Showing posts with label Fourier. Show all posts
Showing posts with label Fourier. Show all posts
Thursday, June 20, 2013
2.23
Assume to the contrary that there does exist an $I \in L^1(\mathbb{R}^d)$ such that:
$$(f*I) = f \hspace{0.25cm} \forall f \in L^1(\mathbb{R}^d)$$
It follows from the latter parts of exercise 21 that:
$$\hat{f}(\xi) = \widehat{(f*I)}(\xi) = \hat{f}(\xi)\hat{I}(\xi)$$
Thus, since $\hat{f}(\xi)$ need not be zero, we have that $\hat{I}(\xi) = 1 \hspace{0.25cm} \forall \xi$. I.e. $\lim_{\xi \to \infty} \hat{I}(\xi) = 1$. This contradicts the Riemann-Lebesgue Lemma. Therefore, $I \notin L^1(\mathbb{R}^d)$.
2.22
This exercise is asking us to prove the Riemann-Lebesgue Lemma. Exactly as the hint prescribes, first observe that $\xi \cdot \xi' = \frac{1}{2}$, and then by the translation invariance of the Lebesgue integral:
$$\hat{f}(\xi) = \int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i (x - \xi') \cdot \xi} \hspace{0.1cm}dx = \ldots$$
$$= \int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i x \cdot \xi} e^{-2\pi i \xi \cdot \xi'} \hspace{0.1cm}dx = -\int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx$$
So we can certainly rewrite $\hat{f}(\xi)$ as:
$$\hat{f}(\xi) = \frac{1}{2} \int_{\mathbb{R}^d}\big(f(x)- f(x - \xi')\big)e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx$$
Now, observe that:
$$\lim_{|\xi| \to \infty} |\hat{f}(\xi)| = \lim_{|\xi| \to \infty} \Bigg|\frac{1}{2} \int_{\mathbb{R}^d}\big(f(x)- f(x - \xi')\big)e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx\Bigg| = \dagger$$
...and thus, by the triangle inequality, and since $\xi' \to 0$ if $|\xi| \to \infty$, it's clear that:
$$\dagger \leq \lim_{\xi' \to 0} \frac{1}{2} \int_{\mathbb{R}^d}\big|f(x)- f(x - \xi')\big| \hspace{0.1cm}dx = 0$$
...from Proposition 2.5 (p. 74).
2.21
Part a.) Since the product of two measurable functions is measurable, it suffices to show that $f(x-y)$ and $g(y)\chi_{\mathbb{R}^d(x)}$ are each measurable in $\mathbb{R}^{2d}$.
Conveniently, since $f$ is measurable on $\mathbb{R}^d$, it follows directly from Proposition 3.9 (p. 86) that $f(x-y)$ is measurable on $\mathbb{R}^{2d}$. Also, since $g$ is measurable on $\mathbb{R}^d$, it follows directly from Corollary 3.7 (P. 85) that $g(y)\chi_{\mathbb{R}^d(x)}$ is measurable on $\mathbb{R}^{2d}$.
Part b.) Since we know $f(x-y)g(y)$ is measurable, by Tonelli's Theorem we have: $$\int_{\mathbb{R}^{2d}} |f(x-y)g(y)| \hspace{0.1cm}d(x,y) \hspace{0.25cm}=\hspace{0.25cm} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y)g(y)| \hspace{0.1cm}dx\hspace{0.1cm}dy$$ ...and from the translation invariance of integration we get: $$\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y)g(y)| \hspace{0.1cm}dx\hspace{0.1cm}dy = \int_{\mathbb{R}^d} |g(y)| \int_{\mathbb{R}^d} |f(x-y)| \hspace{0.1cm}dx\hspace{0.1cm}dy = \ldots$$ $$\ldots = ||f||_{L^1(\mathbb{R}^d)} \int_{\mathbb{R}^d} |g(y)| dy = ||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)} < \infty$$ ...since both $f$ and $g$ are $L^1$.
Part c.) Since $f(x-y)g(y)$ was just shown to be integrable, it follows directly from Fubini's Theorem that for almost every $x \in \mathbb{R}^d$, : $$\int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy < \infty$$ I.e., the convolution: $$(f*g)(x) = \int_{\mathbb{R}^d} f(x-y)g(y)\hspace{0.1cm}dy$$ ...is well-defined for a.e. $x \in \mathbb{R}^d$.
Part d.) Observe that: $$\int_{\mathbb{R}^d} |(f*g)(x)\hspace{0.1cm}| dx = \int_{\mathbb{R}^d} \Bigg| \int_{\mathbb{R}^d}f(x-y)g(y)\hspace{0.1cm}dy \hspace{0.1cm}\Bigg| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{\mathbb{R}^d} \int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy \hspace{0.1cm}dx$$ ...which, by part b, we see: $$||(f*g)||_{L^1(\mathbb{R}^d)} = \int_{\mathbb{R}^d} |(f*g)(x)|\hspace{0.1cm} dx \leq \int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy \hspace{0.1cm}dx = \ldots$$ $$\ldots = ||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)}$$ Now, if $f$ and $g$ are positive functions, $|f(x-y)g(y)|=f(x-y)g(y)$, so equality of $||(f*g)||_{L^1(\mathbb{R}^d)}$ and $||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)}$ follows (again) directly from part b.
Part e.) Let's first check that $\hat{f}(\xi)$ is bounded. Recall that $|e^{i \theta}| = 1 \hspace{0.25cm} \forall \theta \in \mathbb{R}$. Then, observe: $$|\hat{f}(\xi)| = \Bigg| \int_{\mathbb{R}^d} f(x) e^{-2\pi i x \xi} \hspace{0.1cm} dx \Bigg| \leq \ldots $$ $$ \ldots \leq \int_{\mathbb{R}^d} |f(x)||e^{-2\pi i x \xi}| \hspace{0.1cm}dx = \int_{\mathbb{R}^d} |f(x)|\hspace{0.1cm}dx = ||f||_{L^1(\mathbb{R}^d)}$$ Thus, $\hat{f}(\xi)$ is bounded.
Now, let's see if $\hat{f}(\xi)$ is continuous. We begin by observing: $$|\hat{f}(\xi) - \hat{f}(\mu)| = \Bigg| \int_{\mathbb{R}^d} f(x) \big(e^{-2\pi i x \cdot \xi} - e^{-2\pi i x \cdot \mu}\big) \hspace{0.1cm} dx \Bigg| \leq \ldots $$ $$\ldots \leq \int_{\mathbb{R}^d}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx$$ Note that since $f$ is $L^1(\mathbb{R}^d$, for any $\epsilon > 0$ we have that there exists an $R > 0$ such that: $$ \int_{B_R^c} |f(x)| \hspace{0.1cm} dx \leq \frac{\epsilon}{4}$$ (Where $B_R$ is a ball of radius $R$ centered the origin.)
Now, since $\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \leq 2$, we see: $$\int_{\mathbb{R}^d}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx + 2\int_{B_R^c}|f(x)| \hspace{0.1cm}dx$$ From here, require: $$||\xi - \mu|| < \delta = \frac{\epsilon}{8 \pi R ||f||_{L^1(\mathbb{R}^d)}}$$ Now, it should be clear from the Cauchy Schwartz inequality that on $B_R$: $$ |x \cdot (\xi - \mu)| \leq R \delta = \frac{\epsilon}{8 \pi ||f||_{L^1(\mathbb{R}^d)}}$$ Therefore, plugging it all in, we finally see: $$\int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx \leq \ldots$$ $$\leq \int_{B_R}|f(x)| \Big[\big| \cos(2\pi x \cdot (\xi - \mu)) - 1 \big| + \big|\sin(2\pi x \cdot (\xi - \mu))\big|\Big] \hspace{0.1cm}dx$$ $$\leq \int_{B_R}|f(x)| \Big[\big| \cos\big(\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}}\big) - 1 \big| + \big|\sin\big(\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}}\big) \big| \Big]$$ $$\leq \int_{B_R}|f(x)| \Big[\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}} + \frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}} \Big] \hspace{0.1cm}dx \hspace{0.25cm} \leq \ldots$$ $$\ldots \leq \frac{\epsilon ||f||_{L^1(\mathbb{R}^d)}}{2 ||f||_{L^1(\mathbb{R}^d)}} = \frac{\epsilon}{2}$$ Thus, we've just shown, for a sufficiently large $R > 0$: $$||\xi - \mu|| < \delta = \frac{\epsilon}{8 \pi R ||f||_{L^1(\mathbb{R}^d)}} \hspace{0.25cm} \Rightarrow \hspace{0.25cm} |\hat{f}(\xi) - \hat{f}(\mu)| \leq \ldots$$ $$\ldots \leq \int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx + 2\int_{B_R^c}|f(x)| \hspace{0.1cm}dx \hspace{0.25cm} \leq$$ $$\ldots \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ ...as desired.
Finally we want to show: $$\widehat{(f*g)}(\xi) = \hat{f}(\xi)\hat{g}(\xi)$$ Proceed by directly applying Fubini's Theorem: $$\widehat{(f*g)}(\xi) = \int_{\mathbb{R}^d} \Bigg[\int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm}dy \Bigg] e^{-2\pi i \xi x} \hspace{0.1cm} dx = \ldots$$ $$\ldots = \int_{\mathbb{R}^d} \Bigg[\int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm} e^{-2\pi i \xi (x - y + y)} dy \Bigg] \hspace{0.1cm} dx$$ $$\ldots = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \Big(f(x-y) e^{-2\pi i \xi (x-y)} \Big) \Big(g(y) e^{-2\pi i \xi y}\Big) \hspace{0.1cm}dy \hspace{0.1cm} dx$$ $$\ldots = \int_{\mathbb{R}^d} \Big(g(y) e^{-2\pi i \xi y}\Big) \int_{\mathbb{R}^d} \Big(f(x-y) e^{-2\pi i \xi (x-y)} \Big) \hspace{0.1cm}dx \hspace{0.1cm} dy$$ $$\ldots = \hat{f}(\xi)\int_{\mathbb{R}^d} g(y) e^{-2\pi i \xi y} \hspace{0.1cm} dy = \hat{f}(\xi)\hat{g}(\xi)$$ ...as desired.
Conveniently, since $f$ is measurable on $\mathbb{R}^d$, it follows directly from Proposition 3.9 (p. 86) that $f(x-y)$ is measurable on $\mathbb{R}^{2d}$. Also, since $g$ is measurable on $\mathbb{R}^d$, it follows directly from Corollary 3.7 (P. 85) that $g(y)\chi_{\mathbb{R}^d(x)}$ is measurable on $\mathbb{R}^{2d}$.
Part b.) Since we know $f(x-y)g(y)$ is measurable, by Tonelli's Theorem we have: $$\int_{\mathbb{R}^{2d}} |f(x-y)g(y)| \hspace{0.1cm}d(x,y) \hspace{0.25cm}=\hspace{0.25cm} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y)g(y)| \hspace{0.1cm}dx\hspace{0.1cm}dy$$ ...and from the translation invariance of integration we get: $$\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y)g(y)| \hspace{0.1cm}dx\hspace{0.1cm}dy = \int_{\mathbb{R}^d} |g(y)| \int_{\mathbb{R}^d} |f(x-y)| \hspace{0.1cm}dx\hspace{0.1cm}dy = \ldots$$ $$\ldots = ||f||_{L^1(\mathbb{R}^d)} \int_{\mathbb{R}^d} |g(y)| dy = ||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)} < \infty$$ ...since both $f$ and $g$ are $L^1$.
Part c.) Since $f(x-y)g(y)$ was just shown to be integrable, it follows directly from Fubini's Theorem that for almost every $x \in \mathbb{R}^d$, : $$\int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy < \infty$$ I.e., the convolution: $$(f*g)(x) = \int_{\mathbb{R}^d} f(x-y)g(y)\hspace{0.1cm}dy$$ ...is well-defined for a.e. $x \in \mathbb{R}^d$.
Part d.) Observe that: $$\int_{\mathbb{R}^d} |(f*g)(x)\hspace{0.1cm}| dx = \int_{\mathbb{R}^d} \Bigg| \int_{\mathbb{R}^d}f(x-y)g(y)\hspace{0.1cm}dy \hspace{0.1cm}\Bigg| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{\mathbb{R}^d} \int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy \hspace{0.1cm}dx$$ ...which, by part b, we see: $$||(f*g)||_{L^1(\mathbb{R}^d)} = \int_{\mathbb{R}^d} |(f*g)(x)|\hspace{0.1cm} dx \leq \int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy \hspace{0.1cm}dx = \ldots$$ $$\ldots = ||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)}$$ Now, if $f$ and $g$ are positive functions, $|f(x-y)g(y)|=f(x-y)g(y)$, so equality of $||(f*g)||_{L^1(\mathbb{R}^d)}$ and $||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)}$ follows (again) directly from part b.
Part e.) Let's first check that $\hat{f}(\xi)$ is bounded. Recall that $|e^{i \theta}| = 1 \hspace{0.25cm} \forall \theta \in \mathbb{R}$. Then, observe: $$|\hat{f}(\xi)| = \Bigg| \int_{\mathbb{R}^d} f(x) e^{-2\pi i x \xi} \hspace{0.1cm} dx \Bigg| \leq \ldots $$ $$ \ldots \leq \int_{\mathbb{R}^d} |f(x)||e^{-2\pi i x \xi}| \hspace{0.1cm}dx = \int_{\mathbb{R}^d} |f(x)|\hspace{0.1cm}dx = ||f||_{L^1(\mathbb{R}^d)}$$ Thus, $\hat{f}(\xi)$ is bounded.
Now, let's see if $\hat{f}(\xi)$ is continuous. We begin by observing: $$|\hat{f}(\xi) - \hat{f}(\mu)| = \Bigg| \int_{\mathbb{R}^d} f(x) \big(e^{-2\pi i x \cdot \xi} - e^{-2\pi i x \cdot \mu}\big) \hspace{0.1cm} dx \Bigg| \leq \ldots $$ $$\ldots \leq \int_{\mathbb{R}^d}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx$$ Note that since $f$ is $L^1(\mathbb{R}^d$, for any $\epsilon > 0$ we have that there exists an $R > 0$ such that: $$ \int_{B_R^c} |f(x)| \hspace{0.1cm} dx \leq \frac{\epsilon}{4}$$ (Where $B_R$ is a ball of radius $R$ centered the origin.)
Now, since $\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \leq 2$, we see: $$\int_{\mathbb{R}^d}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx + 2\int_{B_R^c}|f(x)| \hspace{0.1cm}dx$$ From here, require: $$||\xi - \mu|| < \delta = \frac{\epsilon}{8 \pi R ||f||_{L^1(\mathbb{R}^d)}}$$ Now, it should be clear from the Cauchy Schwartz inequality that on $B_R$: $$ |x \cdot (\xi - \mu)| \leq R \delta = \frac{\epsilon}{8 \pi ||f||_{L^1(\mathbb{R}^d)}}$$ Therefore, plugging it all in, we finally see: $$\int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx \leq \ldots$$ $$\leq \int_{B_R}|f(x)| \Big[\big| \cos(2\pi x \cdot (\xi - \mu)) - 1 \big| + \big|\sin(2\pi x \cdot (\xi - \mu))\big|\Big] \hspace{0.1cm}dx$$ $$\leq \int_{B_R}|f(x)| \Big[\big| \cos\big(\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}}\big) - 1 \big| + \big|\sin\big(\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}}\big) \big| \Big]$$ $$\leq \int_{B_R}|f(x)| \Big[\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}} + \frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}} \Big] \hspace{0.1cm}dx \hspace{0.25cm} \leq \ldots$$ $$\ldots \leq \frac{\epsilon ||f||_{L^1(\mathbb{R}^d)}}{2 ||f||_{L^1(\mathbb{R}^d)}} = \frac{\epsilon}{2}$$ Thus, we've just shown, for a sufficiently large $R > 0$: $$||\xi - \mu|| < \delta = \frac{\epsilon}{8 \pi R ||f||_{L^1(\mathbb{R}^d)}} \hspace{0.25cm} \Rightarrow \hspace{0.25cm} |\hat{f}(\xi) - \hat{f}(\mu)| \leq \ldots$$ $$\ldots \leq \int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx + 2\int_{B_R^c}|f(x)| \hspace{0.1cm}dx \hspace{0.25cm} \leq$$ $$\ldots \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ ...as desired.
Finally we want to show: $$\widehat{(f*g)}(\xi) = \hat{f}(\xi)\hat{g}(\xi)$$ Proceed by directly applying Fubini's Theorem: $$\widehat{(f*g)}(\xi) = \int_{\mathbb{R}^d} \Bigg[\int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm}dy \Bigg] e^{-2\pi i \xi x} \hspace{0.1cm} dx = \ldots$$ $$\ldots = \int_{\mathbb{R}^d} \Bigg[\int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm} e^{-2\pi i \xi (x - y + y)} dy \Bigg] \hspace{0.1cm} dx$$ $$\ldots = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \Big(f(x-y) e^{-2\pi i \xi (x-y)} \Big) \Big(g(y) e^{-2\pi i \xi y}\Big) \hspace{0.1cm}dy \hspace{0.1cm} dx$$ $$\ldots = \int_{\mathbb{R}^d} \Big(g(y) e^{-2\pi i \xi y}\Big) \int_{\mathbb{R}^d} \Big(f(x-y) e^{-2\pi i \xi (x-y)} \Big) \hspace{0.1cm}dx \hspace{0.1cm} dy$$ $$\ldots = \hat{f}(\xi)\int_{\mathbb{R}^d} g(y) e^{-2\pi i \xi y} \hspace{0.1cm} dy = \hat{f}(\xi)\hat{g}(\xi)$$ ...as desired.
Subscribe to:
Posts (Atom)