Given some $M>0$, consider the cutoff function $\varphi_M:[0,\infty) \to \mathbb{R}$ given by $$ \varphi(x) = \begin{cases} 1 &, 0\le x \le M \\ 0 &, x\ge M+1 \\ M+1 -x &, M<x<M+1 \end{cases} $$ whose graph is a constant $1$ up until $x=M$, then a straight line connecting $(M,1)$ and $(M+1,0)$. It's not very difficult to sit there and work out the definition of Lipschitz to show that $\varphi$ is Lipschitz, but given just the graph, is there something one should be looking at to notice "that function is definitely Lipschitz"?

This cutoff function is something I've been seeing a lot recently. First, let $(V, \langle\cdot,\cdot\rangle)$ be a Hilbert space with $|\cdot|$ the norm induced by the inner product. Given a certain bilinear form $B: V\times V \to V'$ (the dual space of $V$) and then "it is clear that" $\varphi(|x|)B(x,x)$ is Lipschitz, but to me it isn't and moreover trying to work it out algebraically seems very tedious and only results in verifying the claim is true but provides no intuition. I can add in some more details about my bilinear form (it's essentially the $(u\cdot\nabla)v$ term from fluid equations).

  • $\begingroup$ Your cut-off function is piece-wise linear, and it's very easy to show that a linear function is Lipschitz. Piece-wise linear functions are then also Lipschitz, which I think is the intuition you're looking for here $\endgroup$ – postmortes Aug 3 '18 at 6:04

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