Assume that if the real-valued function h(x) is Lipschitz continuous...

I think the answer is no, since the requirements for Lipschitz continuous and epsilon-delta continuous are different.

The reason I'm asking such an odd question is, I made a mistake by writing a proof of the Lipschitz continuity of ##g(h(x))## using the assumption that ##h(x)## is Lipschitz continuous. Because, it turned out I was supposed to use ##f(x)## epsilon-delta continuous. It got me thinking.. would it be possible to get the continuity of ##g(f(x))## using the completed proof about Lipschitz continuity?

What properties are you assuming for the function ##g##?

PeroK said:

What properties are you assuming for the function ##g##?

Ok, I need to tell the truth that I oversimplified the problem statement. Instead of one ##f(x)##, it's actually a finite set of functions ##\{f_1(x),...,f_n(x)\}## which are all Lipschitz continuous (or continuous). ##g## is a finite linear combination of the functions, and defined by ##g=a_1f_1(x)+\cdots+a_nf_n(x)##, with real scalar constants ##a_i##.

docnet said:

Ok, I need to tell the truth that I oversimplified the problem statement. Instead of one ##f(x)##, it's actually a finite set of functions ##\{f_1(x),...,f_n(x)\}## which are all Lipschitz continuous (or continuous). ##g## is a finite linear combination of the functions, and defined by ##g=a_1f_1(x)+\cdots+a_nf_n(x)##, with real scalar constants ##a_i##.

A finite linear combination of Lipschitz continuous functions must be Lipschitz continuous. First prove for the sum of two functions and for multiplication by a scalar, then use induction.

Thank you for your guidance over the years. I've tried what you said (turns out it can work without an induction method, because ##a_i<\infty, \: \forall i \in \{1,...,n\}## and ##n<\infty## makes it possible to define a Lipschitz constant for ##g## as a linear combination of the ##a_i## and the Lipschitz constants ##K_i## for ##f_i##. And I think I have it right.

I was just more curious about whether it's possible to prove the statement '##f_i(x)## are continuous ##\Rightarrow## ##g## is continuous' using the result '##f_i(x)## are Lipschitz ##\Rightarrow## ##g(f_1(x),...,f_n(x))## is Lipschitz', in a more general sense, without involving epsilon and delta quantifiers and (even more generally) without knowing the structure of ##g(f_1(x),...,f_n(x)).##

docnet said:

I was just more curious about whether it's possible to prove the statement '##f_i(x)## are continuous ##\Rightarrow## ##g## is continuous' using the result '##f_i(x)## are Lipschitz ##\Rightarrow## ##g(f_1(x),...,f_n(x))## is Lipschitz', in a more general sense, without involving epsilon and delta quantifiers and (even more generally) without knowing the structure of ##g(f_1(x),...,f_n(x)).##

I don't understand what you are trying to prove here?

I'm not sure if it's possible at all, but what I'm trying to prove is statement 1 (an assumption leading to a conclusion) by using statement 2 (another assumption leading to another conclusion). The statements are related by dealing with the same object ##g(f(x))##, and both statements are about continuity. And statement ##2## uses a stricter definition of continuity (Lipschitz).

docnet said:

I'm not sure if it's possible at all, but what I'm trying to prove is statement 1 (an assumption leading to a conclusion) by using statement 2 (another assumption leading to another conclusion). The statements are related by dealing with the same object ##g(f(x))##, and both statements are about continuity. And statement ##2## uses a stricter definition of continuity (Lipschitz).

Let me try to help. Is it something like this?

If ##f(x)## is Lipschitz continuous and ##g(f(x))## is [blank], then ##g(x)## is [blank].

No it's different from what I'm asking. It's like:

##1.## If we assume that ##f(x)## is Lipschitz, then we can prove that ##g(f(x))## is Lipschitz.

##2.## However, ##f(x)## is not Lipschitz, but only epsilon-delta continuous.

##3.## Without writing an explicit proof using epsilon-delta definitions, and without knowing what the structure of ##g(f(x))## is, what can we say, if anything, about the continuity of ##g(f(x))##?

docnet said:

No it's different from what I'm asking. It's like:

##1.## If we assume that ##f(x)## is Lipschitz, then we can prove that ##g(f(x))## is Lipschitz.

What are the properties of ##g##? That's back to the first question.

docnet said:

##2.## However, ##f(x)## is not Lipschitz, but only epsilon-delta continuous.

What's the context of this assumption on ##f##?

docnet said:

##3.## Without writing an explicit proof using epsilon-delta definitions, and without knowing what the structure of ##g(f(x))## is, what can we say, if anything, about the continuity of ##g(f(x))##?

It depends on the properties we assume for ##g## and ##f##.

Post #3 was actually kind of a tangent from the main topic and doesn't help our case, sorry for posting it. Because, with that specific case of ##g(f(x))## where it's just a linear combination of ##f(x)##, the continuity of ##f(x)=(f_1(x) \: ...\: f_n(x))## alone implies continuity of ##g(f(x))##. But that distracts from the main question. What I'm really curious to know is, whether Lipschitz ##h(x)## leading to Lipschitz ##g(h(x))## can give us the continuity of ##g(f(x))## in the case that ##f(x)## is continuous.

What are the properties of ##g##? That's back to the first question.
The only known property of ##g(h(x))## is that if ##h(x)## is Lipschitz, then ##g(h(x))## is Lipschitz.

What's the context of this assumption on ##f##?
It is given that there exists a valid proof that if ##h(x)## is Lipschitz, then ##g(h(x))## is Lipschitz. We are given ##f(x)## that is only continuous and not Lipschitz, do we know that ##g(f(x))## is continuous?

I realized that the answer could very well be no, there is not enough information to get the continuity of ##g(f(x))##. It would be cool to know the answer.

docnet said:

Ok, I need to tell the truth that I oversimplified the problem statement. Instead of one ##f(x)##, it's actually a finite set of functions ##\{f_1(x),...,f_n(x)\}## which are all Lipschitz continuous (or continuous). ##g## is a finite linear combination of the functions, and defined by ##g=a_1f_1(x)+\cdots+a_nf_n(x)##, with real scalar constants ##a_i##.

"oversimplified" is a massive understatement. I suspect that you need to do a serious review of some basic definitions.

docnet said:

No it's different from what I'm asking. It's like:

##1.## If we assume that ##f(x)## is Lipschitz, then we can prove that ##g(f(x))## is Lipschitz.

##2.## However, ##f(x)## is not Lipschitz, but only epsilon-delta continuous.

##3.## Without writing an explicit proof using epsilon-delta definitions, and without knowing what the structure of ##g(f(x))## is, what can we say, if anything, about the continuity of ##g(f(x))##?

I doubt it. Take, e.g., ##g(x)=\chi(x)##, the characteristic function of the Rationals. Then ##\chi(g(x))## will be the reatriction to a subset, which won't be continuous.

FactChecker said:

"oversimplified" is a massive understatement. I suspect that you need to do a serious review of some basic definitions.

#3 was a tangent, sorry. I wish I could remove it to avoid being distracting, because I already think my initial question was asked in a confusing way. I'm sorry to say, but I think my actual question (please read #11 for clarification) is more difficult to answer than you're giving credit for.

##h(x)## and ##f(x)## are real valued Lipschitz continuous and continuous functions respectively. ##g(f(x))## is a function composition.

WWGD said:

I doubt it. Take, e.g., ##g(x)=\chi(x)##, the characteristic function of the Rationals. Then ##\chi(g(x))## will be the reatriction to a subset, which won't be continuous.

Yes, but isn't the characteristic function of the rationals already discontinuous to begin with? My instinct says the composition of a discontinuous function with itself (which your notation implies) would be discontinuous.

Sorry, I think you misunderstood my question due to my confusing wording and for using the same letter ##f## for Lipschitz continuous and continuous functions. Sorry about that. I clarified my question in #11 and made a small change to the original post to make my question less confusing.

just a comment on: "My instinct says the composition of a discontinuous function with itself (which your notation implies) would be discontinuous."

what about f(x) = x if x is irrational, and f(x) = -x if x is rational? then isn't f discontinuous everywhere except x=0; while f(f(x)) = x for all x, is continuous everywhere?

Or plain old ##\chi(x)##, Characteristic function of the Rationals, where ##\chi(\chi(x))==1##, since both ##1,0## are Rational.

docnet said:

Yes, but isn't the characteristic function of the rationals already discontinuous to begin with? My instinct says the composition of a discontinuous function with itself (which your notation implies) would be discontinuous.

As you can see from the posts of @mathwonk and @WWGD above, this is a very bad place to rely on "instinct". You should not trust instinct too much until you are so familiar with the subject and related proofs that they have become part of your nature. If your instinct includes 99% of the proof, with only routine details to fill in, then it's ok.

FactChecker said:

As you can see from the posts of @mathwonk and @WWGD above, this is a very bad place to rely on "instinct". You should not trust instinct too much until you are so familiar with the subject and related proofs that they have become part of your nature. If your instinct includes 99% of the proof, with only routine details to fill in, then it's ok.

Or until they sweat, bleed Rudin through their pores. ;).

FactChecker said:

As you can see from the posts of @mathwonk and @WWGD above, this is a very bad place to rely on "instinct". You should not trust instinct too much until you are so familiar with the subject and related proofs that they have become part of your nature. If your instinct includes 99% of the proof, with only routine details to fill in, then it's ok.

I agree with your statement, instinct could never replace a rigorous proof. I don't hesitate to admit I'm wrong when the facts are showed. I'll even say that I still know relatively little about real-valued functions in light of the breadth of knowledge that's out there. I had a difficult time with real analysis when I took the class not long ago.

mathwonk said:

just a comment on: "My instinct says the composition of a discontinuous function with itself (which your notation implies) would be discontinuous."

what about f(x) = x if x is irrational, and f(x) = -x if x is rational? then isn't f discontinuous everywhere except x=0; while f(f(x)) = x for all x, is continuous everywhere?

That's a good example that disproves my instinct. From now on, I'll be more careful to rely on logic and not instinct. That also has me wondering, could there be a continuous function ##f## such that that ##f\circ f## is discontinuous? Any examples are welcome :)

At the same time, I'd like to steer the thread back to my original question if that's ok. Not because it's inherently more important than what other members want to say, or that my grades rest upon it, but it's the reason the thread was started in the first place. If no one understands my question, maybe it means I'm crazy, so I should abandon my studies and get a license in massage therapy.

##g## is real-valued such that for any real-valued Lipschitz continuous ##h##, ##g\circ h## is Lipschitz continuous. Could you think of a real valued continuous ##f##, such that for some ##g## that meets the above requirement, ##g\circ f## is discontinuous?

Seeing the members' examples makes me think that someone must know such a pair of ##f## and ##g##.

I also posted this question in the discussion board of my Stochastic Analysis class, and I'm afraid that the TAs will not think kindly of me because it has little to do with the class material.

Re fof, let ##f : X \rightarrow X ; U## open in## X## *
* We could map ##X: \rightarrow Y##, with ## X \subset Y##, but I think this is enough.
What's ##(f o f)^{-1}(U)##?

the basic intuition is that good properties are preserved by composition but not their absence.