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.

2.19

First, noting the definition of our set $E_\alpha$: $$E_\alpha = \big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \big| \hspace{0.25cm} |f(x)| > \alpha \big\rbrace$$ Observe that $E_\alpha$ is certainly measurable. Now, consider the integration: $$ \int_0^\infty m(E_\alpha) \hspace{0.1cm} d\alpha = \int_0^\infty \int_{E_\alpha} dx \hspace{0.1cm} d\alpha$$ Since constant functions over measurable sets are measurable, we can apply Tonelli's Theorem to see: $$\int_0^\infty \int_{E_\alpha} dx \hspace{0.1cm} d\alpha = \int_{E_\alpha} \int_0^\infty d\alpha \hspace{0.1cm} dx = \int_{\mathbb{R}^d} \int_0^\infty \chi_{[0,|f(x)|)} d\alpha \hspace{0.1cm} dx$$ Next, notice from the definition of $E_\alpha$ that: $$\int_{\mathbb{R}^d} \int_0^\infty \chi_{[0,|f(x)|)} d\alpha \hspace{0.1cm} dx = \int_{\mathbb{R}^d} m\big([0,|f(x)|)\big) \hspace{0.1cm} dx = \int_{\mathbb{R}^d} |f(x)| dx$$ ...as desired.

2.5

Part a.) By the definition of the infimum, we have $\forall \epsilon > 0 \hspace{0.25cm} \exists z \in F$ such that $|x - z| < \delta(x) + \epsilon$. Thus, $$\delta(y) \leq |y - z| \leq |y - x| + |x - z| \leq |y - x| + \delta(x) + \epsilon$$ Thus, since $\epsilon$ can be arbitrarily small, we have: $$\delta(y) - \delta(x) \leq |y-x|$$ Now, repeat this argument for $y$, and we have: $$-|y-x| \leq \delta(y) - \delta(x) \leq |y-x| \hspace{0.25cm} \Rightarrow |\delta(y) - \delta(x)| \leq |y-x|$$

Part b.) First observe that if $y \in F$, then $\delta(y) = 0$. Then, certainly: $$ I(x) = \int_F \frac{\delta(y)}{|x-y|^2} dy + \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy = \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy$$ Next, notice that since $F^c$ is an open set, for any $x \in F^c$ we know $\exists r > 0$ such that $B_{r}(x) \subset F^c$, and for any $y \in B_{r}(x)$, $\exists M > 0$ such that $\delta(y) \geq M$. (We can always do this by simply choosing $r$ small enough to fit the ball in $F^c$, then using the ball of radius $\frac{r}{2}$.) Therefore, we have: $$ \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \geq M \int_{B_{r}(x)} \frac{1}{|x-y|^2} dy$$ Now, just observe that we can just center $x$ at the origin, and our calculation becomes: $$M\int_{B_{r}(x)} \frac{1}{|x-y|^2} dy = \lim_{b \to 0} \int_b^\infty \frac{1}{y^2}dy = \infty$$ ...as desired.

Part c.) As the hint prescribes, if we can show $I(x)$ is $L^1(F)$ then we have that $I$ must be almost everywhere finite. Indeed, first notice (from part b) that: $$\int_F I(x) dx = \int_{F} \int_{\mathbb{R}} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx = \int_{F} \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx$$ Before I can use Tonelli's Theorem to swap the limits of integration, I need to first check that the function: $$h(x,y) = \chi_{(F \times F^c)}\frac{\delta(y)}{|x-y|^2}$$ ...is measurable. Certainly, $\chi_{(F \times F^c)}$ is measurable since $F$ and $F^c$ are Borel. Next notice that the sets: $$ \lbrace y \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} \delta(y) > \alpha \rbrace $$ are measurable, since they're just the union of $F$ and countably-many closed intervals. Lastly, the function $\frac{1}{|x-y|^2}$ is obviously measurable since it's continuous almost everywhere. Thus, since the product of measurable functions are measurable, we have $h(x,y)$ is measurable.

Thus, it follows from Tonelli's theorem that: $$\int_{F} \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx = \int_{F^c} \int_{F} \frac{\delta(y)}{|x-y|^2} dx \hspace{0.1cm} dy = \ldots$$ $$ \ldots = \int_{F^c} \delta(y) \int_{F} \frac{1}{|x-y|^2} dx \hspace{0.1cm} dy $$ Now, this part I found to be kind of tricky (at first). Observe that: $$F \subset \lbrace x \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} |x - y| \geq \delta(y) \rbrace$$ Then we have: $$\int_{F} \frac{1}{|x-y|^2} dx \leq 2\int_{\delta(y)}^\infty \frac{1}{x^2} dx = \frac{2}{\delta(y)}$$ Finally, we have: $$\int_{F^c} \delta(y) \int_{F} \frac{1}{|x-y|^2} dx \hspace{0.1cm} dy \leq 2\int_{F^c} \delta(y) \int_{\delta(y)}^\infty \frac{1}{x^2} dx \hspace{0.1cm} dy \leq 2 m(F^c) $$ ...and since $m(F^c) < \infty$ we have $I$ must be $L^1 (F)$.

2.4

It suffices to consider $f$ to be a non-negative $L^1$ function. It's not at first apparent that $g(x)$ is an $L^1$ function. First, define the set: $$T(b) = \lbrace (x,t) \in \mathbb{R} \times \mathbb{R} \hspace{0.25cm} \big|\hspace{0.25cm} 0 < x \leq t, \hspace{0.25cm} x \leq t \leq b \rbrace$$ (It's just the triangle formed by taking the square $[0,b] \times [0,b]$, bisecting it with the line $x = t$, and taking the top half.) Next, define the function: $$F(x,t) = \frac{f(t)}{t} \chi_{T(b)}$$ Certainly, $F$ is measurable, since $\chi_{T(b)}$, $f$, and $\frac{1}{t}$ are measurable functions. Next, observe that:

$$\int_{[0,b]} g(x) dx = \int_{[0,b]} \int_{[x,b]} \frac{f(t)}{t}dtdx = \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{f(t)}{t} \chi_{T(b)} dt dx $$ From here, Tonelli's Theorem gives, that since $F$ is measurable: $$ \int_{[0,b]} \int_{[x,b]} \frac{f(t)}{t}dtdx = \int_{[0,b]} \int_{[0,t]} \frac{f(t)}{t} dx dt = \int_{[0,b]} f(t) dt$$ Thus, since $f \in L^1$, we have: $$\int_{[0,b]} g(x) dx = \int_{[0,b]} f(t) dt < \infty \hspace{0.25cm} \Rightarrow g \in L^1$$ ...as desired.