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Chapter 8: Problem 7

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

Short Answer

Expert verified
The convolution \(f * g\) is \(C^k\).

Step by step solution

01

Understand Convolution

The convolution of two functions \(f\) and \(g\) is defined as \[ (f * g)(x) = \int_{\mathbb{R}^n} f(y)g(x-y)\,dy. \]This operation involves integrating one function at shifted points of the other.
02

Analyze the Given Functions

Given \(f\) is locally integrable on \(\mathbb{R}^{n}\), it suggests that \(f\) can be integrated over any compact subset of \(\mathbb{R}^{n}\). The function \(g\), on the other hand, belongs to \(C^k\) class, meaning it has continuous derivatives up to order \(k\), and has compact support.
03

Examine Compact Support Property

\(g\)'s compact support means that outside some bounded set, \(g(x) = 0\). Consequently, the convolution \((f * g)(x)\) only involves integration over a compact set where \(g(x-y) eq 0\). This restricts the region of integration and simplifies analyzing smoothness.
04

Differentiate the Convolution

To show \(f * g \in C^k,\) assess each derivative up to order \(k\):\[ D^{\alpha}(f * g)(x) = \int_{\mathbb{R}^n} f(y)D^{\alpha}_x g(x-y)\,dy,\]where \(D^{\alpha}_x\) denotes a partial derivative of \(g(x-y)\) of order \(\alpha\). Each \(D^{\alpha}_x g\) is continuous due to \(g \in C^k.\)
05

Utilize the Dominated Convergence Theorem

To ensure continuity, apply the Dominated Convergence Theorem. As \(D^{\alpha}_x g(x-y)\) is continuous and \(f\) is integrable locally, pointwise limits of the derivatives yield a continuous \(D^{\alpha}(f * g)\). This confirms \(f * g\) is continuously differentiable up to order \(k\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Locally Integrable Functions
A function is considered "locally integrable" if it can be integrated over any compact subset of the space. This means that when you restrict the domain to a bounded region, the function's integral exists and is finite. In other words:
  • For a function to be locally integrable on \(\mathbb{R}^{n}\), you should be able to calculate its integral over small, bounded parts of \(\mathbb{R}^{n}\).
  • Being locally integrable is less restrictive than being globally integrable across the entire space, which requires the integral to be finite on \(\mathbb{R}^{n}\).
  • An example of a locally integrable function could be \(f(x) = \frac{1}{x}\) on \(\mathbb{R}\) since it can't be integrated over all of \(\mathbb{R}\), but it can be integrated over intervals that don鈥檛 include zero.
This property is key for convolutions because it ensures that the calculation of the integral in the convolution operation is manageable on any compact subset.
Continuous Derivatives
When we say that a function has continuous derivatives up to order \(k\), denoted by \(C^{k}\), it means the function has derivatives that are smooth and do not have sudden jumps or discontinuities.
  • Each derivative up to \(k\) is continuous, helping in analyzing the smoothness of functions in both pure and applied contexts.
  • This property is crucial for keeping the convolution operation stable and ensuring it yields a function with similar smoothness if:
    • The function \(g\), involved in the convolution, belongs to this \(C^{k}\) class.
    • Each time you differentiate the convolution, the operations respect continuity, building upon the foundation of \(g\)鈥檚 smoothness.
  • Having continuous derivatives allows mathematicians to apply differentiation under the integral sign, making more advanced analysis possible.
In practice, functions like \(e^{-x^2}\) are in \(C^{\infty}\), meaning they have derivatives of all orders that are continuous.
Compact Support
A compact support in function theory refers to a function vanishing outside a bounded set. This characteristic has much significance due to:
  • The function being nonzero only within a specified bounded region, simplifying the analysis involved in integration or convolution.
  • It brings about the convenient property that many operations, including convolution, can be limited to a finite region, reducing complexity.
  • For the function \(g\) with compact support:
    • It ensures that only a limited part of \(f\) affects the convolution result.
    • This is significant in real-life applications where only the behavior in a confined region matters, such as signal processing or data compression.
In easy terms, when a function has compact support, you only need to "worry" about its behavior within a small, bounded area, making certain calculations much simpler and practical.
Dominated Convergence Theorem
The Dominated Convergence Theorem is a powerful tool in analysis used for establishing the interchangeability of limits and integrals. This theorem states that if a sequence of functions \(f_n\) converges pointwise to \(f\), and each \(f_n\) is dominated by some integrable function \(g\), then:
  • The integral of \(f_n\) tends towards the integral of \(f\) as \(n\) approaches infinity.
  • This allows us to swap the limit with integration, which is key in convolutions for proving continuity.
  • In practical terms, it provides a rigorous foundation for dealing with the continuity of convolutions, ensuring that integrals behave well under the limit processes involved when differentiating the convolution.
  • For the convolution \((f * g)(x)\), it certifies the continuous nature of the derivatives up to the desired order, guaranteeing \(f * g \in C^{k}\).
By applying the Dominated Convergence Theorem, mathematical assurance is gained that operations on such functions are legitimate and valid even when making analytical leaps like differentiating under the integral.

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Most popular questions from this chapter

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, \(\widehat{\phi}_{k}=(-i)^{k} \phi_{k}\).

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 $$ \int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} d x=\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} $$

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 \(f\) ?

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.)

(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.)
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