Let $f:\mathbb R^n\to \mathbb R$ be a symmetric convex $C^2$ function defined by $(\lambda_1,\cdots,\lambda_n)\mapsto f(\lambda_1,\cdots,\lambda_n)$.
Define $F:Sym(n)\cong\mathbb R^{\frac{n(n+1)}{2}}\to \mathbb R$ by $F(M)=f(\lambda_1,\cdots,\lambda_n)$ where $Sym(n)$ is the set of $n\times n$ symmetric matrices and $\lambda_i$'s are the eigenvalues of $M$. How to show that $F$ is also $C^2$ and convex?
Thoughts: If $f$ is analytic, then its Taylor series is symmetric and we can group the homogeneous terms, which can be represented by elementary symmetric polynomials. They are just coefficients of the characteristic polynomial, which depends analytically on the entries of $M$. However, I do not know how to weaken the assumption to $C^\infty$ or even just $C^2$.
I do not know if this general statement is true, but I am in fact considering $f=\exp\left(-A\sum_i\arctan\lambda_i\right)$ which appears in solving the Dirichlet problem for the Lagrangian phase operator. This function is convex if $\sum_i\arctan(\lambda_i)\ge(n-2)\pi/2+\delta$ for some $A(\delta)>>0$.