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.

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.

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.

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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}