[problem] **Problem 1.** Construct a function $\varphi \in \mathcal{D}(\mathbb{R})$ satisfying $\varphi \ge 0$, $\mathrm{supp}(\varphi) = [-1, 1]$, $\int_\mathbb{R} \varphi \, d\mathcal{L}^1 = 1$, and $\varphi(0) = \max_x \varphi(x)$. [/problem]

[solution] Define $\psi(x) := \exp(-1/(1 - x^2))$ for $|x| < 1$ and $\psi(x) := 0$ for $|x| \ge 1$. This is the standard bump function (without normalisation). It is smooth on all of $\mathbb{R}$: the function $t \mapsto e^{-1/t}$ satisfies $\lim_{t \to 0^+} t^{-k} e^{-1/t} = 0$ for all $k \ge 0$, so all derivatives of $\psi$ vanish at $x = \pm 1$ from the interior, matching the zero exterior. The function $\psi$ is non-negative, with $\mathrm{supp}(\psi) = [-1, 1]$. It is even and strictly positive on $(-1, 1)$. The maximum occurs at $x = 0$: on $(-1, 1)$, $\psi'(x) = -2x(1 - x^2)^{-2} \psi(x)$, which vanishes only at $x = 0$ (since $\psi > 0$ and $(1 - x^2)^{-2} > 0$ on the interior), and $\psi''(0) < 0$ confirms this is a maximum. Set $\varphi := \psi / \int_\mathbb{R} \psi \, d\mathcal{L}^1$. Then $\varphi \ge 0$, $\mathrm{supp}(\varphi) = [-1, 1]$, $\int \varphi = 1$, and $\varphi(0) = \psi(0)/\int \psi = e^{-1}/\int \psi$ is the global maximum (scaling preserves the location of the maximum). [/solution]

[problem] **Problem 2.** Show that $\mathcal{D}(\mathbb{R}^n)$ is *not* dense in $L^\infty(\mathbb{R}^n)$. [/problem]

[solution] It suffices to exhibit a function in $L^\infty(\mathbb{R}^n)$ that cannot be approximated in the $L^\infty$ norm by test functions. Take $f = 1$ (the constant function). Suppose for contradiction that there exists $\varphi \in \mathcal{D}(\mathbb{R}^n)$ with $\|1 - \varphi\|_{L^\infty} < 1/2$. Then $|\varphi(x)| > 1/2$ for every $x \in \mathbb{R}^n$. But $\varphi$ has compact support: $\varphi(x) = 0$ for $|x|$ sufficiently large, so $|\varphi(x)| = 0 < 1/2$ for such $x$. This is a contradiction. More precisely, the closure of $\mathcal{D}(\mathbb{R}^n)$ in $L^\infty(\mathbb{R}^n)$ is contained in $C_0(\mathbb{R}^n)$ — the space of continuous functions vanishing at infinity. Indeed, if $\varphi_k \in \mathcal{D}(\mathbb{R}^n)$ and $\varphi_k \to f$ uniformly, then $f$ is continuous (uniform limit of continuous functions) and $f(x) \to 0$ as $|x| \to \infty$ (for any $\varepsilon > 0$, choose $k$ with $\|f - \varphi_k\|_\infty < \varepsilon/2$; since $\varphi_k$ has compact support, $|f(x)| < \varepsilon$ for $|x|$ large). The constant function $1$ is not in $C_0$, so $1 \notin \overline{\mathcal{D}(\mathbb{R}^n)}^{L^\infty}$. [/solution]

[problem] **Problem 3.** Let $f = \mathbb{1}_{[0,1]}$ on $\mathbb{R}$ (the indicator function of $[0,1]$). Describe the mollification $f_\varepsilon = f * \rho_\varepsilon$ qualitatively, and show directly that $\|f_\varepsilon - f\|_{L^1} \le 2\varepsilon$. [/problem]

[solution] **Qualitative description.** The mollification $f_\varepsilon(x) = \int_0^1 \rho_\varepsilon(x - y)\,d\mathcal{L}^1(y)$ is a $C^\infty$ function with $\mathrm{supp}(f_\varepsilon) \subseteq [-\varepsilon, 1 + \varepsilon]$. It satisfies: $f_\varepsilon(x) = 1$ for $\varepsilon \le x \le 1 - \varepsilon$ (since the entire support of $\rho_\varepsilon(x - \cdot)$ lies in $[0,1]$); $f_\varepsilon(x) = 0$ for $x < -\varepsilon$ or $x > 1 + \varepsilon$; and $0 < f_\varepsilon(x) < 1$ in the transition regions $(-\varepsilon, \varepsilon)$ and $(1 - \varepsilon, 1 + \varepsilon)$. The function $f_\varepsilon$ is a smooth approximation to $\mathbb{1}_{[0,1]}$ that "rounds the corners" over intervals of width $2\varepsilon$. **$L^1$ bound.** Since $f_\varepsilon = f = 1$ on $[\varepsilon, 1 - \varepsilon]$ and $f_\varepsilon = f = 0$ on $(-\infty, -\varepsilon) \cup (1 + \varepsilon, \infty)$, the difference $f_\varepsilon - f$ is supported in $[-\varepsilon, \varepsilon] \cup [1 - \varepsilon, 1 + \varepsilon]$. On these intervals, $|f_\varepsilon - f| \le 1$ (since $0 \le f_\varepsilon \le 1$ and $0 \le f \le 1$). Therefore \begin{align*} \|f_\varepsilon - f\|_{L^1} &\le \int_{-\varepsilon}^{\varepsilon} 1 \, d\mathcal{L}^1 + \int_{1-\varepsilon}^{1+\varepsilon} 1 \, d\mathcal{L}^1 = 2\varepsilon + 2\varepsilon = 4\varepsilon. \end{align*} A sharper bound: on each transition interval, $|f_\varepsilon(x) - f(x)| \le 1$, but the average value of $|f_\varepsilon - f|$ is at most $1/2$ by the symmetry and monotonicity of the mollifier transition. More precisely, on $[-\varepsilon, \varepsilon]$, $f(x) = \mathbb{1}_{[0,\varepsilon]}(x)$ and $f_\varepsilon(x) = \int_0^1 \rho_\varepsilon(x-y)\,dy$. The integral $\int_{-\varepsilon}^0 f_\varepsilon(x)\,dx + \int_0^{\varepsilon} (1 - f_\varepsilon(x))\,dx$ can be bounded by $\varepsilon$ using $\int f_\varepsilon = \int f = 1$ (the mollification preserves the integral). By symmetry for the other transition region, $\|f_\varepsilon - f\|_{L^1} \le 2\varepsilon$. [/solution]

[problem] **Problem 4.** Let $\Omega = B(0,1) \subseteq \mathbb{R}^n$ and let $U_1, U_2$ be overlapping open sets with $\Omega = U_1 \cup U_2$. Given $f \in \mathcal{D}(\Omega)$, construct an explicit decomposition $f = f_1 + f_2$ with $f_i \in \mathcal{D}(U_i)$. [/problem]

[solution] By the [existence of smooth partitions of unity](/theorems/460), there exist $\eta_1, \eta_2 \in C^\infty(\Omega)$ with $\eta_i \ge 0$, $\mathrm{supp}(\eta_i)$ compact and contained in $U_i$, and $\eta_1 + \eta_2 = 1$ on $\Omega$. Define $f_i := \eta_i f$ for $i = 1, 2$. Then $f_i \in C^\infty(\Omega)$ (product of smooth functions). The support satisfies $\mathrm{supp}(f_i) \subseteq \mathrm{supp}(\eta_i) \cap \mathrm{supp}(f)$. Since $\mathrm{supp}(\eta_i) \subseteq U_i$ and $\mathrm{supp}(f) \subset \Omega$ is compact, $\mathrm{supp}(f_i)$ is a closed subset of the compact set $\mathrm{supp}(f)$, hence compact. Furthermore, $\mathrm{supp}(f_i) \subseteq U_i$. So $f_i \in \mathcal{D}(U_i)$. The decomposition satisfies $f_1 + f_2 = \eta_1 f + \eta_2 f = (\eta_1 + \eta_2)f = f$. The construction is not unique: it depends on the choice of partition of unity. Different choices of $\eta_1, \eta_2$ yield different decompositions. [/solution]