All four parts are proved by working in Fourier space. The [Fourier transform](/page/Fourier%20Transform) of $\phi_\varepsilon$ is $\hat\phi_\varepsilon(\xi) = \hat\phi(\varepsilon\xi)$, and the key properties are: $\hat\phi(0) = \int_{\mathbb{R}^n} \phi\,d\mathcal{L}^n = 1$; $|\hat\phi(\varepsilon\xi)| \le \|\phi\|_{L^1} = 1$ for all $\xi$; and $\hat\phi$ is smooth with all [derivatives](/page/Derivative) bounded (since $\phi \in C^\infty_c$). **Step 1: Proof of (i).** [claim:Mollifier Approximation Rate] $\|J_\varepsilon f - f\|_{H^r} \le C\,\varepsilon^{s-r}\,\|f\|_{H^s}$ for all $f \in H^s(\mathbb{R}^n)$ and $0 \le r \le s$ with $s - r \le 1$. [/claim] [proof] Since $\widehat{J_\varepsilon f}(\xi) = \hat\phi(\varepsilon\xi)\,\hat{f}(\xi)$, the [Parseval Identity](/theorems/248) gives \begin{align*} \|J_\varepsilon f - f\|_{H^r}^2 &= \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} |\hat\phi(\varepsilon\xi) - 1|^2\,\langle\xi\rangle^{2r}\,|\hat{f}(\xi)|^2\,d\mathcal{L}^n(\xi). \end{align*} Write $\theta := s - r \in [0, 1]$. We bound the factor $|\hat\phi(\varepsilon\xi) - 1|$ using two estimates: (a) the [mean value theorem](/theorems/186) gives $|\hat\phi(\varepsilon\xi) - 1| \le C_1\,\varepsilon\,|\xi|$, since $\hat\phi(0) = 1$ and $\hat\phi$ is $C^1$ (as $\phi \in C^\infty_c$ implies $\hat\phi \in \mathcal{S}$ by the [Fourier Transform As Automorphism Of Schwartz Space](/theorems/228)), with $C_1 := \sup_{\eta \in \mathbb{R}^n} |\nabla\hat\phi(\eta)| < \infty$; (b) the triangle inequality gives $|\hat\phi(\varepsilon\xi) - 1| \le 2$. Interpolating: for any $a, b \ge 0$ and $\theta \in [0,1]$, $\min(a, b) \le a^\theta\,b^{1-\theta}$. Applying this with $a = C_1\varepsilon|\xi|$ and $b = 2$: \begin{align*} |\hat\phi(\varepsilon\xi) - 1| &\le (C_1\varepsilon|\xi|)^\theta \cdot 2^{1-\theta} = C\,\varepsilon^\theta\,|\xi|^\theta \le C\,\varepsilon^\theta\,\langle\xi\rangle^\theta, \end{align*} where $C := C_1^\theta \cdot 2^{1-\theta}$. Therefore: \begin{align*} \|J_\varepsilon f - f\|_{H^r}^2 &\le \frac{C^2\,\varepsilon^{2\theta}}{(2\pi)^n}\int_{\mathbb{R}^n} \langle\xi\rangle^{2\theta}\,\langle\xi\rangle^{2r}\,|\hat{f}(\xi)|^2\,d\mathcal{L}^n(\xi) = C^2\,\varepsilon^{2(s-r)}\,\|f\|_{H^s}^2. \end{align*} [/proof] **Step 2: Proof of (ii).** [claim:Mollifier Derivative Cost] $\|J_\varepsilon f\|_{H^k} \le C_k\,\varepsilon^{-k}\,\|f\|_{L^2}$ for all $f \in L^2(\mathbb{R}^n)$, $k \ge 0$, and $\varepsilon \in (0,1]$. [/claim] [proof] By the [Parseval Identity](/theorems/248): \begin{align*} \|J_\varepsilon f\|_{H^k}^2 &= \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \langle\xi\rangle^{2k}\,|\hat\phi(\varepsilon\xi)|^2\,|\hat{f}(\xi)|^2\,d\mathcal{L}^n(\xi). \end{align*} Since $\phi \in C^\infty_c$, the function $\hat\phi$ is a Schwartz function by the [Fourier Transform As Automorphism Of Schwartz Space](/theorems/228), so $|\hat\phi(\eta)| \le C_m\,\langle\eta\rangle^{-m}$ for any $m \ge 0$. Taking $m = k$: $|\hat\phi(\varepsilon\xi)| \le C_k\,\langle\varepsilon\xi\rangle^{-k}$. For $\varepsilon \in (0,1]$, the inequality $1 + \varepsilon^2|\xi|^2 \ge \varepsilon^2(1 + |\xi|^2)$ holds (since $1 \ge \varepsilon^2$), giving $\langle\varepsilon\xi\rangle \ge \varepsilon\,\langle\xi\rangle$. Therefore: \begin{align*} \langle\xi\rangle^k\,|\hat\phi(\varepsilon\xi)| &\le C_k\,\frac{\langle\xi\rangle^k}{\langle\varepsilon\xi\rangle^k} \le C_k\,\frac{\langle\xi\rangle^k}{\varepsilon^k\,\langle\xi\rangle^k} = C_k\,\varepsilon^{-k}. \end{align*} Substituting: \begin{align*} \|J_\varepsilon f\|_{H^k}^2 &\le \frac{C_k^2\,\varepsilon^{-2k}}{(2\pi)^n}\int_{\mathbb{R}^n} |\hat{f}(\xi)|^2\,d\mathcal{L}^n(\xi) = C_k^2\,\varepsilon^{-2k}\,\|f\|_{L^2}^2. \end{align*} [/proof] **Step 3: Proof of (iii).** [claim:Self Adjointness Of [Mollification](/page/Standard%20Mollifier)] $\int_{\mathbb{R}^n} (J_\varepsilon f)\,g\,d\mathcal{L}^n = \int_{\mathbb{R}^n} f\,(J_\varepsilon g)\,d\mathcal{L}^n$ for all $f, g \in L^2(\mathbb{R}^n)$. [/claim] [proof] By definition, $(J_\varepsilon f)(x) = \int_{\mathbb{R}^n} \phi_\varepsilon(x - y)\,f(y)\,d\mathcal{L}^n(y)$. Therefore: \begin{align*} \int_{\mathbb{R}^n} (J_\varepsilon f)(x)\,g(x)\,d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \phi_\varepsilon(x - y)\,f(y)\,g(x)\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(x). \end{align*} The integrand is in $L^1(\mathbb{R}^n \times \mathbb{R}^n)$ since $\phi_\varepsilon \in L^1$, $f \in L^2$, and $g \in L^2$ (by [Young's convolution inequality](/theorems/463), $J_\varepsilon f \in L^2$ with $\|J_\varepsilon f\|_{L^2} \le \|f\|_{L^2}$, so the product $(J_\varepsilon f)\,g \in L^1$ by the [Cauchy-Schwarz Inequality](/theorems/432)). Fubini's theorem gives: \begin{align*} &= \int_{\mathbb{R}^n} f(y)\Bigl(\int_{\mathbb{R}^n} \phi_\varepsilon(x - y)\,g(x)\,d\mathcal{L}^n(x)\Bigr)\,d\mathcal{L}^n(y). \end{align*} Since $\phi(-x) = \phi(x)$ by hypothesis, we have $\phi_\varepsilon(x - y) = \phi_\varepsilon(y - x)$, so the inner [integral](/page/Integral) equals $(J_\varepsilon g)(y)$. Hence the expression equals $\int_{\mathbb{R}^n} f(y)\,(J_\varepsilon g)(y)\,d\mathcal{L}^n(y)$. [/proof] **Step 4: Proof of (iv).** [claim:Mollification Commutes With Derivatives] $J_\varepsilon(\partial^\alpha f) = \partial^\alpha(J_\varepsilon f)$ for all multi-indices $\alpha$ and $f \in H^{|\alpha|}(\mathbb{R}^n)$. [/claim] [proof] On the Fourier side, $\widehat{J_\varepsilon(\partial^\alpha f)}(\xi) = \hat\phi(\varepsilon\xi)\,(i\xi)^\alpha\,\hat{f}(\xi)$ by the [Fourier Differentiation Rule](/theorems/249) and the [convolution](/page/Convolution) property. Similarly, $\widehat{\partial^\alpha(J_\varepsilon f)}(\xi) = (i\xi)^\alpha\,\hat\phi(\varepsilon\xi)\,\hat{f}(\xi)$. Since pointwise multiplication of [functions](/page/Function) commutes, both sides have the same Fourier transform, so they are equal in $L^2$ by the injectivity of the Fourier transform (from the [Plancherel Identity](/theorems/529)). [/proof]