= Solution
Let $g$ be a <bounded function> with $P_fg=0$. A <bounded density tilt> realizes this direction:
$$
f_t(u)=f(u)(1+tg(u)),\qquad |t|\|g\|_\infty<1.
$$
These are nonnegative <probability density functions>, since $\int f_t=1+tP_fg=1$. The uniform <Taylor expansion> of the square root gives
$$
\sqrt{f_t}-\sqrt f-\frac t2g\sqrt f
=\sqrt f\,O(t^2g^2).
$$
The squared <L2 norm> of this remainder is $O(t^4P_fg^4)=o(t^2)$, since $g$ is bounded. Thus the <statistical path> is <differentiable in quadratic mean> with <score function> $g$.
Conversely every <score function> is centered by the <mean-zero score identity under quadratic-mean differentiability>. Choosing all these <bounded density tilts> therefore gives the <statistical tangent set>
$$
\boxed{\dot{\mathcal P}_f=\{g:\ g\text{ is bounded and measurable},\ P_fg=0\}.}
$$
This is a valid choice of <statistical tangent set>; it does not assert that every possible <score function> in the unrestricted density model is bounded.
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