[problem] Compute $\mathbb{1}_{[0,1]} * \mathbb{1}_{[0,1]}$ explicitly on $\mathbb{R}$. [/problem]

[solution] **Step 1: Set up the integral.** \begin{align*} (\mathbb{1}_{[0,1]} * \mathbb{1}_{[0,1]})(x) = \int_{\mathbb{R}} \mathbb{1}_{[0,1]}(x - y) \, \mathbb{1}_{[0,1]}(y) \, d\mathcal{L}^1(y) = \mathcal{L}^1\!\big([0,1] \cap [x-1, x]\big). \end{align*} **Step 2: Case analysis.** The intersection $[0,1] \cap [x-1, x]$ is nonempty when $x - 1 \leq 1$ and $x \geq 0$, i.e., $x \in [0, 2]$. For $0 \leq x \leq 1$: the intersection is $[0, x]$, which has length $x$. For $1 \leq x \leq 2$: the intersection is $[x-1, 1]$, which has length $2 - x$. **Step 3: Result.** \begin{align*} (\mathbb{1}_{[0,1]} * \mathbb{1}_{[0,1]})(x) = \begin{cases} x & \text{if } 0 \leq x \leq 1, \\ 2 - x & \text{if } 1 \leq x \leq 2, \\ 0 & \text{otherwise.} \end{cases} \end{align*} This is the **tent function** (or hat function), supported on $[0, 2] = [0,1] + [0,1]$, peaking at $x = 1$ with value $1$. Note that both inputs are discontinuous (indicator functions), but the output is continuous (and piecewise linear) — convolution has smoothed the jumps. It is not $C^1$, however: the corners at $x = 0, 1, 2$ reflect the fact that both inputs are only $L^\infty$, not $C^1$. Convolving again with $\mathbb{1}_{[0,1]}$ would produce a $C^1$ function (piecewise quadratic on $[0, 3]$), and iterating $k$ times gives a $C^{k-1}$ function. [/solution]

[problem] Prove that $L^1(\mathbb{R}^n)$ has no convolution identity: there is no $e \in L^1(\mathbb{R}^n)$ such that $f * e = f$ for all $f \in L^1(\mathbb{R}^n)$. [/problem]

[solution] **Step 1: Fourier transform of a convolution identity.** Suppose for contradiction that $e \in L^1(\mathbb{R}^n)$ satisfies $f * e = f$ for all $f \in L^1$. By the [Convolution Theorem](/theorems/250), taking Fourier transforms: $\hat{f}(\xi) \hat{e}(\xi) = \hat{f}(\xi)$ for all $\xi \in \mathbb{R}^n$ and all $f \in L^1$. **Step 2: Force $\hat{e} \equiv 1$.** For each $\xi_0 \in \mathbb{R}^n$, choose $f \in L^1$ with $\hat{f}(\xi_0) \neq 0$ (e.g., $f(x) = e^{-|x|^2}$, whose Fourier transform is a Gaussian, everywhere nonzero). Then $\hat{f}(\xi_0)(\hat{e}(\xi_0) - 1) = 0$ forces $\hat{e}(\xi_0) = 1$. Since $\xi_0$ was arbitrary, $\hat{e}(\xi) = 1$ for all $\xi$. **Step 3: Contradiction via Riemann-Lebesgue.** By the [Riemann-Lebesgue Lemma](/theorems/245), $\hat{e}(\xi) \to 0$ as $|\xi| \to \infty$ for any $e \in L^1(\mathbb{R}^n)$. But $\hat{e} \equiv 1$ does not decay. This is a contradiction. **Step 4: Interpretation.** The Dirac delta $\delta_0$ satisfies $f * \delta_0 = f$ for all $f$, and $\hat{\delta}_0 = 1$. But $\delta_0$ is a distribution, not an $L^1$ function. The approximate identities $\rho_\varepsilon$ approach $\delta_0$ as $\varepsilon \to 0$, providing "approximate" convolution identities in $L^1$ without an exact one. [/solution]

[problem] Let $f, g: \mathbb{R} \to \mathbb{R}$ with $\operatorname{supp}(f) = [-1, 1]$ and $\operatorname{supp}(g) = [0, 3]$. Without computing $f * g$ explicitly, determine the smallest interval that is guaranteed to contain $\operatorname{supp}(f * g)$. [/problem]

[solution] By the [Support of Convolution](/theorems/588), $\operatorname{supp}(f * g) \subseteq \overline{\operatorname{supp}(f) + \operatorname{supp}(g)}$. Since both supports are compact, the closure is unnecessary: \begin{align*} \operatorname{supp}(f * g) \subseteq [-1, 1] + [0, 3] = [-1, 4]. \end{align*} The Minkowski sum is $\{a + b : a \in [-1,1], \, b \in [0,3]\} = [\min(-1+0), \max(1+3)] = [-1, 4]$. This is the best possible bound without further information about $f$ and $g$: if $f$ and $g$ are both strictly positive on the interiors of their supports, then $f * g$ is strictly positive on the interior $(-1, 4)$ and the inclusion is an equality. [/solution]