Proof that this set is/isnt convex

Question

Ive got this trigonometric polynomial:

And with this in mind I need to show that this set is/isn't convex:

Following this, I need to prove that this function is convex:

I am completely stuck, any help would be appreciated.

Answer 1

I wanted to give a complete answer to the question now that I have figured it out, however, credits to TSF for starting me off with the first part.

a. \begin{aligned} & \forall x, y \in \Omega \Rightarrow T(x, w) > 0, T(y, w) > 0 \\\\ & \lambda \in [0, 1] \\\\ & T(\lambda x + (1-\lambda)y, w) \\\\ & = (\lambda x_1 + (1-\lambda)y_1) + ... + (\lambda x_n + (1-\lambda)y_n) \cos((n-1)w) \\\\ & = \lambda T(x, w) + (1 - \lambda)T(y, w) \\\\ & > 0 \text{ -> as shown in the first line} \\\\ & \text{meaning that all points between x and y are in the set and it is therefore convex.} \end{aligned}

b. \begin{aligned} & \forall x, y \in \Omega \Rightarrow T(x, w) > 0, T(y, w) > 0 \\\\ & \lambda \in [0, 1] \\\\ & - \intop_{0}^{2 \pi} \log T(\lambda x + (1-\lambda) y, w) dw \\\\ & = \intop_{0}^{2 \pi} -(\log (T(\lambda x, w) + T(1-\lambda) y, w))) dw \text{ -> we proved this equality in the first part} \\\\ & \leq \intop_{0}^{2 \pi} -\log T(\lambda x, w) -\log T((1-\lambda) y, w) dw \text{ -> -log is convex since log is concave} \\\\ & = - \intop_{0}^{2 \pi} \log T(\lambda x, w) dw - \intop_{0}^{2 \pi} \log T((1-\lambda) y, w) dw \\\\ & \text {this means that all points in between are in the set above the function and so the function is convex} \end{aligned}