Is the following function convex F(t;C) in C. Where $F(t;C) = {\exp({\sum_{k = 0}^p {{C_k}\cos (2\pi tk)} })}$

Question

Prove that the following function is convex F(t;C) in C.

Where $F(t;C) = \frac{1}{{{e^{\sum\limits_{k = 0}^p {{C_k}\cos (2\pi tk)} }}}}$ and $ - \frac{1}{2} \le t \le \frac{1}{2}$.

I tried the convexity definition assuming the domain set is a convex set.

I used the definition let ${C_1} \in D$ and ${C_1} \in D$ where $D$ is the convex domain then we need to prove

that $F(t;\lambda {C_1} + (1 - \lambda ){C_2}) \le \lambda F(t;{C_1}) + (1 - \lambda )(F(t;{C_2})$,

where $\lambda \in [0,1]$ in matrix notation I let ${T_k} = [\begin{array}{*{20}{c}} 1&{\cos (2\pi t)}&{\cos (4\pi t)\begin{array}{*{20}{c}} .&.&{\cos (2p\pi t){]^T}} \end{array}} \end{array}$

and ${C_1} = [\begin{array}{*{20}{c}} {{c_0}}&{{c_1}}&{{c_2}\begin{array}{*{20}{c}} .&.&{{c_p}]} \end{array}} \end{array}$ I need to prove $\frac{1}{{{e^{\lambda {C_1}^T.{T_k} + (1 - \lambda ){C_2}^T.{T_k}}}}} \le \frac{\lambda }{{{e^{{C_1}^T.{T_k}}}}} + \frac{{(1 - \lambda )}}{{{e^{{C_2}^T.{T_k}}}}}$.

Any help will be appreciated thank you all.

Answer 1

I did solve the problem and just in case somebody else run into the same problem this is how I solved it. We start $$w(t;\textbf{C}) = \frac{1}{{\exp \left( {\sum\limits_{k = 0}^p {{c_k}\cos (2\pi kt)} } \right)}}$$ where $ - \frac{1}{2} \le t \le \frac{1}{2}$. Before we take the derivative we will do some abbreviation let $$\textbf{C} = \left[ {\begin{array}{*{20}{l}} {{c_0}}&{{c_1}}&{{c_2}}&.&.&{{c_p}} \end{array}} \right]$$ $${\textbf{F}_k} = \left[ {\begin{array}{*{20}{l}} 1&{\cos (2\pi t)}&{\cos (4\pi t)}&.&.&{\cos (2p\pi t)} \end{array}} \right]$$ then we can write the function as following $$w(t;\textbf{C}) = \frac{1}{{\exp \left( {\textbf{C} \cdot \textbf{F}_k^T} \right)}}$$ taking the first derivative (Gradient) $$\frac{{\partial w}}{{\partial {C_m}}} = \frac{{ - \cos (2\pi mt)}}{{\exp \left( {\textbf{C} \cdot F_k^T} \right)}}$$ and taking the second derivative (Hessian) $$\frac{{\partial w}}{{\partial {C_m}\partial {C_n}}} = \frac{{\cos (2\pi mt)\cos (2\pi nt)}}{{\exp \left( {\textbf{C} \cdot F_k^T} \right)}}$$ To write the hessian matrix notice $$\frac{{\partial w}}{{\partial {C_m}\partial {C_n}}} = \left\{ {\begin{array}{*{20}{l}} {\frac{{\cos {{(2\pi mt)}^2}}}{{\exp \left( {C \cdot F_k^T} \right)}}}&{m = n}\\ {\frac{{\cos (2\pi mt)\cos (2\pi nt)}}{{\exp \left( {\textbf{C} \cdot F_k^T} \right)}}}&{m \ne n} \end{array}} \right.$$ let $$P = \exp \left( { - \textbf{C} \cdot \textbf{F}_k^T} \right)$$ writing the Hessian matrix we have $$H = \frac{{\partial w}}{{\partial {C_m}\partial {C_n}}} = \left( {\begin{array}{*{20}{c}} P& \ldots &{\cos (2\pi t)\cos (2p\pi t)P}\\ \vdots & \ddots & \vdots \\ {\cos (2\pi t)\cos (2p\pi t)P}& \cdots &{\cos {{(2p\pi t)}^2}P} \end{array}} \right)$$ where the diagonal of the matrix is the following $$Dig = \left[ {\begin{array}{*{20}{l}} P&{\cos {{(2\pi t)}^2}P}&.&.&{\cos {{(2p\pi t)}^2}P} \end{array}} \right]$$ we can see that the diagonal is all positive, also notice that the matrix is real and symmetric. By using the pivots of the matrix for different order you will notice that all the pivots are non negative hence the Hessian is a positive semi definite matrix, which indicate that $w$ is a convex function which mean it satisfies the following, Let $C_1 \in D$, and $C_2 \in D$ where $D$ is assumed to be a convex set (Domain) then the following is true $$w(t;\lambda {\textbf{C}_1} + (1 - \lambda ){\textbf{C}_2}) \le \lambda w(t;{\textbf{C}_1}) + (1 - \lambda )w(t;{\textbf{C}_2})$$ that is $$\frac{1}{{\exp \left( {\left[ {\lambda {\textbf{C}_1} + (1 - \lambda ){\textbf{C}_2}} \right] \cdot F_k^T} \right)}} \le \frac{\lambda }{{\exp \left( {{\textbf{C}_1} \cdot F_k^T} \right)}} + \frac{{(1 - \lambda )}}{{\exp \left( {{\textbf{C}_2} \cdot F_k^T} \right)}}$$