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This is a rather long problem and I'm having difficulties with a very specific part.

The question starts with a function $\phi\in L_{1}(\mathbb{R})$ which vanishes outside of $(-\pi,\pi)$.

Then for each $n\geq 1$, $\phi_{n}$ is defined to be the $2\pi$-periodic function by $\phi_{n}(t) = n\phi(nt)$ for $t\in (\frac{-\pi}{n}, \frac{\pi}{n})$ and $\phi_{n}(t) = 0$ for $t\in [-\pi,-\frac{\pi}{n})\cup(\frac{\pi}{n}, \pi)$.

The first step was to show that $\{\phi_{n}\}$ is a summability kernel (done).

Next I needed to show that if $f\in L_{1}(\mathbb{T})$, and if $\phi$ above happens to be in $C^{\infty}(\mathbb{R})$, then for each $n\geq 1$, $f*\phi_{n}\in C^{\infty}(\mathbb{R})$ (done) and $supp(f*\phi_{n}) \subset supp(f) + supp(\phi_{n})$.

That is if $(f*\phi_{n})(t)\neq 0$, then $t = t_{1} + t_{2}$ for some $t_{1}, t_{2}$, where $f(t_{1})\neq 0$ and $\phi_{n}(t_{2})\neq 0$.

It is this last part that I bothers me.

UPDATE:

This almost seems too easy to be true. So that's why I'm a little sheepish about it.

Let $t\in supp(f*\phi_{n})$. Since $f*\phi_{n}\neq 0$, we have $\int_{-\pi}^{\pi}f(t-s)\phi_{n}(s)ds \neq 0$. Thus $f(t-s')\phi_{n}(s')\neq 0$ for at least one $s'\in[-\pi,\pi)$. Then $t = (t-s')+ s'$ and $t-s'\in supp(f)$ and $s'\in supp(\phi_{n})$. Thus $supp(f*\phi_{n}) \subset supp(f) + supp(\phi_{n})$.

Is this reasonable or am I oversimplifying the problem?

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up vote 6 down vote accepted

If the support of $f$ is $(t_1, t_2)$ and the support of $g$ is $(t_3, t_4)$, then $$ (f*g)(\tau) = \int_{-\infty}^{\infty} f(t)g(\tau - t) \mathrm dt = \int_{t_1}^{t_2} f(t)g(\tau - t) \mathrm dt. $$ The argument of $g(\tau - t)$ decreases from $\tau - t_1$ to $\tau - t_2$ as $t$ increases from $t_1$ to $t_2$, and so the integrand is $0$ if $\tau$ is such that $\tau - t_2 > t_4$ or $t_3 > \tau - t_1$. In other words, $(f*g)(\tau) = 0$ if $\tau$ is outside the interval $(t_1 + t_3, t_2 + t_4)$.

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I think our arguments are similar, but you argue by contrapositive? I think it seems too simple because the rest of the problems from this course are all nightmarish. But this isn't too short I guess since it was just one small part of a bigger problem. –  Kyle Schlitt Oct 7 '11 at 3:34
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