Trace distance of two classical-quantum state with hashing

Let's say I have a classical-quantum(cq) state $$\rho_{XE}$$, where the classical part $$(X)$$ is orthogonal. It's trace distance from another uniform density operator is defined to be:

$$\frac{1}{2}||\rho_{XE} - \rho_U \otimes \rho_E||_1.$$

Now I apply a hash function $$f$$ from $$F$$ distributed according to probability $$p_f$$ on the first register and get a new state: $$\rho_{F(X)E}:= \sum_f p_f \; \rho_{f(X)E}.$$ Then I notice that its trace distance has the following upper bound:

$$\frac{1}{2}||\rho_{F(X)E} - \rho_U \otimes \rho_E|| \le \epsilon .$$ Now, from this upper bound, what can I infer for the first trace distance, i.e. without the hash function? Would the following be true?

$$\frac{1}{2}||\rho_{XE} - \rho_U \otimes \rho_E|| \le \epsilon.$$

Thanks!

No, this is not possible. The existence of such a hash function requires the (smooth) min-entropy of the initial state to be large enough but does not depend on its trace distance from a uniform state. For the simplest example possible let's forget about the side information $$E$$ and just focus on the $$X$$ system. The basic idea is that we can always pad $$X$$ with extra information such that the min-entropy stays constant but the distance from uniform grows.
Suppose $$X$$ is a random bit string of length $$n$$ such that the first bit $$x_1$$ is chosen uniformly at random and then $$x_2, \dots, x_n$$ are all $$0$$. Now the min-entropy is $$- \log \max_{(x_1,\dots,x_n)} p(x_1,\dots, x_n) = 1$$. Using, for example, Lemma 1 from  there is a hashing procedure such that we can extract 1 bit (i.e. $$F(X)$$ is a single bit) and $$\frac12 \| F(X) - U_1\|_1 \leq 1/2$$ where $$U_k$$ denotes a uniformly distributed random variable over $$k$$ bits. But now we can calculate \begin{aligned} \frac12 \| X - U_n\|_1 &= \frac12\sum_{(x_1,\dots, x_n) \in \{0,1\}^n} |p(x_1,\dots,x_n) - 2^{-n}| \\ &= \frac12\left(2 |\frac12 - 2^{-n}| + 2(2^{n-1} - 1)|2^{-n}| \right) \\ &= 1 - 2^{1-n}. \end{aligned} Thus as $$n\rightarrow \infty$$ we get $$\frac12 \| X - U_n\|_1 \rightarrow 1$$ but it can always be hashed down to a single bit that is distance $$1/2$$ away from uniform. By considering larger uniform sequences at the beginning of $$X$$ you should be able to make this distance grow even further, i.e. the hashed tends to perfectly uniform but the non-hashed is almost perfectly distinguishable.