Finite linear combination of continuous functions is 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.

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.

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##.