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Define the norm $\Vert \cdot \Vert_\text{Lip}:X\to [0,\infty)$ to be the least Lipschitz constant of $f$. Prove that $(X,\Vert \cdot > \Vert_\text{Lip})$ is a Banach space.

I have already shown in the standard way that there is a pointwise limit, (called $f$) that is likely Lipschitz, but I am struggling on how to impose the Lipschitz condition. I thought I could use an $\frac{\epsilon}{3}$ type argument (as used when showing uniform continuity) before letting $\epsilon \to 0$, but obviously our $\epsilon$ will depend on some $n$, and this is just a fudge.

Thanks for your help.

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  • $\begingroup$ The pointwise limit of a sequence of functions with a common Lipschitz constant $L$ is also Lipschitz continuous with constant $L$. Pick arbitrary $x\neq y$ and look at $\frac{\lvert f(y) - f(x)\rvert}{\lvert y-x\rvert}$. $\endgroup$ – Daniel Fischer Oct 27 '16 at 12:40

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