Answer to Question #280100 in Real Analysis for abc

Question

Let 𝑓: (0,1) → ℝ be a bounded continuous function. Show that 𝑔(𝑥) = 𝑥(1 − 𝑥)𝑓(𝑥) is uniformly continuous. 

Answer 1

$|g(x)-g(x_0)|=|x(1-x)f(x)-x_0(1-x_0)f(x_0)|=$

$=|x(1-x)f(x)-x(1-x)f(x_0)+x(1-x)f(x_0)-x_0(1-x_0)f(x_0)|\le$

$\le |x(1-x)||f(x)-f(x_0)|+|fx_0||x(1-x)-x_0(1-x_0)|$

Since f(x) is bounded we have that $f(x)\le M$ where M is some number. Then:

$|x(1-x)||f(x)-f(x_0)|+|fx_0||x(1-x)-x_0(1-x_0)|\le$

$\le 2M|x(1-x)|+M|x(1-x)-x_0(1-x_0)|$

find maximum of $x(1-x)$ :

$x=1/2$

then:

$2M|x(1-x)|+M|x(1-x)-x_0(1-x_0)|<2M\cdot1/4+M\cdot1/4=3M/4$

$\delta=4\varepsilon/3M$

Since $\delta$ does not depend on x₀, g(x) is uniformly continuous.