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.