While I was reading a Convex Optimization book, I found an example, which is attached below. What is confused me is that how the authors derived/concluded that the hyperbolic cone is the image inverse of f(x).
2026-04-07 09:22:10.1775553730
Finding the inverse image of a function
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Let $C = \{(x, t) | z^Tz \le t^2 | t \ge 0\}$ and $K := \{x|x^TPx \le (c^Tx)^2, c^Tx \ge 0\}$. It's straightforward to see that $C$ has the representation $C = \{(z, t) | \|z\|_2 \le t, t \ge 0\} = \text{epi}(\|.\|_2) \cap (\mathbb R^n \times \mathbb R_+)$, and is thus convex.
It should be clear that $f$ is affine, and it's a rather straightforward exercise to show that the pre-image of a convex set under an affine function is again convex. So we just need to show that $K = f^{-1}C$. Now, noting that $ x^TPx = (P^{1/2}x)^TP^{1/2}x = ||P^{1/2}x||_2^2$ (since $P \in \mathbb S^n_+$), we get
$$ K = \{x|\|P^{1/2}x\|_2 \le c^Tx, c^Tx \ge 0\} = \{x | f(x) \in C\} = f^{-1}C,$$
and we're done.