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Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continuous function. Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)\geq x X\}}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq x X\}}]$.

My goal is to prove that $g$ is Lipschitz-continous on $[0,E[f(X)]]$.

So far I have, for $\epsilon>0$ \begin{align} g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}. \end{align}

I am not sure what to do with that.

Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continuous function. Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq 1\}}]$.

My goal is to prove that $g$ is Lipschitz-continous on $[0,E[f(X)]]$.

So far I have, for $\epsilon$ \begin{align} g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq 1\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq 1\}}]}. \end{align}

Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continous function. Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)\geq x X\}}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq x X\}}]$.

My goal is to prove that $g$ is Lipschitz-continous on $[0,E[f(X)]]$.

So far I have, for $\epsilon>0$ \begin{align} g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}. \end{align}

I am not sure what to do with that.

Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continuous function. Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)\geq x X\}}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq x X\}}]$.

My goal is to prove that $g$ is Lipschitz-continous on $[0,E[f(X)]]$.

So far I have, for $\epsilon>0$ \begin{align} g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}. \end{align}

I am not sure what to do with that.

Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continous function. Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)\geq x X\}}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq x X\}}]$.

My goal is to prove that $g$ is Lipschitz-continous on $[0,\frac{E[f(X)]}{1+E[X]}]$.

So far I have, for $\epsilon>0$ \begin{align} g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}. \end{align}

I am not sure what to do with that.

Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continous function. Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)\geq x X\}}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq x X\}}]$.

My goal is to prove that $g$ is Lipschitz-continous on $[0,E[f(X)]]$.

So far I have, for $\epsilon>0$ \begin{align} g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq (x+\epsilon)X\}}]}. \end{align}

I am not sure what to do with that.