Consider the step function $\Delta: \mathbb{R}\rightarrow [0,1]$ $$ \Delta(x;\lambda,\mu)\equiv \sum_{j=1}^J \lambda_j\times 1\{\mu_j\leq x\} $$ where
$\lambda\equiv (\lambda_1,...,\lambda_J)$ is a parameter
$\mu\equiv (\mu_1,...,\mu_J)$ is a parameter
$J=3$
$\lambda_j\geq 0$ $\forall j$; $\sum_{j=1}^J \lambda_j=1$
$\mu_j\in \mathbb{R}$ $\forall j$; $\mu_1<...<\mu_J$
$1\{\cdot\}$ is an indicator function taking value $1$ if the condition inside is satisfied and zero otherwise
Claim: If $\lambda_1\times \lambda_2\times \lambda_3\neq 0$ and $\frac{\mu_2-\mu_1}{\mu_3-\mu_2}\neq 1$, then $\Delta(\cdot; \lambda,\mu)$ cannot be such that $\Delta(x;\lambda,\mu)=1-\Delta(-x;\lambda,\mu)$.
Question: Could you help me to show this claim? Even just the intuition would be OK.
Also, is there a way to generalise this claim to any $J$?
Some thoughts: one way to show the claim is to prove that
"If $\Delta(x;\lambda,\mu)=1-\Delta(-x;\lambda,\mu)$ $\forall x$, then $\lambda_1\times \lambda_2\times \lambda_3= 0$ or $\frac{\mu_2-\mu_1}{\mu_3-\mu_2}= 1$."
I have tried to picture in my mind the cases in which we can have $\Delta(x;\lambda,\mu)=1-\Delta(-x;\lambda,\mu)$ $\forall x$ when $J=3$. Given that when $J=3$ we can have 4 steps at most, I ended up with 2 cases only:
1) $\lambda_j=\lambda_k=0$, $\lambda_h=1$, $\mu_h=0$
2) $\lambda_j=0$, $\lambda_k=\lambda_h=\frac{1}{2}$, $\mu_k=-\mu_h$
These two cases definitively satisfy $\lambda_1\times \lambda_2\times \lambda_3=0$. I can't picture a case implying $\frac{\mu_2-\mu_1}{\mu_3-\mu_2}= 1$.
Hint: Since all $\lambda_j$ are nonzero and $\mu_3-\mu_2\neq \mu_2-\mu_1$ it should be easy to see that the function $$\Delta(x,\lambda,\mu)=\begin{cases}0,x<\mu_1\\\lambda_1,\mu_1\leq x<\mu_2\\ \lambda_1+\lambda_2,\mu_2\leq x<\mu_3\\ 1=\lambda_1+\lambda_2+\lambda_3,\mu_3\leq x \end{cases}$$ doesn´t fulfill $\Delta(\delta,\lambda,\mu)=1-\Delta(-\delta,\lambda,\mu)$ for all $\delta>0$.