Htykuut

Question

Let $$f(z) = sum_{k=0}^{n}c_{k}z^{k}$$ be a polynomial, where $c_{k}inmathbb{C}$.

Prove that

$$int_{-1}^{1}|f(x)|^{2}dxleqpiint_{0}^{2pi}|f(e^{itheta})|^{2}frac{dtheta}{2pi} = pisum_{k=0}^{n}|c_{k}|^{2}$$

I tried letting $c_{k} = a_{k}+ib_{k}$ for real $a_{k},b_{k}$ and got to $$int_{-1}^{1}|f(x)|^{2}dx = int_{-1}^{1}left(sum_{k=0}^{n}a^{k}x^{k}right)^{2}+left(sum_{k=0}^{n}b^{k}x^{k}right)^{2}dx$$ but I'm not sure how to go from here.
Am I on the right track (if so can you outline the next few steps?) or should I be approaching this another way?

What I know;

• In the previous part of the question I proved for real $c_{k}$
  $$int_{-1}^{1}[f(x)]^{2}dxleqpiint_{0}^{2pi}|f(e^{itheta})|^{2}frac{dtheta}{2pi} = pisum_{k=0}^{n}c_{k}^{2}$$ using the Cauchy Theorem and two integrals around the uppper/lower unit semicircles:

Proof (if helpful) Consider the counter-clockwise contour from traversing the real axis from $(-1,0)$ to $(1,0)$, and then a semicircle of radius 1 from $(1,0)$ to $(-1,0)$. By the Cauchy Theorem the integral is $0$ as $f$ is holomorphic everywhere, so splitting the integrals into one along the real axis and one along the arc gives:
$$int_{-1}^{1}[f(x)]^{2}dx = -int_{0}^{pi}ie^{itheta}[f(e^{itheta})]^{2}dtheta$$

Now since $f$ is real valued if $c_{k}inmathbb{R}$, $|f|^{2}geq 0$, so $$int_{-1}^{1}[f(x)]^{2}dx = left|int_{-1}^{1}[f(x)]^{2}dxright| = left|-int_{0}^{pi}ie^{itheta}[f(e^{itheta})]^{2}dthetaright|leq int_{0}^{pi}|f(e^{itheta})|^{2}dtheta$$

Similarly by considering the contour traversing the real axis from $(1,0)$ to $(-1,0)$ and then along the semicircle of radius 1 from $(-1,0)$ to $(1,0)$, we get
$$int_{-1}^{1} [f(x)]^{2}dxleq int_{pi}^{2pi}|f(e^{itheta})|^{2}dtheta $$
Adding the two then gives the equality
$$int_{-1}^{1}[f(x)]^{2}dxleqpiint_{0}^{2pi}|f(e^{itheta})|^{2}frac{dtheta}{2pi} = pisum_{k=0}^{n}c_{k}^{2}$$
- I know how to prove the equality $$piint_{0}^{2pi}|f(e^{itheta})|^{2}frac{dtheta}{2pi} = pisum_{k=0}^{n}|c_{k}|^{2}$$

Here is the full question for reference; (note I only need help for the second part)

Write $c_k=a_k+text{i}b_k$ for $k=0,1,2,ldots,n$, where $a_k,b_kinmathbb{R}$. Then, $$f(z)=g(z)+text{i},h(z),,$$
where $g(z):=sumlimits_{k=0}^n,a_k,z^k$ and $h(z):=sumlimits_{k=0}^n,b_k,z^k$ are polynomials with real coefficients.

Note that, for $xinmathbb{R}$,
$$big|f(x)big|^2=big(g(x)big)^2+big(h(x)big)^2,.$$
Applying Part (a) of the question, you get
$$int_{-1}^{+1},big(g(x)big)^2,text{d}xleq pi,sum_{k=0}^n,a_k^2$$
and
$$int_{-1}^{+1},big(h(x)big)^2,text{d}xleq pi,sum_{k=0}^n,b_k^2,.$$
Adding the two inequalities above yields
$$int_{-1}^{+1},big|f(x)big|^2,text{d}xleq pi,sum_{k=0}^n,|c_k|^2,.$$