if $f^2(x)$ is uniformly continuous then f is uniformly continuous

Question

this is the real question.1I saw this question like if f is from R to infinity and continuous if g(x)=$f^2(x)$ if g(x) is uniformly continuous so is f(x) $f^2(x)=f(x)*f(x)$

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My Thought: 1)for sequence $x_n$ and $y_n$ are like $d(x_n,y_n)$ tends to 0 then $d(g(x_n),g(y_n))$ tends to 0 as g is uniformly continuous .considering usual metric of R(real line) $|g(x)-g(y)|=|f^2(x) -f^2(y)|=|f(x)-f(y)|*|f(x)+f(y)|$ from here how can I show that $|f(x)-f(y)|$ tends to 0
2)first I thought about $\sqrt(x)$ , as it is uniformly continuous and so is g(x) so $\sqrt(g(x))$ will also be uniformly continuous.But as $\sqrt(x)$ is not defined on whole R so I cannot conclude f is uniformly continuous on whole R.
or there is any counter example.
pleas

Answer 1

With your point 2 you have proven that $|f|$ is uniformly continuous.

Let $\epsilon>0$. Let $\delta>0$ such that for all $x,y\in\mathbb{R}$, $|x-y|<\delta$, $||f(x)|-|f(y)||<\frac{\epsilon}{2}$.

Let $x,y\in\mathbb{R}$, $|x-y|<\delta$.

If $f(x)$ and $f(y)$ have the same sign, then $|f(x)-f(y)| = ||f(x)|-|f(y)||<\frac{\epsilon}{2} < \epsilon$.

If $f(x)$ and $f(y)$ have different signs, then by continuity of $f$ there is $z \in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

Answer 2

Composition of two uniformly continuous is uniformly continuous . Now take $g(x)=f^2(x)$ and $h(x)=\sqrt(x)$ it is also uniformly continuous .Now think $h(g(x))$???

Answer 3

Following your second idea, you need to show that if $f$ is continuous and $\sqrt{g}=|f|$ is uniformly continuous, then $f$ is also uniformly continuous. To get some intuition about the situation, come up with an example of a function $f$ such that $|f|$ is continuous, but $f$ is not. From this, you should see why it might be natural to divide the domain into two parts: $\{|f(x)| \ge \varepsilon\}$ and $\{|f(x)| < \varepsilon\}$.

Answer 4

$f^2$ is uniformly cont $\implies (|x-y|<\delta\implies|f^2(x)-f^2(y)|<\epsilon) \implies |f^2(x)-f^2(y)|<{\epsilon\over\delta}|x-y|\implies {|f^2(x)-f^2(y)|\over|x-y|}<{\epsilon\over\delta}\implies|2f(x)f'(x)|<{\epsilon\over\delta}\implies |f'(x)|<{k\over |f(x)|}<\frac k {inf \,f}$ If $inf f\ne 0$, we are done since Derivative is bounded implies function is uniformly continous.
Now comes the really non rigourous part. Let $K=\{x||f(x)|<1\}$. Since $f$ is cont, $f'(x)<M\,\,\forall x\in K$ for some $M\in \mathbb N,$ also $f'(x)<k\,\,\forall x\in\mathbb R\cap K^c$
Thus, $f'$ is bounded and hence $f$ in uniformly continuous