Chapter 8

Let \(\eta(t)=e^{-1 / t}\) for \(t>0, \eta(t)=0\) for \(t \leq 0\). a. For \(k \in \mathbb{N}\) and \(t>0, \eta^{(k)}(t)=P_{k}(1 / t) e^{-1 / t}\) where \(P_{k}\) is a polynomial of degree \(2 k\). b. \(\eta^{(k)}(0)\) exists and equals zero for all \(k \in \mathbb{N}\).

If \(f\) is locally integrable on \(\mathbb{R}^{n}\) and \(g \in C^{k}\) has compact support, then \(f * g \in C^{k}\).

(Wirtinger's Inequality) If \(f \in C^{1}([a, b])\) and \(f(a)=f(b)=0\), then $$ \int_{a}^{b}|f(x)|^{2} d x \leq\left(\frac{b-a}{\pi}\right)^{2} \int_{a}^{b}\left|f^{\prime}(x)\right|^{2} d x . $$ (By a change of variable it suffices to assume \(a=0, b=\frac{1}{2}\). Extend \(f\) to \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) by setting \(f(-x)=-f(x)\), and then extend \(f\) to be periodic on \(\mathbb{R}\). Check that \(f\), thus extended, is in \(C^{1}(T)\) and apply the Parseval identity.)

Let \(\operatorname{sinc} x=(\sin \pi x) / \pi x(\operatorname{sinc} 0=1)\). a. If \(a>0, \widehat{\chi}_{|-a, a|}(x)=\chi_{[a, a]}^{V}(x)=2 a \operatorname{sinc} 2 a x\). b. Let \(\mathcal{H}_{a}=\left\\{f \in L^{2}: \hat{f}(\xi)=0\right.\) (a.c.) for \(\left.|\xi|>a\right\\}\). Then \(\mathcal{T}\) is a Hilbert space and \(\\{\sqrt{2 a}\) sinc \((2 a x-k): k \in Z\\}\) is an orthonormal basis for \(\mathcal{H}\). c. (The Sampling Theorem) If \(f \in \mathcal{T}_{a}\), then \(f \in C_{0}\) (after modification on a null set), and \(f(x)=\sum_{-\infty}^{\infty} f(k / 2 a) \operatorname{sinc}(2 a x-k)\), where the series converges both uniformly and in \(L^{2}\). (In the terminology of signal analysis, a signal of bandwidth \(2 a\) is completely determined by sampling its values at a sequence of points \(\\{k / 2 a\\}\) whose spacing is the reciprocal of the bandwidth.)

Let \(\operatorname{sinc} x=(\sin \pi x) / \pi x(\operatorname{sinc} 0=1)\). a. If \(a>0, \widehat{\chi}_{|-a, a|}(x)=\chi_{[a, a]}^{V}(x)=2 a \operatorname{sinc} 2 a x\). b. Let \(\mathcal{H}_{a}=\left\\{f \in L^{2}: \hat{f}(\xi)=0\right.\) (a.c.) for \(\left.|\xi|>a\right\\}\). Then \(\mathcal{T}\) is a Hilbert space and \(\\{\sqrt{2 a}\) sinc \((2 a x-k): k \in Z\\}\) is an orthonormal basis for \(\mathcal{H}\). c. (The Sampling Theorem) If \(f \in \mathcal{T}_{a}\), then \(f \in C_{0}\) (after modification on a null set), and \(f(x)=\sum_{-\infty}^{\infty} f(k / 2 a) \operatorname{sinc}(2 a x-k)\), where the series converges both uniformly and in \(L^{2}\). (In the terminology of signal analysis, a signal of bandwidth \(2 a\) is completely determined by sampling its values at a sequence of points \(\\{k / 2 a\\}\) whose spacing is the reciprocal of the bandwidth.)

Given \(a>0\), let \(f(x)=e^{-2 \pi x} x^{a-1}\) for \(x>0\) and \(f(x)=0\) for \(x \leq 0\). a. \(f \in L^{1}\), and \(f \in L^{2}\) if \(a>\frac{1}{2}\). b. \(\hat{f}(\xi)=\Gamma(a)[(2 \pi)(1+i \xi)]^{-a}\). (Here we are using the branch of \(z^{a}\) in the right half plane that is positive when \(z\) is positive. Cauchy's theorem may be used to justify the complex substitution \(y=(1+i \xi) x\) in the integral defining \(\widehat{f_{0}}\) ) c. If \(a, b>\frac{1}{2}\) then $$ \int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} d x=\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} $$

Given \(a>0\), let \(f(x)=e^{-2 \pi x} x^{a-1}\) for \(x>0\) and \(f(x)=0\) for \(x \leq 0\). a. \(f \in L^{1}\), and \(f \in L^{2}\) if \(a>\frac{1}{2}\). b. \(\hat{f}(\xi)=\Gamma(a)[(2 \pi)(1+i \xi)]^{-a}\). (Here we are using the branch of \(z^{a}\) in the right half plane that is positive when \(z\) is positive. Cauchy's theorem may be used to justify the complex substitution \(y=(1+i \xi) x\) in the integral defining \(\widehat{f_{0}}\) ) c. If \(a, b>\frac{1}{2}\) then $$ \int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} d x=\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} $$

22\. Since \(F\) commutes with rotations, the Fourier transform of a radial function is radial; that is, if \(F \in L^{1}\left(\mathbb{R}^{n}\right)\) and \(F(x)=f(|x|)\), then \(\widehat{F}(\xi)=g(|\xi|)\), where \(f\) and \(g\) are related as follows. a. Let \(J(\xi)=\int_{S} e^{i x \xi} d \sigma(x)\) where \(\sigma\) is surface measure on the unit sphere \(S\) in \(\mathbb{R}^{n}\) (Theorem 2.49). Then \(J\) is radial - say, \(J(\xi)=j(|\xi|)\) - and \(g(\rho)=\int_{0}^{\infty} j(2 \pi r \rho) f(r) r^{n-1} d r .\) b. \(J\) satisfies \(\sum_{1}^{n} \partial_{k}^{2} J+J=0\). c. \(j\) satisfies \(\rho j^{\prime \prime}(\rho)+(n-1) j^{\prime}(\rho)+\rho j(\rho)=0\). (This equation is a variant of Bessel's equation. The function \(j\) is completely determined by the fact that it is a solution of this equation, is smooth at \(\rho=0\), and satisfies \(j(0)=\sigma(S)=\) \(2 \pi^{n / 2} / \Gamma(n / 2)\). In fact, \(j(\rho)=(2 \pi)^{n / 2} \rho^{(2-n) / 2} J_{(n-2) / 2}(\rho)\) where \(J_{\alpha}\) is the Bessel function of the first kind of order \(\alpha .)\) d. If \(n=3, j(\rho)=4 \pi \rho^{-1} \sin \rho\). (Set \(f(\rho)=\rho j(\rho)\) and use (c) to show that \(f^{\prime \prime}+f=0\). Alternatively, use spherical coordinates to compute the integral defining \(J(0,0, \rho)\) directly.) (use induction on \(k\) ), and in particular, $$ h_{k}(x)=\frac{(-1)^{k}}{\left[\pi^{1 / 2} 2^{k} k !\right]^{1 / 2}} e^{x^{2} / 2}\left(\frac{d}{d x}\right)^{k} e^{-x^{2}} . $$ f. Let \(H_{k}(x)=e^{x^{2} / 2} h_{k}(x)\). Then \(H_{k}\) is a polynomial of degree \(k\), called the \(k\) th normalized Hermite polynomial. The linear span of \(H_{0}, \ldots, H_{m}\) is the set of all polynomials of degree \(\leq m\). (The kth Hermite polynomial as usually defined is \(\left[\pi^{1 / 2} 2^{k} k !\right]^{1 / 2} H_{k}\).) g. \(\left\\{h_{k}\right\\}_{0}^{\infty}\) is an orthonormal basis for \(L^{2}(\mathbb{R})\). (Suppose \(f \perp h_{k}\) for all \(k\), and let \(g(x)=f(x) e^{-x^{2} / 2}\). Show that \(\widehat{g}=0\) by expanding \(e^{-2 \pi i \xi \cdot x}\) in its Maclaurin series and using (f).) h. Define \(A: L^{2} \rightarrow L^{2}\) by \(A f(x)=(2 \pi)^{1 / 4} f(x \sqrt{2 \pi})\), and define \(\tilde{f}=\) \(A^{-1} \mathcal{F} A f\) for \(f \in L^{2}\). Then \(A\) is unitary and \(\widetilde{f}(\xi)=(2 \pi)^{-1 / 2} \int f(x) e^{-i \xi x} d x\). Moreover, \(\widetilde{T f}=-i T(\widetilde{f})\) for \(f \in \mathcal{S}\), and \(\widetilde{h}_{0}=h_{0}\); hence \(\widetilde{h}_{k}=(-i)^{k} h_{k}\). Therefore, if \(\phi_{k}=A h_{k} .\left\\{\phi_{k}\right\\}_{0}^{\infty}\) is an orthonormal basis for \(L^{2}\) consisting of eigenfunctions for \(\mathcal{F}\); namely, \(\widehat{\phi}_{k}=(-i)^{k} \phi_{k}\).

Let \(\phi(x)=e^{-|x| / 2}\) on \(\mathbb{R}\). Use the Fourier transform to derive the solution \(u=f * \phi\) of the differential equation \(u-u^{\prime \prime}=f\), and then check directly that it works. What hypotheses are needed on \(f\) ?