$f:R\to [0,\infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?
My attempt :
$|f^2(x)-f^2(y)|<\epsilon$ for $|x-y|<\delta$
then
$|f(x)-f(y)<\epsilon/|f(x)+f(y)|$ for $|x-y|<\delta$
My problem is that how to control above difference as f may be 0 at both x and y
SO how to show above is uniformly continuous
Any help will be appreciated
Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $\sqrt.$ composed with $f^{2}$.
To follow up on Martin R's comment:
Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $g\circ h$ is uniformly continuous on $I$.
Proof: Just chain the definitions. For any $\epsilon>0$, there is some $\delta>0$ such that $|g(y_1)-g(y_2)| < \epsilon$ whenever $|y_1 - y_2| < \delta$ for $y_1,y_2\in h(I)$. Then, for any $\delta > 0$, there is some $\gamma > 0$ such that $|h(x_1)-h(x_2)| < \delta$ whenever $|x_1-x_2| < \gamma$ for $x_1,x_2\in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <\epsilon$ whenever $|x_1-x_2| <\gamma$ for $x_1,x_2\in I$. That's the definition of uniform continuity, and we're done.
Next, we show that $g(y)=\sqrt{y}$ is uniformly continuous on $[0,\infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|\sqrt{y_1}-\sqrt{y_2}|\le \sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < \epsilon^2$, $|\sqrt{y_1}-\sqrt{y_2}| <\epsilon$.
Finally, the original problem has $f=g\circ h$, where $g$ is the square root function on $[0,\infty)$ and $h(x)=f^2(x)$ on $\mathbb{R}$. Since $h$ is uniformly continuous on its domain $\mathbb{R}$ and $g$ is uniformly continuous on $h(\mathbb{R})$, the composition is uniformly continuous on its domain $\mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.
I think the result holds for all $f\in C(\mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.
