For any $n\in\mathbb{N}$, let $\{0,1\}^n$ denote the set of $\{0,1\}$-strings of length $n$. For $n\in\mathbb{N}$ let $$\{0,1\}^* = \bigcup_{k\in\mathbb{N}}\{0,1\}^k \text{ and } \{0,1\}^{\leq n} = \bigcup_{k\leq n}\{0,1\}^k.$$
Note that $|\{0,1\}^{\leq n}| = 2^{n+1} -1$.
Cryptographic functions $h$ map $\{0,1\}^*$ to $\{0,1\}^n$ for some fixed $n\in\mathbb{N}$. Note that in general, for a cryptographic hash function $h:\{0,1\}^*\to \{0,1\}^n$ it is not known whether $\text{im}(h) = \{0,1\}^n$ where $\text{im}(h) = \{h(x): x\in\{0,1\}^*\}$ is the image of $h$.
A cryptographic hash function $h:\{0,1\}^*\to \{0,1\}^n$ is said to be weakly uniform if for every hash value there are infinitely many collisions, i.e. $h^{-1}(\{h(x)\})$ is infinite for every $x\in\{0,1\}^*$. (The pidgeonhole principle only implies the existence of one $x\in\{0,1\}^*$ such that $h^{-1}(\{h(x)\})$ is infinite.)
A cryptographic hash function $h:\{0,1\}^*\to \{0,1\}^n$ is said to be strongly uniform if the number of pre-images for each possible hash value approaches $1/|\text{im(h)}|$ as the input length grows longer, or, formally, if for every $x\in \{0,1\}^*$ we have: $$\lim \sup_{n\to\infty} \frac{|h^{-1}(\{h(x)\})\cap \{0,1\}^{\leq n}|}{2^{n+1}-1} =\frac{1}{|\text{im}(h)|}.$$
Question. Of all the hash functions known to be cryptographically secure, is there one of which we know whether it is weakly, or even strongly, uniform?