High School Finite linear combination of continuous functions is continuous?

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The discussion centers on whether a finite linear combination of Lipschitz continuous functions, specifically the expression aG(t) + bH(t), remains Lipschitz continuous. Participants highlight a misunderstanding regarding the definition of Lipschitz continuity, noting that the incorrect definition permits unbounded growth, contrary to the intended limitation of function speed. The correct approach involves using the triangle inequality to derive a suitable constant K3 for the linear combination. Clarifications are made about the implications of the wrong definition, particularly in cases where the functions yield the same output for different inputs. Overall, the conversation emphasizes the importance of accurate definitions in mathematical proofs.
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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.
 
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##.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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