I am trying to solve the following problem.
Consider the Euclidean space $\mathbb{E}:=\mathbb{R}^n$. For a sequence of positive numbers $\{\alpha_k\}_{k=1}^n$ define the function $\varphi:\mathbb{E}\to\mathbb{R}$ by
$$\varphi(x):=\sum_{k=1}^n \alpha_kx_k^2,\,\,\,\,\,\,\,x=(x_1,...,x_n)^T\in\mathbb{E}$$
(a) Show that $\varphi$ is convex.
(b) Compute the conjugate function $\varphi^*$.
(c) Compute the second conjugate and show (explicitly) that $\varphi^{**}=\varphi$.
Part(a) is straight forward. For part (b) I got the following answer
$$\varphi^*(f)=\begin{cases}\frac{1}{4}\sum_{k=1}^n \frac{1}{k}f_k^2\,\,\,\,\,\,\text{ if }\sum_{k=1}^n \frac{1}{k}f_k^2<\infty\\ +\infty \,\,\,\,\,\,\,\,\text{ if otherwise}\end{cases}$$
Not sure whether I got it right or not? Any help would be highly appreciated.