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The following is a step in a proof: $f$ is a Lipschitz function from $E$ to $F$ where $E$ is a finite-dimensional Banach space and $F$ an arbitrary Banach space. $\phi\geq 0$ is a $C^\infty$ function with compact support, $\int\phi=1$, $\phi(x)=\phi(-x)$. The function $g(z)=\int f(z+x)\phi(x)dx$ is defined. The claim is that $||g||_{\text{Lip}}\leq||f||_{\text{Lip}}$.

I see why this would be true with some additional condition on $\phi$ (say, $||\phi||_{L^2}\leq 1$), but it's not clear to me why the given conditions are sufficient to control the Lipschitz constant of $g$.

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up vote 4 down vote accepted

Recall Bochner's theorem: A measurable function $h: \Omega \to F$ is integrable if and only if the scalar function $\|h\|$ is integrable. This is a straightforward consequence of the usual dominated convergence theorem (approximate $h$ by simple functions).

This readily implies that $\|\int h \| \leq \int \|h\|$, see e.g. Section 2.3 in Ryan, Introduction to tensor products of Banach spaces, Springer Monographs in Mathematics, 2002.

Using this, you have \[ \Vert g(z) - g(z')\Vert \leq \int \Vert f(z+x) - f(z'+x)\Vert \phi(x)\,dx \leq \|f\|_{\text{Lip}} \cdot \|z - z'\| \int \phi \, dx, \] so $\Vert g\Vert_{\text{Lip}} \leq \Vert f \Vert_{\text{Lip}}$ as desired.

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Ah, you're just a bit earlier than me :) +1. – Jonas Teuwen Feb 17 '11 at 22:38
@Jonas T: Sorry :) – t.b. Feb 17 '11 at 22:42

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