[problem] **Problem 1.** Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \max(0, 1 - |x|)$ (the "hat function"). Compute the first and second distributional derivatives of $T_f$. [/problem]

[solution] **Step 1: Identify the structure of $f$.** The function $f$ is continuous, piecewise linear, with slope $+1$ on $(-1, 0)$, slope $-1$ on $(0, 1)$, and identically zero outside $[-1, 1]$. It has corners (non-differentiable points) at $x = -1$, $x = 0$, and $x = 1$, but no jump discontinuities. **Step 2: First distributional derivative.** Since $f$ is continuous and piecewise $C^1$, the distributional derivative equals the regular distribution generated by the piecewise classical derivative (no delta contributions from continuous functions). The classical derivative is \begin{align*} f'(x) &= \begin{cases} 1 & \text{if } -1 < x < 0, \\ -1 & \text{if } 0 < x < 1, \\ 0 & \text{otherwise}, \end{cases} \end{align*} which equals $\mathbb{1}_{(-1,0)}(x) - \mathbb{1}_{(0,1)}(x)$ almost everywhere. To verify: for any $\varphi \in \mathcal{D}(\mathbb{R})$, \begin{align*} (\partial T_f)(\varphi) &= -\int_{-\infty}^\infty f(x) \varphi'(x) \, d\mathcal{L}^1(x) = -\int_{-1}^0 (1+x)\varphi'(x) \, d\mathcal{L}^1(x) - \int_0^1 (1-x)\varphi'(x) \, d\mathcal{L}^1(x). \end{align*} Integrating by parts on each interval (using that $f(-1) = f(1) = 0$ and $f(0) = 1$): \begin{align*} -\int_{-1}^0 (1+x)\varphi' \, d\mathcal{L}^1 &= -(1+x)\varphi\Big|_{-1}^0 + \int_{-1}^0 \varphi \, d\mathcal{L}^1 = -\varphi(0) + \int_{-1}^0 \varphi \, d\mathcal{L}^1, \\ -\int_0^1 (1-x)\varphi' \, d\mathcal{L}^1 &= -(1-x)\varphi\Big|_0^1 + \int_0^1 (-1)\varphi \, d\mathcal{L}^1 = \varphi(0) - \int_0^1 \varphi \, d\mathcal{L}^1. \end{align*} Adding: $(\partial T_f)(\varphi) = \int_{-1}^0 \varphi \, d\mathcal{L}^1 - \int_0^1 \varphi \, d\mathcal{L}^1 = \int_{-\infty}^\infty (\mathbb{1}_{(-1,0)} - \mathbb{1}_{(0,1)}) \varphi \, d\mathcal{L}^1 = T_{f'}(\varphi)$. **Step 3: Second distributional derivative.** The function $f' = \mathbb{1}_{(-1,0)} - \mathbb{1}_{(0,1)}$ is piecewise constant with jump discontinuities: it jumps from $0$ to $1$ at $x = -1$, from $1$ to $-1$ at $x = 0$, and from $-1$ to $0$ at $x = 1$. By the jump formula for piecewise $C^1$ functions, $\partial T_{f'} = T_{(f')'} + [f']_{-1}\delta_{-1} + [f']_0\delta_0 + [f']_1\delta_1$. Since $(f')' = 0$ almost everywhere (the function is piecewise constant), and the jumps are $[f']_{-1} = 1$, $[f']_0 = -2$, $[f']_1 = 1$: \begin{align*} \partial^2 T_f &= \delta_{-1} - 2\delta_0 + \delta_1. \end{align*} The second derivative of the hat function is a discrete measure: a positive delta at each end and a negative delta of double weight at the peak. This is the distributional analogue of the second difference operator $(\delta_{-1} - 2\delta_0 + \delta_1)(\varphi) = \varphi(-1) - 2\varphi(0) + \varphi(1)$, which approximates $\varphi''(0)$ — consistent with the hat function being the simplest piecewise linear finite element basis function. [/solution]

[problem] **Problem 2.** Let $\Omega = (0, 1) \subseteq \mathbb{R}$ and define the distribution $T \in \mathcal{D}'(\Omega)$ by \begin{align*} T: \mathcal{D}(\Omega) &\to \mathbb{R} \\ \varphi &\mapsto \sum_{k=1}^\infty \varphi'(1/k). \end{align*} 1. Verify that $T$ is a well-defined distribution. 2. Show that $T$ is singular (not a regular distribution). 3. Determine $\mathrm{supp}(T)$. [/problem]

[solution] **Step 1: Well-definedness.** Let $\varphi \in \mathcal{D}(\Omega)$. Since $\mathrm{supp}(\varphi) \subset (0,1)$ is compact, there exists $\varepsilon > 0$ such that $\mathrm{supp}(\varphi) \subseteq [\varepsilon, 1 - \varepsilon]$. For all $k$ with $1/k < \varepsilon$ (i.e. $k > 1/\varepsilon$), the point $1/k$ lies outside $\mathrm{supp}(\varphi)$, so $\varphi'(1/k) = 0$. Hence the sum is finite — it has at most $\lfloor 1/\varepsilon \rfloor$ nonzero terms — and $T(\varphi)$ is a well-defined real number. Linearity is immediate from linearity of differentiation and finite sums. For continuity, suppose $\varphi_j \to 0$ in $\mathcal{D}(\Omega)$ with all supports in a compact set $K \subset (0, 1)$. Then $K \subseteq [\varepsilon, 1 - \varepsilon]$ for some $\varepsilon > 0$, and the sum $T(\varphi_j) = \sum_{k: 1/k \in K} \varphi_j'(1/k)$ has a fixed finite number of terms (independent of $j$). Since $\varphi_j' \to 0$ uniformly, each term $\varphi_j'(1/k) \to 0$, so $T(\varphi_j) \to 0$. **Step 2: Singularity.** Suppose for contradiction that $T = T_f$ for some $f \in L^1_\mathrm{loc}(\Omega)$, so that $\int_0^1 f \varphi \, d\mathcal{L}^1 = \sum_k \varphi'(1/k)$ for all $\varphi \in \mathcal{D}(\Omega)$. Choose $\varphi \in \mathcal{D}(\Omega)$ supported in $(1/3, 1/2)$ with $\varphi' \not\equiv 0$. The only $k$ with $1/k \in (1/3, 1/2)$ are $k = 3$ (giving $1/3$, on the boundary) — but $\mathrm{supp}(\varphi) \subset (1/3, 1/2)$ is open, so $\varphi(1/3) = \varphi'(1/3) = 0$, and $T(\varphi) = 0$. Thus $\int_{1/3}^{1/2} f \varphi \, d\mathcal{L}^1 = 0$ for all such $\varphi$, which forces $f = 0$ a.e. on $(1/3, 1/2)$. By the same argument applied to intervals $(1/(k+1), 1/k)$ for each $k \geq 2$, we get $f = 0$ a.e. on each such interval, hence $f = 0$ a.e. on $(0, 1)$. But then $T = T_0 = 0$, which is false since $T(\varphi) = \varphi'(1)$ for any $\varphi$ with support near $x = 1$ and away from all other $1/k$. This contradicts $T = T_f$, so $T$ is singular. **Step 3: Support.** From Step 1, $T(\varphi) = 0$ whenever $\mathrm{supp}(\varphi)$ is disjoint from the set $\{1/k : k \in \mathbb{N}\}$. The closure of this set in $(0,1)$ is $\{0\} \cup \{1/k : k \in \mathbb{N}\}$, but $0 \notin \Omega = (0, 1)$, so the closure within $\Omega$ is $\{1/k : k \in \mathbb{N}\}$ (which is closed in $(0,1)$ since its only accumulation point, $0$, is outside $\Omega$). If $x_0 \notin \{1/k : k \in \mathbb{N}\}$ and $x_0 \in (0, 1)$, then $x_0$ has a neighbourhood $U \subseteq (0,1)$ disjoint from $\{1/k\}$, so $T(\varphi) = 0$ for all $\varphi \in \mathcal{D}(U)$. Conversely, for each $1/k$, there exist test functions supported near $1/k$ with $\varphi'(1/k) \neq 0$. Therefore $\mathrm{supp}(T) = \{1/k : k \in \mathbb{N}\}$. [/solution]