Suppose that we have a sequence of matrices $\{\Sigma_n\in\mathbb{R}^{n\times n}\}_{n=1}^\infty$ with uniformly bounded spectral norm, that is to say, the sequence $\{\|\Sigma_n\|\}_{n=1}^\infty$ is uniformly bounded. It worth noted that the dimensions of $\Sigma_n$ increase as $n$ increase.
Now we consider a scalar function $f:\mathbb{R}\to\mathbb{R}$, and for any matrix $A=\left(a_{ij}\right)_{n\times n}$, we denote $f(A)$ as the matrix that $f$ operates on each element of $A$ individually, that is $$f(A):=\left(f(a_{ij})\right)_{n\times n}.$$ My question is under what conditions can we obtain that the sequence $\{\|f(\Sigma_n)\|\}_{n=1}^\infty$ is also uniformly bounded?
Actually, I faced this problem in a specific case that $\{\Sigma_n\in\mathbb{R}^{n\times n}\}_{n=1}^\infty$ is a sequence of symmetric positive definite matrix with uniformly bounded spectral norm, and all diagonal elements equal to $1$ as well as all non-diagonal elements $\in(-1,1)$. The function is $\arcsin(\cdot)$, and I want to show that the sequence $\{\|\arcsin(\Sigma_n)\|\}_{n=1}^\infty$ is uniformly bounded. Here, although $\arcsin(\cdot)$ is a bounded function, I have tried several ways in different equivalent definitions of spectral norm, but still unsolved.
I would be very grateful if you can solve the general problem or the specific case above!