Controlling the Lipschitz norm of the limit of a sequence of functions

Consider the Fréchet space $\Omega = C(\mathbb R^d)$ of real-valued continuous functions equipped with the seminorms $$\|f\|_D := \sup_{x,y \in D} \left\{ |f(x)|, \tfrac{|f(x)-f(y)|}{|x-y|} \right\}, \qquad \mathrm{for~compact~} D \subseteq \mathbb R^d.$$ That is, $\|f\|_D$ is the larger of the supremum norm and the Lipschitz constant of $f$ over $D$.

Let $D \subseteq \mathbb R^d$ be compact, and let $f \in \Omega$. Consider a sequence $D_n$ of compact sets and a sequence of functions $f_n$ so that:

• $D_n \to D$ in the Hausdorff topology,
• $f_n \to f$ in $\Omega$, and
• There exists a value $h$ so that $\|f_n\|_{D_n} \le h$ for all $n$.

Is it the case that $\|f\|_D \le h$?

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This is what I would say if you posted the question to www.artofproblemsolving.com ... "Have you tried proving it? How did you start? What went wrong?" –  Gerald Edgar Dec 9 '10 at 15:28
In what sense does $f_n\to f$? Uniform convergence? –  Willie Wong Dec 9 '10 at 15:49
I agree with Gerald. You can work this out yourself. And, Willie, I don't think it matters what kind of convergence it is. –  Deane Yang Dec 9 '10 at 16:14
Oh... yeah, you are right. I wasn't paying enough attention. –  Willie Wong Dec 9 '10 at 16:33
Thanks, guys. As you can tell, my brain is fried at the end of this semester. Sometimes you need people saying, "This is easy, solve it yourself," to realize that this is easy and you can solve it yourself. –  Tom LaGatta Dec 10 '10 at 3:41