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I am having a bit of difficulty with this proof.

I know $|x-a| < \delta \implies |f(x) - f(a)| < \epsilon$.

I know $f$ is only continuous at a, so do I need to prove f is continuous at $f(a)$ before moving on with the proof? If yes, I'm not sure how.

I have the first line $|f(x) - f(f(a))|$, but I'm not sure how to prove, in a general sense, how this is continuous.

Even bypassing proving the f(f(a)) continuity, the bulk of the problem is still an issue: Proving $|x-a| \implies f(f(x)) - f(f(a)) < \epsilon$.

Any help would be appreciated!

Nimu Basak
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Ann Watson
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  • I understand that you define $f\cdot f$ as the composition $f· f(x) = f(f(x))$? Hint: what if you replace $x$ by $f(x)$ and $a$ by $f(a)$ in your first equation ;) – LL 3.14 Apr 25 '20 at 04:55
  • In general you will need to use continuity of $f$ at all points of the domain, to show the composition of $f$ with itself is continuous at all points of the domain. Why do you say "I know $f$ is only continuous at $a$"? – hardmath Apr 25 '20 at 04:55
  • @hardmath My apologies, I left that condition out of the original question. As it was given to me, f is only defined at a. – Ann Watson Apr 25 '20 at 04:58
  • $f$ is only defined on the point $a$ ? Are you sure ? Then $|f(f(a))-f(f(a))| = 0$ ... – LL 3.14 Apr 25 '20 at 05:02
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    There is a big difference between saying $f$ is only defined at $a$ and saying it is only continuous at $a$. In any case, if $f$ is only defined at $a$, then the composition of $f$ with itself is not defined unless $f(a) = a$. A function defined only at one point is not really very interesting from the standpoint of continuity there. – hardmath Apr 25 '20 at 05:03
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    Are you sure $f\cdot f(x)$ is not $f(x)f(x)$ ? – LL 3.14 Apr 25 '20 at 05:04
  • Indeed, we're talking about $f\cdot f = f^2$, not the composition. You would need an assumption about continuity at $f(a)$, as you surmised, to do something with $f\circ f$. – Ted Shifrin Apr 25 '20 at 05:20

2 Answers2

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The statement doesn't hold always...

See the example $f:\mathbb{R} \rightarrow \mathbb{R}$: $$ f(x)=\begin{cases} x+2 &\text { when }x\in (-\infty,4) \\ 10 &\text { when }x\in [4,\infty] \end{cases} $$ Then: $$ (f\circ f)(x) = \begin{cases} x+4 &\text { when }x\in (-\infty,2) \\ 10 &\text{ when } x\in [2,\infty]\end{cases} $$ Clearly $f $ is continuous at $2$, but $f\circ f$ is not continuous at $2$.

Nimu Basak
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Some notes to help the OP sort out what the real problem is (to be deleted once the Question is appropriately clarified).

If $f$ were only defined at $x=a$, it would not make much sense to speak of it being continuous. The definition of continuity at $x=a$ calls for $f$ to be defined on a deleted neighborhood of $a$.

Furthermore if $f(f(a))$ is defined, then $f(a)$ has to belong to the domain of $f$. Unless we knew $f$ was continuous at $f(a)$ as well as at $x=a$, we would have no reason to expect $f\circ f$ to be continuous at $x=a$. Counterexamples are easily constructed.

Possibly $f\cdot f$ is the product of $f$ with itself. In this case, as @LL3.14 suggests, $f$ could be defined everywhere and $f\cdot f$ would be defined everywhere (on the real numbers), and continuity of $f$ at $x=a$ would imply continuity of $f\cdot f$ at $x=a$.

hardmath
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