# $$L^2$$ and Fourier Analysis

## Fall 2020.5 #real_analysis/qual/completed

Suppose $$\varphi\in L^1({\mathbb{R}})$$ with \begin{align*} \int \varphi(x) \, dx = \alpha .\end{align*} For each $$\delta > 0$$ and $$f\in L^1({\mathbb{R}})$$, define \begin{align*} A_\delta f(x) \coloneqq\int f(x-y) \delta^{-1} \varphi\qty{\delta^{-1} y}\, dy .\end{align*}

• Prove that for all $$\delta > 0$$, \begin{align*} {\left\lVert {A_\delta f} \right\rVert}_1 \leq {\left\lVert {\varphi} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 .\end{align*}

• Prove that \begin{align*} A_\delta f \to \alpha f \text{ in } L^1({\mathbb{R}}) {\quad \operatorname{as} \quad} \delta\to 0^+ .\end{align*}

Hint: you may use without proof the fact that for all $$f\in L^1({\mathbb{R}})$$, \begin{align*} \lim_{y\to 0} \int_{\mathbb{R}}{\left\lvert {f(x-y) - f(x)} \right\rvert}\, dx = 0 .\end{align*}

See Folland 8.14.

This is a direct application of Fubini-Tonelli: \begin{align*} {\left\lVert {A_\delta f} \right\rVert} &\coloneqq\int {\left\lvert { \int f(x-y)\delta^{-1}\varphi(\delta^{-1}y)\,dy} \right\rvert} \,dx\\ &\leq \int \int {\left\lvert {f(x-y)\delta^{-1}\varphi(\delta^{-1}y)} \right\rvert} \,dy\,dx\\ &\overset{FT}{=} \int \int {\left\lvert { f(x-y) } \right\rvert} \cdot {\left\lvert {\delta^{-1}\varphi(\delta^{-1}y)} \right\rvert} \,dx\,dy\\ &= \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y) } \right\rvert} \qty{ \int {\left\lvert { f(x-y) } \right\rvert}\,dx} \,dy\\ &= \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y)} \right\rvert}\cdot {\left\lVert {f} \right\rVert} \,dy\\ &= {\left\lVert {f} \right\rVert} \cdot \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y) } \right\rvert} \,dy\\ &= {\left\lVert {f} \right\rVert} \cdot {\left\lVert {\varphi} \right\rVert} .\end{align*} Here we’ve used translation and dilation invariance of the Lebesgue integral.

Write $$\phi_\delta(y) \coloneqq\delta^{-1}\phi(\delta^{-1}y)$$, then \begin{align*} {\left\lVert {A_\delta f - \alpha f} \right\rVert}_1 &\coloneqq\int {\left\lvert {A_\delta f(x) - \alpha f(x) } \right\rvert} \,dx\\ &= \int {\left\lvert { \int {f(x-y) \phi_\delta(y) } \,dy- \alpha f(x) } \right\rvert}\,dx\\ &= \int {\left\lvert { \int { \tau_y f (x) \phi_\delta(y) } \,dy- \int f(x) \phi_\delta(y) \,dy} \right\rvert}\,dx\\ &\leq \int\int {\left\lvert {\tau_y f(x) - f(x)} \right\rvert}\cdot {\left\lvert {\phi_\delta(y)} \right\rvert} \,dy\,dx\\ &= \int\int {\left\lvert {\tau_y f(x) - f(x)} \right\rvert}\cdot {\left\lvert {\phi_\delta(y)} \right\rvert} \,dx\,dy\\ &= \int{\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}_1 \,dy ,\end{align*} where the interchange of integration order is justified by Tonelli since the integrands are positive. The goal is to now make this small when $$\delta$$ is small

# One way to do this immediately: make a change of variables $$y=tz$$ to get \begin{align*} {\left\lVert {A_\delta f - \alpha f} \right\rVert}_1 \leq \int {{\left\lvert {\phi(z)} \right\rvert}} {\left\lVert {\tau_{tz}f -f} \right\rVert}_1 \,dz ,\end{align*} use that $${\left\lVert {\tau_{tz} f- f} \right\rVert}_1 \leq 2{\left\lVert {f} \right\rVert}_1 < \infty$$ by the triangle inequality and apply the DCT:  \begin{align*} \lim_{t\to 0} \int {{\left\lvert {\phi(z)} \right\rvert}} \cdot {\left\lVert {\tau_{tz}f -f} \right\rVert}_1 ,dz

\int {{\left\lvert {\phi(z)} \right\rvert}} \lim_{t\to 0} {\left\lVert {\tau_{tz}f -f} \right\rVert}_1 ,dz = 0 .\end{align*} {=html}

More directly, use continuity in $$L^1$$ (as per the hint) to pick a $$h>0$$ such that \begin{align*} {\left\lVert {\tau_y f - f} \right\rVert}< {\varepsilon}\quad \text{ for } y\in A \coloneqq\left\{{y{~\mathrel{\Big\vert}~}{\left\lvert {y} \right\rvert} \leq h}\right\} .\end{align*} Now choose $$\delta_0 \gg 1$$ large enough so that \begin{align*} \int_{A^c}{\left\lvert {\phi_\delta(y)} \right\rvert}\,dy< {\varepsilon}\quad \text{ for all }\delta > \delta_0 .\end{align*} Now  \begin{align*} \int_{\mathbb{R}}{\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}1 ,dy &= \int_A {\left\lvert {\phi\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}_1 ,dy

• \int_{A^c} {\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}1 ,dy\ &\leq \int_A {\left\lvert {\phi\delta(y)} \right\rvert}\cdot {\varepsilon},dy
• \int_{A^c} {\left\lvert {\phi_\delta(y)} \right\rvert} \cdot 2{\left\lVert {f} \right\rVert}1 ,dy\ &\leq {\varepsilon}{\left\lVert {\phi\delta} \right\rVert}_1 + 2{\varepsilon}{\left\lVert {f} \right\rVert}_1 \ &\longrightarrow 0 .\end{align*} {=html}

## Spring 2020.6 #real_analysis/qual/completed

• Show that \begin{align*} L^2([0, 1]) \subseteq L^1([0, 1]) {\quad \operatorname{and} \quad} \ell^1({\mathbb{Z}}) \subseteq \ell^2({\mathbb{Z}}) .\end{align*}

• For $$f\in L^1([0, 1])$$ define \begin{align*} \widehat{f}(n) \coloneqq\int _0^1 f(x) e^{-2\pi i n x} \, dx .\end{align*} Prove that if $$f\in L^1([0, 1])$$ and $$\left\{{\widehat{f}(n)}\right\} \in \ell^1({\mathbb{Z}})$$ then \begin{align*} S_N f(x) \coloneqq\sum_{{\left\lvert {n} \right\rvert} \leq N} \widehat{f} (n) e^{2 \pi i n x} .\end{align*} converges uniformly on $$[0, 1]$$ to a continuous function $$g$$ such that $$g = f$$ almost everywhere.

Hint: One approach is to argue that if $$f\in L^1([0, 1])$$ with $$\left\{{\widehat{f} (n)}\right\} \in \ell^1({\mathbb{Z}})$$ then $$f\in L^2([0, 1])$$.

From Neil:

• $$\widehat{f}$$ in $$\ell^1$$ ensures that $$S_N$$ converges uniformly to something, call it $$g$$.
• $$\widehat{f} \in\ell^1$$ Implies $$\widehat{f}\in \ell^2$$ which (by characterization of an o.n.b.) implies $$f$$ is in $$L^2$$ (Parseval) and (again by characterization of an o.n.b.) that $$S_N$$ converges to $$f$$ in $$L^2$$ (and hence a subsequence converges to f a.e.)
• By uniqueness of limits $$f=g$$.

Other stuff:

• For $$e_n(x) \coloneqq e^{2\pi i n x}$$, the set $$\left\{{e_n}\right\}$$ is an orthonormal basis for $$L^2([0, 1])$$.
• For any orthonormal sequence in a Hilbert space, we have Bessel’s inequality: \begin{align*} \sum_{k=1}^{\infty}\left|\left\langle x, e_{k}\right\rangle\right|^{2} \leq\|x\|^{2} .\end{align*}
• When $$\left\{{e_n}\right\}$$ is a basis, the above is an equality (Parseval)
• Arguing uniform convergence: since $$\left\{{\widehat{f}(n)}\right\} \in \ell^1({\mathbb{Z}})$$, we should be able to apply the $$M$$ test.

Claim: if $$f\in L^1[0, 1]$$ and $$\widehat{f}\in \ell^1({\mathbb{Z}})$$, then $$S_Nf \to f$$ uniformly.

• Since $$\widehat{f}\in \ell^1({\mathbb{Z}})$$, we have $$S_Nf\to g$$ uniformly for some continuous $$g$$ by the $$M{\hbox{-}}$$test.

• Now consider $$\widehat{g}$$. We have \begin{align*} \widehat{g}(n) = \int_0^1 \sum_m \qty{\widehat{f}(m)e_m(x)}e_{-n}(x) \,dx= \widehat{f}(n) ,\end{align*} using that $$\int_0^1 e_n(x)\,dx= \chi_{n=0}$$.

• We’ll now show $$f-g= 0$$ a.e. by mollifying against an approximate identity $$\varphi\in L^1$$, setting \begin{align*} \varphi_{\varepsilon}(x) \coloneqq{\varepsilon}^{-1}\varphi({\varepsilon}^{-1}x) \in L^1[0, 1] .\end{align*}

• A computation: \begin{align*} \widehat{f\ast\varphi_{\varepsilon}}(n) &= \widehat{f}\cdot \widehat{\varphi_{\varepsilon}}(n) \\ &= \widehat{g}\cdot \widehat{\varphi_{\varepsilon}}(n) \\ &= \widehat{g\ast\varphi_{\varepsilon}}(n) ,\end{align*} so \begin{align*} \widehat{(f-g)\ast\varphi_{\varepsilon}} = 0 \quad \forall n \implies (f-g)\ast\varphi_{\varepsilon}\equiv 0 ,\end{align*} using that $$(f-g)\ast\varphi_{\varepsilon}\in L^2$$ and $$\left\{{e_n}\right\}$$ for a complete orthonormal basis of $$L^2$$.

• Now use that $$(f-g)\ast\varphi_{\varepsilon}\to (f-g)$$ in $$L^1$$ and $$(f-g)\ast\varphi_{\varepsilon}\equiv 0$$ to conclude $$f-g = 0$$ a.e.

$$\ell^1({\mathbb{Z}}) \subseteq \ell^2({\mathbb{Z}})$$.

• Set $$\mathbf{c} = \left\{{c_k {~\mathrel{\Big\vert}~}k\in {\mathbb{Z}}}\right\} \in \ell^1({\mathbb{Z}})$$.
• It suffices to show that if $$\sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert} < \infty$$ then $$\sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert}^2 < \infty$$.
• Let $$S = \left\{{c_k {~\mathrel{\Big\vert}~}{\left\lvert {c_k} \right\rvert} \leq 1}\right\}$$, then $$c_k \in S \implies {\left\lvert {c_k} \right\rvert}^2 \leq {\left\lvert {c_k} \right\rvert}$$
• Claim: $$S^c$$ can only contain finitely many elements, all of which are finite.
• If not, either $$S^c \coloneqq\left\{{c_j}\right\}_{j=1}^\infty$$ is infinite with every $${\left\lvert {c_j} \right\rvert} > 1$$, which forces \begin{align*}\sum_{c_k\in S^c} {\left\lvert {c_k} \right\rvert} = \sum_{j=1}^\infty {\left\lvert {c_j} \right\rvert} > \sum_{j=1}^\infty 1 = \infty.\end{align*}
• If any $$c_j = \infty$$, then $$\sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert} \geq c_j = \infty$$.
• So $$S^c$$ is a finite set of finite integers, let $$N = \max \left\{{{\left\lvert {c_j} \right\rvert}^2 {~\mathrel{\Big\vert}~}c_j \in S^c}\right\} < \infty$$.
• Rewrite the sum \begin{align*} \sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert}^2 &= \sum_{c_k\in S} {\left\lvert {c_k} \right\rvert}^2 + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq \sum_{c_k\in S} {\left\lvert {c_k} \right\rvert} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq \sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \quad\text{since the ${\left\lvert {c_k} \right\rvert}$ are all positive} \\ &= {\left\lVert {\mathbf{c}} \right\rVert}_{\ell^1} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq {\left\lVert {\mathbf{c}} \right\rVert}_{\ell^1} + {\left\lvert {S^c} \right\rvert}\cdot N \\ &< \infty .\end{align*}

$$L^2([0, 1]) \subseteq L^1([0, 1])$$.

• It suffices to show that $$\int {\left\lvert {f} \right\rvert}^2 < \infty \implies \int {\left\lvert {f} \right\rvert} < \infty$$.

• Define $$S = \left\{{x\in [0, 1] {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} \leq 1}\right\}$$, then $$x\in S^c \implies {\left\lvert {f(x)} \right\rvert}^2 \geq {\left\lvert {f(x)} \right\rvert}$$.

• Break up the integral: \begin{align*} \int_{\mathbb{R}}{\left\lvert {f} \right\rvert} &= \int_S {\left\lvert {f} \right\rvert} + \int_{S^c} {\left\lvert {f} \right\rvert} \\ &\leq \int_S {\left\lvert {f} \right\rvert} + \int_{S^c} {\left\lvert {f} \right\rvert}^2 \\ &\leq \int_S {\left\lvert {f} \right\rvert} + {\left\lVert {f} \right\rVert}_2 \\ &\leq \sup_{x\in S}\left\{{{\left\lvert {f(x)} \right\rvert}}\right\} \cdot \mu(S) + {\left\lVert {f} \right\rVert}_2 \\ &= 1 \cdot \mu(S) + {\left\lVert {f} \right\rVert}_2 \quad\text{by definition of } S\\ &\leq 1 \cdot \mu([0, 1]) + {\left\lVert {f} \right\rVert}_2 \quad\text{since } S\subseteq [0, 1] \\ &= 1 + {\left\lVert {f} \right\rVert}_2 \\ &< \infty .\end{align*}

Note: this proof shows $$L^2(X) \subseteq L^1(X)$$ whenever $$\mu(X) < \infty$$.

• First, $$S_Nf$$ converges in $${\mathcal{H}}$$ to something, say $$g \coloneqq\lim_{n\to\infty} S_n f$$, since \begin{align*} {\left\lVert {g - S_Nf} \right\rVert} = {\left\lVert {\sum_{{\left\lvert {n} \right\rvert} \geq N} \widehat{f} (n) e_n(x) } \right\rVert} \leq \sum_{{\left\lvert {n} \right\rvert} \geq N } {\left\lvert {\widehat{f}(n)} \right\rvert} \overset{N\to\infty}\longrightarrow 0 ,\end{align*} where the last term is the tail of a convergent sum since $$\left\{{\widehat{f}(n)}\right\} \in \ell^1$$.

• This also shows that $$S_N\to g$$ uniformly.

• $$g$$ is continuous, as the uniform limit of continuous functions.

• Showing that $$g = f$$ a.e.: it suffices to show that $$S_N$$ converges to $$f$$ in $$L^p$$ for some $$p$$, since this will provide a subsequence that converges to $$f$$ a.e..

• Claim: $$\widehat{f}\in \ell^1 \subseteq \ell^2$$ implies that $$f \in L^2$$. This follows from Parseval: \begin{align*} \infty > {\left\lVert {\widehat{f}} \right\rVert}_{\ell^2}^2 = \sum_{n\in {\mathbb{Z}}} {\left\lvert {\widehat{f}(n)} \right\rvert}^2 = \int_0^1 {\left\lvert {f(z)} \right\rvert}^2 \,dz = {\left\lVert {f} \right\rVert}_{L^2}^2 .\end{align*}

• Claim: $$S_N\to f$$ in $$L^2$$.

• This follows from the fact that $$\left\{{e_n}\right\}_{n\in {\mathbb{Z}}}$$ is a complete orthonormal basis, so $$f = \sum {\left\langle {f},~{e_n} \right\rangle}e_n$$ uniquely, recognizing $$\widehat{f}(n) = {\left\langle {f},~{e_n} \right\rangle}$$, and writing \begin{align*} f = \sum_{n} {\left\langle {f},~{e_n} \right\rangle}e_n = \sum_n \widehat{f}(n) e_n \coloneqq\lim_{N\to\infty }S_N f .\end{align*}
• So a subsequence $$\left\{{S_{N_k}}\right\}_{k\geq 0}$$ converges to $$f$$ a.e.. Since $$S_N\to g$$ a.e., $$f=g$$ a.e. by uniqueness of limits.

## Fall 2017.5 #real_analysis/qual/completed

Let $$\varphi$$ be a compactly supported smooth function that vanishes outside of an interval $$[-N, N]$$ such that $$\int _{{\mathbb{R}}} \varphi(x) \, dx = 1$$.

For $$f\in L^1({\mathbb{R}})$$, define \begin{align*} K_{j}(x) \coloneqq j \varphi(j x), \qquad f \ast K_{j}(x) \coloneqq\int_{{\mathbb{R}}} f(x-y) K_{j}(y) \, dy \end{align*} and prove the following:

• Each $$f\ast K_j$$ is smooth and compactly supported.

• \begin{align*} \lim _{j \to \infty} {\left\lVert {f * K_{j}-f} \right\rVert}_{1} = 0 \end{align*}

Hint: \begin{align*} \lim _{y \to 0} \int _{{\mathbb{R}}} |f(x-y)-f(x)| dy = 0 \end{align*}

\todo[inline]{Add concepts.}

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• ?

Part a

Lemma: If $$\varphi \in C_c^1$$, then $$(f \ast \varphi)' = f \ast \varphi'$$ almost everywhere.

Silly Proof:

\begin{align*} \mathcal{F}( (f \ast \varphi)' ) &= 2\pi i \xi ~\mathcal{F}(f\ast \varphi) \\ &= 2\pi i \xi ~ \mathcal{F}(f) ~ \mathcal{F}(\varphi) \\ &= \mathcal{F}(f) \cdot \left( 2\pi i \xi ~\mathcal{F}(\varphi)\right) \\ &= \mathcal{F}(f) \cdot \mathcal{F}(\varphi') \\ &= \mathcal{F}(f\ast \varphi') .\end{align*}

Actual proof:

\begin{align*} (f\ast \varphi)'(x) &= (\varphi\ast f)'(x) \\ &= \lim_{h\to 0} \frac{(\varphi\ast f)'(x+h) - (\varphi\ast f)'(x)}{h} \\ &= \lim_{h\to 0} \int \frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y) \\ &\overset{DCT}= \int \lim_{h\to 0} \frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y) \\ &= \int \varphi'(x-y) f(y) \\ &= (\varphi' \ast f)(x) \\ &= (f \ast \varphi')(x) .\end{align*}

To see that the DCT is justified, we can apply the MVT on the interval $$[0, h]$$ to $$f$$ to obtain

\begin{align*} \frac{\varphi(x + h - y) - \varphi(x - y)}{h} &= \varphi'(c) \quad c\in [0, h] ,\end{align*}

and since $$\varphi'$$ is continuous and compactly supported, $$\varphi'$$ is bounded by some $$M < \infty$$ by the extreme value theorem and thus \begin{align*} \int {\left\lvert {\frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y)} \right\rvert} &= \int {\left\lvert {\varphi'(c) f(y)} \right\rvert} \\ &\leq \int {\left\lvert {M} \right\rvert}{\left\lvert {f} \right\rvert} \\ &= {\left\lvert {M} \right\rvert} \int {\left\lvert {f} \right\rvert} < \infty ,\end{align*}

since $$f\in L^1$$ by assumption, so we can take $$g\coloneqq{\left\lvert {M} \right\rvert} {\left\lvert {f} \right\rvert}$$ as the dominating function.

Applying this theorem infinitely many times shows that $$f\ast \varphi$$ is smooth.

To see that $$f\ast \varphi$$ is compactly supported, approximate $$f$$ by a continuous compactly supported function $$h$$, so $${\left\lVert {h - f} \right\rVert}_1 \overset{L^1}\to 0$$.

Now let $$g_x(y) = \varphi(x-y)$$, and note that $$\mathrm{supp}(g) = x - \mathrm{supp}(\varphi)$$ which is still compact.

But since $$\mathrm{supp}(h)$$ is bounded, there is some $$N$$ such that \begin{align*} {\left\lvert {x} \right\rvert} > N \implies A_x\coloneqq\mathrm{supp}(h) \cap\mathrm{supp}(g_x) = \emptyset \end{align*}

and thus \begin{align*} (h\ast \varphi)(x) &= \int_{\mathbb{R}}\varphi(x-y) h(y)~dy \\ &= \int_{A_x} g_x(y) h(y) \\ &= 0 ,\end{align*}

so $$\left\{{x {~\mathrel{\Big\vert}~}f\ast g(x) = 0}\right\}$$ is open, and its complement is closed and bounded and thus compact.

Part b

\begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - f(x)} \right\rvert}~dx \\ &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - \int f(x) K_j(y) ~ dy} \right\rvert}~dx \\ &= \int {\left\lvert {\int ( f(x-y) - f(x) ) K_j(y) ~dy } \right\rvert} ~dx \\ &\leq \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} ~ dy~dx \\ &\overset{FT}= \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} \mathbf{~ dx~dy}\\ &= \int {\left\lvert {K_j(y)} \right\rvert} \left( \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} ~ dx\right) ~dy \\ &= \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy .\end{align*}

We now split the integral up into pieces.

• Chose $$\delta$$ small enough such that $${\left\lvert {y} \right\rvert} < \delta \implies {\left\lVert {f - \tau_y f} \right\rVert}_1 < \varepsilon$$ by continuity of translation in $$L^1$$, and

• Since $$\varphi$$ is compactly supported, choose $$J$$ large enough such that \begin{align*} j > J \implies \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} ~dy = \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {j\varphi(jy)} \right\rvert} = 0 \end{align*}

Then  \begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &\leq \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}1 ~dy \ &= \int{{\left\lvert {y} \right\rvert} < \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy

• \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}1 ~dy \ &= \varepsilon \int{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} + 0 \ &\leq \varepsilon(1) \to 0 .\end{align*} {=html}

## Spring 2017.5 #real_analysis/qual/work

Let $$f, g \in L^2({\mathbb{R}})$$. Prove that the formula \begin{align*} h(x) \coloneqq\int _{-\infty}^{\infty} f(t) g(x-t) \, dt \end{align*} defines a uniformly continuous function $$h$$ on $${\mathbb{R}}$$.

## Spring 2015.6 #real_analysis/qual/work

Let $$f \in L^1({\mathbb{R}})$$ and $$g$$ be a bounded measurable function on $${\mathbb{R}}$$.

• Show that the convolution $$f\ast g$$ is well-defined, bounded, and uniformly continuous on $${\mathbb{R}}$$.
• Prove that one further assumes that $$g \in C^1({\mathbb{R}})$$ with bounded derivative, then $$f\ast g \in C^1({\mathbb{R}})$$ and \begin{align*} \frac{d}{d x}(f * g)=f *\left(\frac{d}{d x} g\right) \end{align*}

## Fall 2014.5 #real_analysis/qual/work

• Let $$f \in C_c^0({\mathbb{R}}^n)$$, and show \begin{align*} \lim _{t \to 0} \int_{{\mathbb{R}}^n} |f(x+t) - f(x)| \, dx = 0 .\end{align*}

• Extend the above result to $$f\in L^1({\mathbb{R}}^n)$$ and show that \begin{align*} f\in L^1({\mathbb{R}}^n), \quad g\in L^\infty({\mathbb{R}}^n) \quad \implies f \ast g \text{ is bounded and uniformly continuous. } \end{align*}