Influence function and sensitivity of the median

Since $I$ is open, $f$ is continuous at $m$, and $f(m)>0$, choose $\delta >0$ and $c>0$ so that $[m-\delta ,m+\delta ]\subset I$ and $f(u)\ge c$ throughout this closed interval. For $s<t$ in it, the density identity gives $F(t)-F(s)={\int }_s^{t}f(u)du\ge c(t-s)>0$. Thus $F$ is continuous and strictly increasing there, with $F(m)=1\mathrm{/}2$.

Suppose first that $x<m$, and shrink $\delta$ so that $m-\delta >x$. For sufficiently small $ϵ>0$, the number $r_{ϵ}=(1\mathrm{/}2-ϵ)\mathrm{/}(1-ϵ)$ lies strictly between $F(m-\delta )$ and $1\mathrm{/}2$. Continuity and strict increase give a unique $q_{ϵ}\in (m-\delta ,m)$ with $F(q_{ϵ})=r_{ϵ}$. Since $r_{ϵ}\uparrow 1\mathrm{/}2$ as $ϵ\downarrow 0$, for every $0<\eta <\delta$ eventually $r_{ϵ}>F(m-\eta )$, so $m-\eta <q_{ϵ}<m$ and $q_{ϵ}\to m$. Since $q_{ϵ}>x$, the mixture cdf at $q_{ϵ}$ is $(1-ϵ)F(q_{ϵ})+ϵ=1\mathrm{/}2$. For $t<x$, the contamination term is zero and, by monotonicity and $F(m)=1\mathrm{/}2$, the mixture cdf is $(1-ϵ)F(t)\le (1-ϵ)\mathrm{/}2<1\mathrm{/}2$. At $t=x$, monotonicity gives $F(x)\le F(m-\delta )<r_{ϵ}$, so $(1-ϵ)F(x)+ϵ<(1-ϵ)r_{ϵ}+ϵ=1\mathrm{/}2$. For $x<t<q_{ϵ}$, the contamination term is $ϵ$ and $F(t)<F(q_{ϵ})=r_{ϵ}$: below the local interval this follows from monotonicity and $F(t)\le F(m-\delta )<F(q_{ϵ})$, and inside it from strict increase. Therefore the mixture cdf at such $t$ is $(1-ϵ)F(t)+ϵ<(1-ϵ)r_{ϵ}+ϵ=1\mathrm{/}2$. Together with the cases $t<x$ and $t=x$, this puts the mixture cdf below $1\mathrm{/}2$ for every $t<q_{ϵ}$, while it equals $1\mathrm{/}2$ at $q_{ϵ}$, so [F2] gives $m(F_{ϵ,x})=q_{ϵ}$.

If $x>m$, shrink $\delta$ so that $m+\delta <x$. For small $ϵ>0$, $s_{ϵ}=1\mathrm{/}(2(1-ϵ))$ lies strictly between $1\mathrm{/}2$ and $F(m+\delta )$. There is a unique $q_{ϵ}\in (m,m+\delta )$ with $F(q_{ϵ})=s_{ϵ}$. Since $s_{ϵ}\downarrow 1\mathrm{/}2$ as $ϵ\downarrow 0$, for every $0<\eta <\delta$ eventually $s_{ϵ}<F(m+\eta )$, so $m<q_{ϵ}<m+\eta$ and $q_{ϵ}\to m$. Now $q_{ϵ}<x$, so the mixture cdf at $q_{ϵ}$ is $(1-ϵ)F(q_{ϵ})=1\mathrm{/}2$. For every $t<q_{ϵ}$, the contamination term is zero; if $m\le t<q_{ϵ}$, strict increase on the local interval gives $F(t)<F(q_{ϵ})$, and if $t<m$ then monotonicity gives $F(t)\le F(m)<F(q_{ϵ})$. Thus the mixture cdf is below $1\mathrm{/}2$ for every $t<q_{ϵ}$, while it equals $1\mathrm{/}2$ at $q_{ϵ}$, so [F2] gives $m(F_{ϵ,x})=q_{ϵ}$.

In the case $x<m$, step 2.1 gives $F(q_{ϵ})-F(m)=-ϵ\mathrm{/}[2(1-ϵ)]$ and $q_{ϵ}\to m$. The density identity and continuity of $f$ imply $[F(q_{ϵ})-F(m)]\mathrm{/}(q_{ϵ}-m)\to f(m)$. Dividing these two relations yields $(q_{ϵ}-m)\mathrm{/}ϵ\to -1\mathrm{/}(2f(m))$.

In the case $x>m$, step 2.2 gives $F(q_{ϵ})-F(m)=ϵ\mathrm{/}[2(1-ϵ)]$ and $q_{ϵ}\to m$. The same density quotient tends to $f(m)$, hence $(q_{ϵ}-m)\mathrm{/}ϵ\to +1\mathrm{/}(2f(m))$.

If $x=m$, continuity of $F$ at $m$ gives $F_{ϵ,m}(m-)=(1-ϵ)\mathrm{/}2<1\mathrm{/}2$ and $F_{ϵ,m}(m)=(1+ϵ)\mathrm{/}2>1\mathrm{/}2$. Monotonicity then puts every $t<m$ below the median threshold, while $m$ itself is above it, so the lower median remains exactly $m$ and its right derivative is $0$. By [F3] and steps 3.1–3.2, these three cases give the displayed pointwise influence function.

The absolute influence value is $1\mathrm{/}(2f(m))$ for every $x\mathrm{\ne }m$ and zero at $m$. Both sides of $m$ are nonempty in $\mathbb{R}$, so the supremum over all directions is $1\mathrm{/}(2f(m))$ by [F4]. Since $F$ has a density near $m$, $F\{m\}=0$; therefore the conventional expression $(1\mathrm{/}2-\mathbf{1}\{x\le m\})\mathrm{/}f(m)$ agrees with this pointwise derivative $F$-almost everywhere, though its value at $x=m$ is not the literal right derivative.