I've been learning complex analysis recently, there's a question about the estimation of polynomials on the real line using the knowledge of complex analysis.
Suppose $f(x)={c_0}+{c_1}x+{c_2}x^2+...+{c_n}x^n$ is a polynomial.If all the parameters are real numbers,then prove:$$\int_{-1}^1 f(x)^2dx \leqslant \pi\int_0^{2\pi}|f(e^{i\theta})|^2\frac{d\theta}{2\pi} = \pi \sum_{k=0}^n c_k^2 $$
I know that the last equaiton is the Parseval's identity which is also right for many other functions more than polynomials.But I'm confuse with the first inequation.From my previous learning ,I guess I need transform the integral interval from real line to the complex plane.I've tried the following operation, $$\int_{-1}^1 f(x)^2dx = \int_0^{2\pi}f(cos\theta)^2sin\theta d\theta$$It's still operated on the real line.I want to know the next step which transforms the integral into the form above.
Applying Cauchy to $f^2$ on the upper semidisk $|z| \le 1, \Im z \ge 0$ one gets $$\int_{-1}^1f^2(x)dx+\int_0^{\pi}ie^{i\theta}f^2(e^{i\theta})d\theta=0$$
Using the triangle inequality gives (the middle equality holding by the positivity of $f^2$ on $[-1,1]$):
$$\int_{-1}^1f^2(x)dx=-\int_0^{\pi}ie^{i\theta}f^2(e^{i\theta})d\theta=|\int_0^{\pi}ie^{i\theta}f^2(e^{i\theta})d\theta|\le \int_0^{\pi}|f(e^{i\theta})|^2d\theta$$
Applying Cauchy on the lower semidisk gives as above
$$\int_{-1}^1f^2(x)dx\le \int_{-\pi}^{0}|f(e^{i\theta})|^2d\theta$$
Adding the two relations we are done!