Does uniform continuity of |f| imply uniform continuity of f?

[Eclair_de_XII]

TL;DR Summary

Define a function $f:D\longrightarrow \mathbb{R}$ such that for some $a\in D$, $f(x)>0$ if $x>a$ and $f(x)\leq 0$ if $x\leq a$. Note: $f(a)=0$ by the IVT. If $|f|$ is uniformly continuous on $D$, then does that mean $f$ is also uniformly continuous?

I'd say yes, it is. Suppose $|f|$ is uniformly continuous on $D$.

Then for all $\epsilon>0$ there is $\delta>0$ (call this $\delta'$) such that if $x,y\in D$, then $||f(x)|-|f(y)||<\epsilon$.

Define sets:

$D^+=\{x\in D: x>a\}$

$D^-=\{x\in D: x<a\}$

Restrict the domain of $f$ to $D^+$; note that $f=|f|$ here so it follows that $f$ is uniformly continuous here. In other words, if you want the distance between the images any two points $x,y\in D^+$ to be as small as possible, you need only ensure that the distance between $x,y$ is at most $\delta'$.

Now restrict the domain of $f$ to $D^-$. Let $\epsilon>0$ and choose $\delta'$. Note that if $x,y\in D^-$, then:

\begin{align}
|f(x)-f(y)|&=&|-|f(x)|-(-|f(y))|\\
&=&|-|f(x)|+|f(y)|| \\
&<&\epsilon
\end{align}

since $|f|$ is uniformly continuous.

Finally, note that $|f|$ is continuous at $x=a$ and moreover, to ensure that the image of a point $x$ is within $\epsilon$ of $f(a)=0$, you can choose an $x$ within $(a-\delta',a+\delta')$, since $|f|$ is uniformly continuous.

Let $\epsilon>0$ and choose $\delta=\delta'>0$. Suppose $x\in D^+ \cup D^-$. Then:

\begin{align}
|f(x)-f(a)|&=&|f(x)|\\
&=&||f|(x)|\\
&<&\epsilon
\end{align}

 

[fresh_42]

Does it even imply continuity?
\begin{align*}
1\text{____________________________}&\\
&\text{_______________________________________}-1
\end{align*}

 

[Eclair_de_XII]

Oh! So $f(x)=1$ if $x<a$ and $f(x)=-1$ if $x\geq a$.

$|f|(x)=1$ for all $x$, so it is uniformly continuous, but $f$ is not even continuous much less uniformly.

No, it doesn't. Thanks.

I think I made a mistake when I assumed that $f$ was continuous when citing the IVT.

Note: $f(a)=0$ by the IVT.

 

[FactChecker]

Does it even imply continuity?
\begin{align*}
1\text{____________________________}&\\
&\text{_______________________________________}-1
\end{align*}

But does this satisfy the stipulation that |f| is uniformly continuous at the point x=a, where f(a)=0? Does D omit a?

 

[fresh_42]

But does this satisfy the stipulation that |f| is uniformly continuous at the point x=a, where f(a)=0? Does D omit a?

Seems I only answered to the title. If $f(a)=0,$ and $|f|$ is uniformly continuous at $a$, then it there probably won't be a significant difference between $f$ and $|f|$ at $a$, since all terms are only locally defined.

 

[pasmith]

But does this satisfy the stipulation that |f| is uniformly continuous at the point x=a, where f(a)=0? Does D omit a?

The problem statement is confused. Consider the first sentence:

Summary:: Define a function $f:D\longrightarrow \mathbb{R}$ such that for some $a\in D$, $f(x)>0$ if $x>a$ and $f(x)\leq 0$ if $x\leq a$.

Here there is no requirement that $f$ be continuous, or that $f(a) = 0$. A function like $$g : x \mapsto \begin{cases} -1 & x \leq a \\ 1 & x > a \end{cases}$$ satisfies the given condition. Note that $|g| = 1$ is uniformly continuous, but $g$ itself is not continuous at $a$.

But then it is said that

Note: $f(a)=0$ by the IVT.

This implies that $f$ is continuous, since if it weren't the IVT simply would not apply. I think the OP needs to clarify whether this is actually part of the problem statement or whether it is a conclusion they have arriced at themselves, and if so their justification for it.

If $|f|$ is uniformly continuous on $D$, then does that mean $f$ is also uniformly continuous?

As the example above shows, you also require continuity of $f$, which was not expressly stated to be the case.

The condition does imply that for $x \leq a$ we have $f = -|f| \leq 0$ which is uniformly continuous if $|f|$ is, and for $x > a$ we have $f = |f| > 0$ which again is uniformly continuous if $|f|$ is. (If $|f|$ is uniformly continuous on $D$ then it is uniformly continuous on any subset of $D$: the same $\delta$ will work as for all of $D$.)

If $f$ is actually restricted to be continuous then by the above argument it is indeed uniformly continuous on $D$.

 

[Eclair_de_XII]

OP needs to clarify whether this is actually part of the problem statement

Sorry. That was sort of an assumption that I made on my own. It was not actually part of the assignment.

If $f$ is uniformly continuous, does it follow that $|f|$ is also uniformly continuous? If $|f|$ is uniformly continuous, does it follow that $f$ is uniformly continuous? Answer the same questions with uniformly continuous replaced with ''continuous''. Explain why.

 

[FactChecker]

After the preceding discussion, what are your answers to the exact assignment questions?

 

[fresh_42]

After the preceding discussion, what are your answers to the exact assignment questions?

If $f(a)=0$ then $||f(x)|-|f(a)||=|f(x)-f(a)|$.

 

[FactChecker]

If $f(a)=0$ then $||f(x)|-|f(a)||=|f(x)-f(a)|$.

I was meaning to ask the OP how he would answer the actual problem as stated in post #7. It is quite different.

 

[Eclair_de_XII]

answers to the exact assignment questions

Yes, uniform continuity of $f$ implies uniform continuity of $|f|$. If $\epsilon>0$, then for some $\delta>0$, if $x,y\in D$, then $|x-y|<\delta$ implies $||f(x)|-|f(y)||\leq |f(x)-f(y)|<\epsilon$.

Conversely, no. Let:

$f(x)= \begin{cases} -1\quad\mathrm{x<0} \\1\quad\mathrm{x\geq0}\end{cases}$
% (Anyone know how to typeset so that this looks cleaner?)

This is the counterexample provided by Mr. fresh_42.Yes, $|f|$ is continuous for the same reasons as stated above.
Same counterexample as above.

This isn't exactly how I'm going to write it out, though.