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

Where $F(t;C) = frac{1}{{{e^{sumlimits_{k = 0}^p {{C_k}cos (2pi tk)} }}}}$ 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 ${T_k} = [begin{array}{*{20}{c}}
1&{cos (2pi t)}&{cos (4pi t)begin{array}{*{20}{c}}
.&.&{cos (2ppi t){]^T}}
end{array}}
end{array}$

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.

I did solve the problem and just in case somebody else run into the same problem this is how I solved it. We start
$$w(t;textbf{C}) = frac{1}{{exp left( {sumlimits_{k = 0}^p {{c_k}cos (2pi kt)} } right)}}$$
where $ - frac{1}{2} le t le frac{1}{2}$. Before we take the derivative we will do some abbreviation let
$$textbf{C} = left[ {begin{array}{*{20}{l}}
{{c_0}}&{{c_1}}&{{c_2}}&.&.&{{c_p}}
end{array}} right]$$
$${textbf{F}_k} = left[ {begin{array}{*{20}{l}}
1&{cos (2pi t)}&{cos (4pi t)}&.&.&{cos (2ppi t)}
end{array}} right]$$
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)
$$frac{{partial w}}{{partial {C_m}}} = frac{{ - cos (2pi mt)}}{{exp left( {textbf{C} cdot F_k^T} right)}}$$ and taking the second derivative (Hessian) $$frac{{partial w}}{{partial {C_m}partial {C_n}}} = frac{{cos (2pi mt)cos (2pi nt)}}{{exp left( {textbf{C} cdot F_k^T} right)}}$$
To write the hessian matrix notice
$$frac{{partial w}}{{partial {C_m}partial {C_n}}} = left{ {begin{array}{*{20}{l}}
{frac{{cos {{(2pi mt)}^2}}}{{exp left( {C cdot F_k^T} right)}}}&{m = n}\
{frac{{cos (2pi mt)cos (2pi nt)}}{{exp left( {textbf{C} cdot F_k^T} right)}}}&{m ne n}
end{array}} right.$$
let
$$P = exp left( { - textbf{C} cdot textbf{F}_k^T} right)$$
writing the Hessian matrix we have
$$H = frac{{partial w}}{{partial {C_m}partial {C_n}}} = left( {begin{array}{*{20}{c}}
P& ldots &{cos (2pi t)cos (2ppi t)P}\
vdots & ddots & vdots \
{cos (2pi t)cos (2ppi t)P}& cdots &{cos {{(2ppi t)}^2}P}
end{array}} right)$$
where the diagonal of the matrix is the following
$$Dig = left[ {begin{array}{*{20}{l}}
P&{cos {{(2pi t)}^2}P}&.&.&{cos {{(2ppi t)}^2}P}
end{array}} right]$$
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
$$frac{1}{{exp left( {left[ {lambda {textbf{C}_1} + (1 - lambda ){textbf{C}_2}} right] cdot F_k^T} right)}} le frac{lambda }{{exp left( {{textbf{C}_1} cdot F_k^T} right)}} + frac{{(1 - lambda )}}{{exp left( {{textbf{C}_2} cdot F_k^T} right)}}$$