Check proof that $|f_{a_1}(x)-f_{a_2}(x)| \leq f'(x) |a_1 - a_2|$

Question

Let $a_1, a_2 \in A$ where A is a compact set. Let $x \rightarrow f_a(x)$ be a measurable function such that $a \rightarrow f_a(x)$ is differentiable for almost every $x$. I have to prove that there exists a measurable function f'(x) such that

$$|f_{a_1}(x)-f_{a_2}(x)| \leq f'(x) |a_1 - a_2|.$$

My approach is the following.

I know that $sup _a sup _x |f_a(x)|/(1 + x^2) \leq M < \infty$.

Hence by the triangle inequality we know that

$$|f_{a_1}(x)-f_{a_2}(x)|/(1 + x^2) \leq |f_{a_1}(x)|/(1 + x^2) + |f_{a_2}(x)|/(1 + x^2) \leq 2M $$

Hence,

$$|f_{a_1}(x)-f_{a_2}(x)| \leq (1 + x^2) 2M$$

But how do I make appear $|a_1 - a_2|$ in the right-hand side?

Answer 1

This is false. A counter-example is $f_a(x)=a^{3/2}\sin (\frac 1 a)$ on $A=[0,1]$. This satisfies the hypothesis and the conclusion fails because the derivative of $f_a$ w.r.t $a$ is unbounded along the sequence $(\frac 1{2n\pi})$.