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I'm self studying (for fun) the book "Functional Analysis, Calculus of Variations and Optimal Control" (by Clarke), and I'd appreciate some feedback for my proposed solution for an exercise from the book (I'm not comfortable with it, for reasons I can't point down).

The question is: Let $\{f_\alpha:S\rightarrow R\}_\alpha$ be a family of functions, each of which is Lipschitz of rank K on S, such that the upper envelope $f(x)=sup_\alpha f_\alpha (x)$ is finite-valued on S. Prove that f is Lipschitz of rank K on S.

My suggested proof: Let ${{x}_{1}},{{x}_{2}}\in S$. For any $\alpha $ we have:

       $f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right)=\left( f\left( {{x}_{1}} \right)-{{f}_{a}}\left( {{x}_{1}} \right) \right)+\left( {{f}_{a}}\left( {{x}_{2}} \right)-f\left( {{x}_{2}} \right) \right)+\left( {{f}_{a}}\left( {{x}_{1}} \right)-{{f}_{a}}\left( {{x}_{2}} \right) \right)$.

It follows from the triangle inequality:

       $\left| f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right) \right|\le \left| f\left( {{x}_{1}} \right)-{{f}_{\alpha }}\left( {{x}_{1}} \right) \right|+\left| {{f}_{a}}\left( {{x}_{2}} \right)-f\left( {{x}_{2}} \right) \right|+\left| {{f}_{a}}\left( {{x}_{1}} \right)-{{f}_{\alpha }}\left( {{x}_{2}} \right) \right|$.

Since the limit superior is sub-additive:

       $\left| f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right) \right|\le \sup_\alpha \left| f\left( {{x}_{1}} \right)-{{f}_{\alpha }}\left( {{x}_{1}} \right) \right|+\sup_\alpha \left| {{f}_{a}}\left( {{x}_{2}} \right)-f\left( {{x}_{2}} \right) \right|+\sup_\alpha \left| {{f}_{a}}\left( {{x}_{1}} \right)-{{f}_{\alpha }}\left( {{x}_{2}} \right) \right|$

But

       ${{\sup }_{\alpha }}\left| {{f}_{a}}\left( {{x}_{1}} \right)-{{f}_{\alpha }}\left( {{x}_{2}} \right) \right|\le K\left\| {{x}_{1}}-{{x}_{2}} \right\|$

and

       ${{\sup }_{\alpha }}\left| f\left( {{x}_{1}} \right)-{{f}_{\alpha }}\left( {{x}_{1}} \right) \right|={{\sup }_{\alpha }}\left| {{f}_{a}}\left( {{x}_{2}} \right)-f\left( {{x}_{2}} \right) \right|=0$,

thus: $\left| f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right) \right|\le K\left\| {{x}_{1}}-{{x}_{2}} \right\|$. Q.E.D.

Is it correct? Excessively inelegant?

Thanks! :)

Borbei
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  • This looks fine – Mohammad W. Alomari Jul 20 '14 at 21:17
  • How do you conclude that ${{\sup }{\alpha }}\left| f\left( {{x}{1}} \right)-{{f}{\alpha }}\left( {{x}{1}} \right) \right|={{\sup }{\alpha }}\left| {{f}{a}}\left( {{x}{2}} \right)-f\left( {{x}{2}} \right) \right|=0$? In general this can't be true. Take $f_\alpha(x) = \alpha$ with $\alpha \in [0,1]$. Then $f(x) = 1$, of course, but $\sup_\alpha |f(x)-f_\alpha(x)| = 1$ for all $x$. – copper.hat Jul 20 '14 at 21:53

1 Answers1

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The proof is incorrect, a comment above indicates why.

Here is another approach:

You have $f_\alpha(x) - f_\alpha(y) \le K \|x-y\|$ for all $\alpha$, $ x,y \in S$. In other words, $f_\alpha(x) \le f_\alpha(y) + K\|x-y\|$. Taking the $\sup$ on the right hand side gives $f_\alpha(x) \le f(y) + K\|x-y\|$, and taking the $\sup $ of the left hand side gives $f(x) \le f(y) + K\|x-y\|$.

Interchanging the roles of $x,y$ gives the desired result.

copper.hat
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