= Solution
For the <bounded density tilt> $f_t=f(1+tg)$, differentiate the polynomial in $t$:
$$
\left.\frac{d}{dt}\int_0^1 f_t(u)^4\,du\right|_{t=0}
=4\int_0^1 f(u)^4g(u)\,du=P_f(4f^3g).
$$
Since every <score function> is centered, the centered representer is
$$
\boxed{\varphi_f(u)=4\left(f(u)^3-\int_0^1 f(v)^4\,dv\right).}
$$
A bounded baseline $f$ makes this a bounded <mean-zero function>, hence an element of $L^2_0(P_f)$. Thus it is the <efficient influence function> for the <density fourth-power functional> relative to the regular <bounded density tilts>. Its squared <L2 norm> is $16\{\int f^7-(\int f^4)^2\}$.
\b[A bounded baseline alone does not make this functional differentiable along every quadratic-mean differentiable path.] The <derivative> above is the intended regular-path answer. To see the need for the qualification, let $f_0=1$ and put $b(s)=6s(1-s)$ on $[0,1]$, extended by zero outside. For $0<|t|<1$, define
$$
f_t(u)=1-t^4+t^{-2}b(u/t^6),\qquad 0\leq u\leq1.
$$
Each $f_t$ is a nonnegative continuous <probability density function>, because its narrow bump has mass $t^4$. Moreover,
$$
\int(\sqrt{f_t}-1)^2\,du\leq\int|f_t-1|\,du\leq2t^4=o(t^2).
$$
It is therefore a <differentiable-in-quadratic-mean path> with zero <score function>. Yet
$$
\int f_t^4\,du\geq t^{-2}\int_0^1b(s)^4\,ds=\frac{72}{35t^2}\longrightarrow\infty.
$$
The <density fourth-power functional> is not even continuous along this <statistical path>, although every $f_t$ is individually bounded. Consequently no <efficient influence function> represents <derivatives> over the unrestricted class of all such paths.
One sufficient additional condition is a common bound $f_t,f\leq M$ for all small $t$. <Taylor expansion> then bounds the fourth-power remainder by $6M^2(f_t-f)^2$, while
$$
\|f_t-f\|_2\leq2\sqrt M\,\|\sqrt{f_t}-\sqrt f\|_2=O(|t|).
$$
Together with the <quadratic-mean to L1 density derivative>, this gives $\psi(P_{f_t})-\psi(P_f)=tP_f(\varphi_fg)+o(|t|)$. Under this local bound, or when the chosen <statistical paths> are the <bounded density tilts>, the boxed <canonical gradient> is fully justified. The counterexample is a <spike obstruction to density-power differentiability>.
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