= Solution
An <M-estimator> based on an <estimating equation> chooses a consistent root of
$$
\frac1n\sum_{i=1}^n\psi(X_i,\widehat\theta)=0.
$$
Its population target $T(F)=\theta_F$ solves $\mathbb E_F\psi(X,\theta_F)=0$. For a scalar parameter, suppose the root is locally unique and
$$
A=-\mathbb E_F[\partial_\theta\psi(X,\theta_F)]\ne0,
\qquad B=\mathbb E_F[\psi(X,\theta_F)^2]<\infty.
$$
For the contaminated distribution $F_\varepsilon=(1-\varepsilon)F+\varepsilon\delta_x$, differentiate its population equation at zero contamination. The <influence function> satisfies
$$
0=\psi(x,\theta_F)-A\operatorname{IF}(x;T,F).
$$
Consequently
$$
\boxed{\operatorname{IF}(x;T,F)=\frac{\psi(x,\theta_F)}{A}.}
$$
The constant of proportionality is independent of $x$. A local expansion of the sample <estimating equation> similarly gives
$$
\sqrt n(\widehat\theta-\theta_F)
=A^{-1}\frac1{\sqrt n}\sum_{i=1}^n\psi(X_i,\theta_F)+o_p(1).
$$
The <central limit theorem> yields <asymptotic normality> with the <sandwich variance of an M-estimator>
$$
\boxed{V(\psi,F)=\mathbb E_F[\operatorname{IF}(X;T,F)^2]=\frac B{A^2}.}
$$
Thus the first-order <variance> of the unscaled estimator is $V/n$. For a vector parameter the same differentiation gives $\operatorname{IF}=A^{-1}\psi$ and $V=A^{-1}BA^{-\mathsf T}$, with $A=-\mathbb E\partial_\theta\psi$ and $B=\mathbb E\psi\psi^{\mathsf T}$.
For the location problem, take the target location to be zero and let $Z$ have the standard <normal distribution>, with density $\phi$ and distribution function $\Phi$. An odd score has $\mathbb E\psi(Z)=0$. For a differentiable score, integration by parts gives
$$
A=\mathbb E\psi'(Z)=\mathbb E[Z\psi(Z)].
$$
The last expression also gives the appropriate derivative of the population equation for bounded monotone scores with jumps, by differentiating the shifted normal density. Assume $A>0$ and put $h=\psi/A$. Then
$$
\mathbb E[Zh(Z)]=1,\qquad |h(x)|\le C,\qquad V=\mathbb E h(Z)^2.
$$
Multiplying a score by a positive constant changes neither its estimator nor this normalized <influence function>.
First establish feasibility. The bound implies
$$
1=\mathbb E[Zh(Z)]\le C\mathbb E|Z|=C\sqrt{2/\pi},
$$
so necessarily $C\ge\sqrt{\pi/2}$. For $C>\sqrt{\pi/2}$, let $K>0$ solve
$$
C=\frac K{p_K},\qquad p_K=2\Phi(K)-1.
$$
This solution is unique: the ratio tends to $\sqrt{\pi/2}$ as $K\downarrow0$, tends to infinity as $K\to\infty$, and has positive derivative because
$$
p_K-2K\phi(K)=2\int_0^K\{\phi(z)-\phi(K)\}\,dz>0.
$$
Take the <Huber score> $\psi_K(x)=\max(-K,\min(x,K))$. Its derivative is one on $(-K,K)$ and zero outside, so $A=p_K$. Hence
$$
h_K(x)=\frac{\psi_K(x)}{p_K}
=\max\left(-C,\min\left(\frac{x}{p_K},C\right)\right).
$$
It is odd, nondecreasing, meets the bound, and satisfies $\mathbb E[Zh_K(Z)]=1$.
To prove optimality, put $a=1/p_K$. For each fixed $x$, the minimum of $u^2-2axu$ over $|u|\le C$ is attained by projecting $ax$ onto $[-C,C]$, which is exactly $h_K(x)$. Therefore every competing normalized <influence function> $h$ satisfies
$$
h(x)^2-h_K(x)^2\ge2ax\{h(x)-h_K(x)\}.
$$
Taking <expectations>, the right-hand side vanishes because both functions satisfy $\mathbb E[Zh(Z)]=1$. Thus $\mathbb E h^2\ge\mathbb E h_K^2$. This proves the <optimal bounded influence function for normal location>, even over the larger class of all bounded normalized functions, and hence over the required odd nondecreasing class. Its minimum <asymptotic variance> is
$$
\boxed{V_K=
\frac{p_K-2K\phi(K)+K^2(1-p_K)}{p_K^2},
\qquad C=\frac K{p_K}.}
$$
The numerator is $\mathbb E[Z^2\mathbf1_{\{|Z|\le K\}}]+K^2\Pr(|Z|>K)$.
At the included boundary $C=\sqrt{\pi/2}$, equality in the feasibility bound forces $h(x)=C\operatorname{sgn}(x)$ almost everywhere. This is the <influence function of the sample median>, with $V=\pi/2$. It is obtained from the rescaled scores $\psi_K/K$ as $K\downarrow0$. \b[The boundary solution is the sign score, or equivalently the median estimator.] Literal substitution of $K=0$ in the clipped score gives the identically zero function and no identifiable <estimating equation>; the printed clipped formula therefore requires this limiting interpretation at equality.
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