<Home>Qualifying Exams><Analysis>Fall 2014 - Problem 4

Fall 2014 - Problem 4

Fourier analysis

Given f∈L2([0,π])f \in L^2\p{\br{0,\pi}}, we say that f∈Gf \in \mathcal{G} if ff admits a representation of the form

f(x)=∑n=0∞cncos⁡nxwith∑n=0∞(1+n2)∣cn∣2<∞.f\p{x} = \sum_{n=0}^\infty c_n\cos{nx} \quad\text{with}\quad \sum_{n=0}^\infty \p{1 + n^2}\abs{c_n}^2 < \infty.

Show that if f∈Gf \in \mathcal{G} and g∈Gg \in \mathcal{G}, then fg∈Gfg \in \mathcal{G}.
Solution.

Write f(x)=∑n=0∞cncos⁡nxf\p{x} = \sum_{n=0}^\infty c_n\cos{nx} and g(x)=∑n=0∞dncos⁡nxg\p{x} = \sum_{n=0}^\infty d_n\cos{nx} with the convergence properties given. Observe that

f(x)g(x)=∑n=0∞∑m=0∞cndmcos⁡nxcos⁡mx=∑n=0∞∑m=0∞cndm2(cos⁡(n+m)x+cos⁡(n−m)x)≕∑k=0∞akcos⁡kx,\begin{aligned} f\p{x}g\p{x} &= \sum_{n=0}^\infty \sum_{m=0}^\infty c_nd_m \cos{nx} \cos{mx} \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{c_nd_m}{2} \p{\cos{\p{n + m}x} + \cos{\p{n - m}x}} \\ &\eqqcolon \sum_{k=0}^\infty a_k\cos{kx}, \end{aligned}

so fgfg can be represented in the right form as long as the sum converges absolutely (and hence may be rearranged). This is indeed the case, as

∑n=0∞∑m=0∞∣cndmcos⁡nxcos⁡mx∣≤(∑n=0∞∣cn∣)(∑m=0∞∣dm∣)=(∑n=0∞1+n2∣cn∣⋅11+n2)(∑m=0∞1+m2∣dm∣⋅11+m2)≤(∑n=0∞(1+n2)∣cn∣2)1/2(∑m=0∞(1+m2)∣dm∣2)1/2∑n=0∞11+n2<∞,\begin{aligned} \sum_{n=0}^\infty \sum_{m=0}^\infty \abs{c_nd_m \cos{nx} \cos{mx}} &\leq \p{\sum_{n=0}^\infty \abs{c_n}} \p{\sum_{m=0}^\infty \abs{d_m}} \\ &= \p{\sum_{n=0}^\infty \sqrt{1 + n^2}\abs{c_n} \cdot \frac{1}{\sqrt{1 + n^2}}} \p{\sum_{m=0}^\infty \sqrt{1 + m^2}\abs{d_m} \cdot \frac{1}{\sqrt{1 + m^2}}} \\ &\leq \p{\sum_{n=0}^\infty \p{1 + n^2}\abs{c_n}^2}^{1/2} \p{\sum_{m=0}^\infty \p{1 + m^2}\abs{d_m}^2}^{1/2} \sum_{n=0}^\infty \frac{1}{1 + n^2} \\ &< \infty, \end{aligned}

so the sum converges absolutely. By absolute convergence and orthogonality, we get

∫0π∣f(x)g(x)∣2 dx=∫0π∣(∑k=0∞akcos⁡kx)(∑ℓ=0∞aℓcos⁡ℓx)∣ dx≤∑k=0∞∫0π∣ak∣2cos⁡2kx dx=π2∑k=0∞∣ak∣2.\begin{aligned} \int_0^\pi \abs{f\p{x}g\p{x}}^2 \,\diff{x} &= \int_0^\pi \abs{\p{\sum_{k=0}^\infty a_k\cos{kx}} \p{\sum_{\ell=0}^\infty a_\ell\cos{\ell x}}} \,\diff{x} \\ &\leq \sum_{k=0}^\infty \int_0^\pi \abs{a_k}^2 \cos^2{kx} \,\diff{x} \\ &= \frac{\pi}{2} \sum_{k=0}^\infty \abs{a_k}^2. \end{aligned}

Hence, if we show that ∑k=0∞(1+k2)∣ak∣2<∞\sum_{k=0}^\infty \p{1 + k^2}\abs{a_k}^2 < \infty, then we will have shown that fg∈L2fg \in L^2 as well. Since we are working with non-negative values, we may re-order the terms:

∑k=0∞(1+k2)∣ak∣2=14∑n=0∞∑m=0∞∣cndm∣2((1+(n+m)2)+(1+(n−m)2))=12∑n=0∞∑m=0∞∣cndm∣2(1+m2+n2)≤12∑n=0∞∑m=0∞∣cndm∣2(1+m2)(1+n2)=12(∑n=0∞(1+n2)∣cn∣2)(∑m=0∞(1+m2)∣dm∣2)<∞,\begin{aligned} \sum_{k=0}^\infty \p{1 + k^2}\abs{a_k}^2 &= \frac{1}{4} \sum_{n=0}^\infty \sum_{m=0}^\infty \abs{c_nd_m}^2\p{\p{1 + \p{n + m}^2} + \p{1 + \p{n - m}^2}} \\ &= \frac{1}{2} \sum_{n=0}^\infty \sum_{m=0}^\infty \abs{c_nd_m}^2 \p{1 + m^2 + n^2} \\ &\leq \frac{1}{2} \sum_{n=0}^\infty \sum_{m=0}^\infty \abs{c_nd_m}^2 \p{1 + m^2}\p{1 + n^2} \\ &= \frac{1}{2} \p{\sum_{n=0}^\infty \p{1 + n^2}\abs{c_n}^2} \p{\sum_{m=0}^\infty \p{1 + m^2}\abs{d_m}^2} \\ &< \infty, \end{aligned}

so fg∈Gfg \in \mathcal{G}.