}$

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

(iii) $2\int_{-1}^{1}|f(x)|^2dx\leq\int_{0}^{2\pi}|f(e^{i\Theta})|^2d\Theta$

Can anyone help?

Answer 1

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?

Answer 2

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):

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

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

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

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?