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The theorem is that for any summability kernel $\{\phi_{n}\}$, if $f\in L_{p}(\mathbb{T}^{d})$, then $||f*\phi_{n} - f||_{p}\rightarrow 0$.

The step that I cannot follow is this:

$$\left\|\frac{1}{2\pi^d}\int_{[-\pi,\pi)^{d}}(f(\cdot - t) - f(\cdot))\phi_{n}(t)dt\right\|_{p}\leq\frac{1}{2\pi^d}\int_{[-\pi,\pi)^{d}}||f(\cdot - t) - f\cdot)||_{p}|\phi_{n}(t)|dt$$

I see how we could get the from the Minkowski Inequality: $$\left\|\frac{1}{2\pi^d}\int_{[-\pi,\pi)^{d}}(f(\cdot - t) - f(\cdot))\phi_{n}(t)dt\right\|_{p}\leq \frac{1}{2\pi^d}\int_{[-\pi,\pi)^{d}}\|(f(\cdot - t) - f(\cdot))\phi_{n}(t)\|_{p}dt,$$ which is not quite what the claim is, however.

roo
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    You were almost there... Just use the fact that $|\lambda g|_p = |\lambda| |g|_p$ with $\lambda = \phi_n(t)$ and $g = f(\cdot-t)-f(\cdot)$ and integrate this over $\mathbb{T}^d$ to see that your last integral equals the one you want. (Then apply the result of this question from yesterday to get the theorem you want) – t.b. Oct 05 '11 at 00:51
  • facepalm Right. $\phi_{n}(t)$ is independant of "$\cdot$", which is the only variable that the $p$ norm cares about.

    Thank you once again. :)

     – roo Oct 05 '11 at 00:55
  • You are welcome :) By the way shouldn't that be $(2\pi)^d$ instead of $2 \pi^d$? – t.b. Oct 05 '11 at 00:58

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