3.3

Part a.) Under the given conditions, we know that $\forall \epsilon > 0$ there exists $r > 0$ such that: $$\frac{m(E \cap [-r,r])}{m([-r,r])} \geq (1-\epsilon)$$ Now, simply observe (by the translation / rotation invariance of the Lebesgue Measure): $$m(-E \cap [-r,r]) = m(E \cap [-r,r]) \geq (1-\epsilon)2r$$ Thus, for $\epsilon < \frac{1}{2}$: $$0 < (1-2\epsilon)2r \leq m(-E \cap E \cap [-r,r])$$ ...and thus, since these sets have positive measure, we can always find a sequence of $x_n's$ that satisfy the condition.

Part b.) It suffices to simply consider the $cE$ case, where $c > 1$. Since: $$\frac{m(cE \cap [-r,r])}{m([-r,r])} = \frac{m(E \cap [-\frac{r}{c}, \frac{r}{c}])}{m([-\frac{r}{c},\frac{r}{c}])}$$ We have: $$m(cE \cap [-r,r]) = \frac{m([-r,r])}{m([-\frac{r}{c},\frac{r}{c}])}m\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big) = \ldots$$ $$\ldots = cm\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big)$$ Thus, like in part a, we have that since $\forall \epsilon > 0$ there exists an $r_0 > 0$ such that $\forall r \in (0,r_0)$: $$m(E \cap [-r,r]) > (1 - \epsilon)2r$$ Therefore, since $\frac{r}{c} < r$, we have: $$m(cE \cap [-r,r]) = cm\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big) > c(1 - \epsilon)2\frac{r}{c} = \ldots$$ $$\ldots = (1 - \epsilon)2r > 0$$ ...when $\epsilon < \frac{1}{2}$. Thus, under these settings, the estimate: $$m(cE \cap E \cap [-r,r]) > (1 - 2\epsilon)2r$$ ...holds, $\implies$ we can choose a sequence of $x_n's$ satisfying the desired condition.

To see that this condition holds for $0 < c < 1$, observe that if we simply define $cE = F$, that: $\frac{1}{c}F = E$, and the statement holds per the previous argument.

Now, since we know it works for $c > 0$, and $c = -1$ (from Part a.), simply combine the arguments to see that this clearly works for $c \in \mathbb{R} \backslash \lbrace 0 \rbrace$.

3.2

Recall the statement from exercise 3.1c, however, instead of: $$\int_\mathbb{R}^d K_\delta (y) = 1$$ Write: ($\dagger$) For some particular $C \in \mathbb{R}$: $$\int_\mathbb{R}^d K_\delta (y) = C$$ The new statement should read: $$ \dagger \hspace{0.25cm} \Rightarrow (f*K_\delta)(x) \to Cf(x) \hspace{0.25cm} \text{as} \hspace{0.25cm} \delta \to 0$$ The argument follows identically to how the $C=1$ case is shown for approximations to the identity. Now, simply consider $C = 0$, and we're done.

3.1

Part a.) We first need to show that $K_\delta(x)$ satisfies $(i),(ii),$ and $(iii)$ listed at the top of page 109. Given $\delta > 0$, and $\phi$ is integrable s.t. $\int_{\mathbb{R}^d} \phi = 1$, we have that: $$\int_{\mathbb{R}^d} K_\delta (x)\hspace{0.1cm}dx = \int_{\mathbb{R}^d} \frac{1}{\delta^d} \phi(x / \delta) \hspace{0.1cm}dx$$ ...and by the dilation property of $L^1$ functions: $$\int_{\mathbb{R}^d} \frac{1}{\delta^d} \phi(x / \delta) \hspace{0.1cm}dx = \int_{\mathbb{R}^d} \frac{\delta^d}{\delta^d} \phi(x) \hspace{0.1cm}dx = \int_{\mathbb{R}^d} \phi(x) \hspace{0.1cm}dx = 1$$ Which satisfies $(i)$. Next, notice: $$\int_{\mathbb{R}^d} |K_\delta (x)| \hspace{0.1cm}dx = \int_{\mathbb{R}^d} |\phi(x)| \hspace{0.1cm}dx = || \phi ||_{L^1} < \infty$$ ...which satisfies $(ii)$. Finally, observe that: $$\int_{B_\mu} |K_\delta (x)| \hspace{0.1cm}dx = \int_{B_\mu} |K_\delta (x)| \hspace{0.1cm}dx = \int_{\mathbb{R}^d} \frac{1}{\delta^d} |\phi(x / \delta)| \chi_{B_\mu}(x) dx $$ ...again by the dilation property of $L^1$ functions: $$ \int_{\mathbb{R}^d} \frac{1}{\delta^d} |\phi(x / \delta)| \chi_{B_\mu}(x) dx = \int_{\mathbb{R}^d} |\phi(x)| \chi_{B_\mu}(\delta x) dx = \int_{B_{\mu / \delta}} |\phi(x)| dx$$ Therefore, given that for any $\mu > 0$, we have $B_{\mu / \delta} \to \mathbb{R}^d$ as $\delta \to 0$, it follows directly by, say the Dominated Convergence Theorem that: $$\lim_{\delta \to 0} \int_{B_\mu^c} |K_\delta (x)| \hspace{0.1cm}dx = 0$$ ...which was $(iii)$.

Part b.) With the added assumptions that $|\phi| \leq M$ where $M > 0$ and $\phi$ is supported on a compact set $S \subset \mathbb{R^d}$, we need to show that $K_\delta (x)$ is an approximation to the identity. (Properties $(ii')$ and $(iii')$) Certainly: $$|K_\delta (x)| = |\frac{1}{\delta^d} \phi(x / \delta)| \leq \frac{1}{\delta^d}M$$ ...satisfying condition $(ii')$. Next, since $S$ is compact, let $\overline{B_r}$ be a ball of radius $r = \max \lbrace \max(S), 1 \rbrace$. $$|K_\delta (x)| \leq \frac{M}{\delta^d} \chi_S(x/\delta) \leq \frac{M}{\delta^d} \chi_{\overline{B_r}}(x/\delta)$$ Now, for $|x| > \delta r$, we have: $$\frac{M \delta}{|x|^{d+1}} \geq \frac{M}{\delta^d} \chi_{\overline{B_r}}(x/\delta) = 0$$ For $0 < |x| \leq \delta r$ we have: $$\frac{M \delta}{|x|^{d+1}} \geq \frac{M \delta}{|r\delta|^{d+1}} \geq \frac{M}{\delta^d}$$ ...satisfying condition $(iii')$.

Part c.) First, since $\int_{\mathbb{R}^d} K_\delta (y) dy = 1$, observe that: $$f(x) = \int_{\mathbb{R}^d}f(x) K_\delta (y) dy$$ So it now follows directly that: $$||(f*K_\delta) - f||_{L^1} = \int_{\mathbb{R}^d} \Bigg| \int_{\mathbb{R}^d} f(x - y)K_{\delta}(y)dy - f(x) \Bigg| \hspace{0.1cm}dx = \ldots$$ $$\ldots = \int_{\mathbb{R}^d} \Bigg| \int_{\mathbb{R}^d} \big(f(x - y) - f(x)\big)K_{\delta}(y)dy \Bigg| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy \hspace{0.1cm}dx $$ Now, by Fubini's Theorem and the triangle inequality that $\forall r > 0$: $$\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy \hspace{0.1cm}dx = \ldots $$ $$\int_{\mathbb{R}^d} \Bigg( \int_{B_r(0)} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy + \ldots$$ $$\ldots + \int_{B_r^c(0)} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy \hspace{0.1cm}\Bigg)dx \leq \ldots$$ $$\ldots \leq ||f(x-y) - f(x)||_{L^1} \int_{B_r(0)}|K_{\delta}(y)|dy + \ldots$$ $$\ldots + 2||f||_{L^1}\int_{B_r^c(0)}|K_{\delta}(y)|dy$$ Now, since $f$ is $L^1$, and from property $(ii)$ of good kernals (P.109), $\exists A > 0$ such that $||K_\delta||_{L^1} < A \hspace{0.1cm} \forall \delta$, we know $\forall \epsilon > 0$ there exists an $r > 0$ small enough such that: $$y \in B_r(0) \hspace{0.25cm} \Rightarrow \hspace{0.25cm} ||f(x-y) - f(x)|| < \frac{\epsilon}{2A}$$ And, from property $(iii)$ of good kernals (P.109), we have that $\forall \epsilon > 0 \hspace{0.25cm} \exists \delta > 0$ small enough such that: $$\int_{B_r^c(0)}|K_{\delta}(y)|dy < \frac{\epsilon}{4||f||_{L^1}}$$ Putting everything together, we finally see that if we choose both $\delta, r > 0$ small enough, that: $$||f(x-y) - f(x)||_{L^1} \int_{B_r(0)}|K_{\delta}(y)|dy + \ldots$$ $$\ldots + 2||f||_{L^1}\int_{B_r^c(0)}|K_{\delta}(y)|dy < \ldots$$ $$\ldots < \frac{\epsilon}{2A}\int_{B_r(0)}|K_{\delta}(y)|dy + 2||f||_{L^1} \frac{\epsilon}{4||f||_{L^1}}<\ldots$$ $$\ldots < \frac{\epsilon}{2A} A + \frac{\epsilon}{2} = \epsilon$$ ...as desired.

2.24

Part a.) Given the equation: $$(f*g)(x) = \int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm}dy$$ If we assume $f$ is integrable, and $\exists M \geq 0$ such that $|g| \leq M \hspace{0.1cm} \forall x \in \mathbb{R}^d$, we have: $$\big|(f*g)(x) - (f*g)(z)\big| = \ldots$$ $$\ldots = \Bigg|\int_{\mathbb{R}^d} \bigg(f(x-y) - f(z - y)\bigg)g(y) \hspace{0.1cm}dy\Bigg| \leq \ldots$$ $$\ldots \leq M\int_{\mathbb{R}^d} \bigg|f(x-y) - f(z - y)\bigg| \hspace{0.1cm}dy =\ldots$$ $$\ldots = M\int_{\mathbb{R}^d} \bigg|f(-y) - f((z - x) - y)\bigg| \hspace{0.1cm}dy$$ Thus since $f$ is $L^1(\mathbb{R}^d)$, by Proposition 2.5, $\forall \epsilon > 0,\hspace{0.25cm} \exists \delta$ such that $||z - x|| < \delta \hspace{0.25cm} \Rightarrow ||f(y) - f(y-(z-x))||_{L^1} < \epsilon$. Thus, since $$\big|(f*g)(x) - (f*g)(z)\big| \leq ||f(y) - f(y-(z-x))||_{L^1}$$ The convolution $(f*g)(x)$ must be uniformly continuous.

Part b.) If $f$ and $g$ are both $L^1(\mathbb{R}^d)$, we proved in exercise 21 part d that $(f*g)(x)$ is also $L^1(\mathbb{R}^d)$. Thus, since $(f*g)(x)$ is uniformly continuous, and integrable, we have (by exercise 6 part b) that: $$\lim_{|x| \to \infty} (f*g)(x) = 0$$ ...as desired.

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.