Finite linear combination of continuous functions is continuous?

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Discussion Overview

The discussion revolves around the properties of Lipschitz continuous functions, specifically whether a finite linear combination of such functions remains Lipschitz continuous. Participants explore definitions, mathematical reasoning, and potential approaches to proving the continuity of the linear combination.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that if ##G## and ##H## are Lipschitz continuous, then the expression ##aG(t) + bH(t)##, where ##a## and ##b## are real constants, should also be Lipschitz continuous, but they are struggling to prove it.
  • Another participant identifies a potential error in the definition of Lipschitz continuity referenced in the first post and suggests using the triangle inequality to find a suitable constant ##K_3##.
  • There is a reiteration of the concern regarding the definition of Lipschitz continuity, emphasizing that the incorrect definition allows for unbounded growth, which contradicts the intended limitation of Lipschitz continuity.
  • A later reply points out that the incorrect definition leads to a scenario where no suitable constant ##K## can be found, further complicating the proof of continuity.

Areas of Agreement / Disagreement

Participants express agreement on the identification of an error in the definition of Lipschitz continuity, but the discussion remains unresolved regarding the proof of continuity for the linear combination of functions.

Contextual Notes

Participants have noted limitations in their understanding of the definition of Lipschitz continuity, which affects their ability to prove the continuity of the linear combination. The discussion highlights the dependence on accurate definitions and the implications of incorrect assumptions.

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TL;DR
This isn't a homework problem but a general question that I had.
Is the linear combination of Lipschitz continuous functions also continuous?
##G## and ##H## are real valued Lipschitz continuous functions. There exists a ##K_1,K_2\geq 0## such that for all ##s,t##,
$$(s-t)^2\leq K_1^2 (G(s)-G(t))^2$$
and
$$(s-t)^2\leq K_2^2 (H(s)-H(t))^2.$$
Is ##aG(t)+bH(t)## where ##a,b## are real constants also Lipschitz continuous?
I tried showing this is true and am having difficulty.
$$\begin{align*}
(s-t)^2&\leq K_3^2 (aG(t)+bH(t) -(aG(s)+bH(s)))^2\\
&= K_3^2(aG(t)-aG(s))+(bH(t)-bH(s)))^2\\
&= K_3^2 (a(G(t)-G(s))+b(H(t)-H(s)))^2\\
&=K_3^2(a^2(G(t)-G(s))^2+b^2(H(t)-H(s))^2 +ab(G(t)-G(s))(H(t)-H(s))
\end{align*}$$
I considered choosing ##K=\max\left(\frac{K_1}{a},\frac{K_2}{b} \right)## but realized it wouldn't work. A Little help would go a long way. Thanks.
 
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Oh yikes... it turns out the definition of Lipschitz continuous is wrong in my notes. I also think that the easiest way to find ##K_3## is using the triangle inequality after doing some work using the correct definition.
 
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docnet said:
Oh yikes... it turns out the definition of Lipschitz continuous is wrong in my notes. I also think that the easiest way to find ##K_3## is using the triangle inequality after doing some work using the correct definition.
Yes. Good catch. The definition you have in post #1 allows the function, G, to grow with unlimited speed. The correct definition of "Lipschitz continuous" is intended to do the opposite and limit the speed.
 
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FactChecker said:
Yes. Good catch. The definition you have in post #1 allows the function, G, to grow with unlimited speed. The correct definition of "Lipschitz continuous" is intended to do the opposite and limit the speed.
Noted! Thank you.

The wrong definition is also false, because for ##s##, ##t## and ##G## such that ##|s-t|>0## and ##G(s)=G(t)##, there does not exist a ##K\geq 0## such that ## |s-t|\leq K| G(s)-G(t)| = 0##.
 

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