Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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 5 '11 at 0:51
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 5 '11 at 0:55
You are welcome :) By the way shouldn't that be $(2\pi)^d$ instead of $2 \pi^d$? – t.b. Oct 5 '11 at 0:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.