[problem] **Problem 1.** Let $f \in L^1_{\mathrm{loc}}(\mathbb{R})$ and suppose $T_f = 0$ in $\mathcal{D}'(\mathbb{R})$ on the open interval $(-1, 1)$ — meaning $\int_{-1}^1 f\varphi \, d\mathcal{L}^1 = 0$ for every $\varphi \in \mathcal{D}((-1,1))$. Must $f = 0$ a.e. on $(-1, 1)$? [/problem]

[solution] Yes. The [injectivity of the canonical embedding](/theorems/450) applied to $\Omega = (-1, 1)$ gives: if $T_f = 0$ in $\mathcal{D}'((-1,1))$, then $f = 0$ $\mathcal{L}^1$-a.e. on $(-1, 1)$. The point is that the injectivity theorem holds for *any* non-empty open set $\Omega$, so it can be localised. Even if $f$ is nonzero outside $(-1,1)$, the vanishing of $T_f$ against test functions supported in $(-1,1)$ forces $f$ to vanish a.e. *on that interval*. [/solution]

[problem] **Problem 2.** Let $f \in L^2(\mathbb{R}^n)$. Show that the regular distribution $T_f$ is a tempered distribution, and that the distributional Fourier transform $\widehat{T_f}$ (defined by duality on $\mathcal{S}'(\mathbb{R}^n)$) agrees with the $L^2$ Fourier transform: $\widehat{T_f} = T_{\hat{f}}$, where $\hat{f}$ is the [Plancherel](/theorems/247) extension of the Fourier transform to $L^2$. [/problem]

[solution] **Step 1: $T_f \in \mathcal{S}'(\mathbb{R}^n)$.** Since $f \in L^2(\mathbb{R}^n) \subset L^1_{\mathrm{loc}}(\mathbb{R}^n)$, the distribution $T_f \in \mathcal{D}'(\mathbb{R}^n)$ is well-defined. To show $T_f \in \mathcal{S}'(\mathbb{R}^n)$, we verify the [semi-norm bound](/theorems/456). By Cauchy–Schwarz: $|T_f(\varphi)| = |\int f\varphi \, d\mathcal{L}^n| \le \|f\|_{L^2}\|\varphi\|_{L^2}$. Since $\varphi \in \mathcal{S}(\mathbb{R}^n)$, we have $|\varphi(x)| \le \|\varphi\|_{N,0}(1+|x|)^{-N}$ for any $N > n/2$, and $(1+|x|)^{-N} \in L^2$ when $N > n/2$, giving $\|\varphi\|_{L^2} \le C_N\|\varphi\|_{N,0}$. Thus $|T_f(\varphi)| \le C\|\varphi\|_{N,0}$, confirming $T_f \in \mathcal{S}'$. **Step 2: Compatibility of the two Fourier transforms.** The distributional Fourier transform is defined by $\widehat{T_f}(\varphi) := T_f(\hat{\varphi}) = \int_{\mathbb{R}^n} f(x)\hat{\varphi}(x) \, d\mathcal{L}^n(x)$ for $\varphi \in \mathcal{S}(\mathbb{R}^n)$. We must show this equals $T_{\hat{f}}(\varphi) = \int_{\mathbb{R}^n} \hat{f}(\xi)\varphi(\xi) \, d\mathcal{L}^n(\xi)$. For $f \in L^1 \cap L^2$, both the classical and $L^2$ Fourier transforms agree, and Fubini's theorem gives $\int f(x)\hat{\varphi}(x) \, d\mathcal{L}^n(x) = \int \hat{f}(\xi)\varphi(\xi) \, d\mathcal{L}^n(\xi)$ directly. For general $f \in L^2$, approximate by $f_k := f \cdot \mathbb{1}_{B(0,k)} \in L^1 \cap L^2$. Then $f_k \to f$ in $L^2$, so $\hat{f}_k \to \hat{f}$ in $L^2$ by [Plancherel](/theorems/247). For each $\varphi \in \mathcal{S}$: \begin{align*} \widehat{T_f}(\varphi) &= \int f\hat{\varphi} \, d\mathcal{L}^n = \lim_{k \to \infty} \int f_k \hat{\varphi} \, d\mathcal{L}^n = \lim_{k \to \infty} \int \hat{f}_k \varphi \, d\mathcal{L}^n = \int \hat{f}\varphi \, d\mathcal{L}^n = T_{\hat{f}}(\varphi), \end{align*} where the first limit uses $f_k \to f$ in $L^2$ and $\hat{\varphi} \in \mathcal{S} \subset L^2$, the middle equality uses Fubini for $f_k \in L^1 \cap L^2$, and the last limit uses $\hat{f}_k \to \hat{f}$ in $L^2$ and $\varphi \in L^2$. [/solution]

[problem] **Problem 3.** Give an example of a distribution $u \in \mathcal{D}'(\mathbb{R})$ such that $u$ is singular but the distributional derivative $u' \in \mathcal{D}'(\mathbb{R})$ is regular. [/problem]

[solution] Let $u = \delta_0$, the Dirac delta at the origin. This is singular (proved in the example above). Its distributional derivative $\delta_0'$ is defined by $\delta_0'(\varphi) = -\varphi'(0)$. This is also singular (it has order $1$, but more directly: no $L^1_{\mathrm{loc}}$ function can reproduce $-\varphi'(0)$ by integration). So this does not work. Instead, consider $u = T_H$ where $H(x) = \mathbb{1}_{[0,\infty)}(x)$ is the Heaviside function. Then $u$ is a regular distribution. Its distributional derivative $u' = \delta_0$ is singular. This gives the reverse — a regular distribution with singular derivative. The problem asks for the other direction. Take $u(x) = \log|x|$ on $\mathbb{R}$. This is locally integrable (the singularity $\log|x|$ is integrable near $0$ in one dimension), so $T_{\log|\cdot|} \in \mathcal{D}'(\mathbb{R})$ is regular. Its distributional derivative is $T_{\log|\cdot|}'(\varphi) = -\int \log|x| \, \varphi'(x) \, d\mathcal{L}^1(x) = \int \frac{\varphi(x)}{x} \, d\mathcal{L}^1(x)$ (integrating by parts, using $(\log|x|)' = 1/x$ for $x \ne 0$), but $1/x$ is not locally integrable near $0$. The derivative is the **principal value distribution** $\mathrm{p.v.}(1/x)$, defined by $\mathrm{p.v.}(1/x)(\varphi) = \lim_{\varepsilon \to 0^+}\int_{|x|>\varepsilon} \varphi(x)/x \, d\mathcal{L}^1(x)$. This is singular: $1/x \notin L^1_{\mathrm{loc}}(\mathbb{R})$. So actually the problem as stated asks for the *reverse*: $u$ singular, $u'$ regular. Consider $u = \mathrm{sgn}(x) = 2H(x) - 1$. Then $u$ is regular (it is bounded, hence locally integrable). And $u' = 2\delta_0$, which is singular. Still wrong direction. Consider instead: $u = \delta_0 * H$ does not help either. Let us construct directly. Define $u \in \mathcal{D}'(\mathbb{R})$ by $u = \sum_{k=1}^\infty 2^{-k} \delta_{1/k}$. This is a distribution (the sum converges in $\mathcal{D}'$ since any compact set contains only finitely many of the points $1/k$). It is singular — it is a countable sum of point masses, which is a measure that is not absolutely continuous with respect to $\mathcal{L}^1$. Its distributional derivative is $u' = \sum_{k=1}^\infty 2^{-k}\delta_{1/k}'$, which is also singular. A simpler approach: the problem has no solution in the stated direction. If $u$ is singular and $u'$ is regular, say $u' = T_g$ for some $g \in L^1_{\mathrm{loc}}(\mathbb{R})$, then (at least on any bounded interval) $u = T_G + c\delta_0 + \ldots$ where $G$ is an antiderivative of $g$ — but this needs care. In fact, if $u' = T_g$ with $g \in L^1_{\mathrm{loc}}$, then on any bounded interval $u$ equals $T_G$ for some antiderivative $G(x) = \int_a^x g(t) \, dt + c$, which is absolutely continuous, hence locally integrable, so $u$ is regular on every bounded interval, and therefore regular globally. This is a contradiction: **if $u'$ is regular, then $u$ is regular.** So no such example exists. The correct answer is: **no such $u$ exists.** If the distributional derivative $u'$ is a regular distribution $T_g$ with $g \in L^1_{\mathrm{loc}}(\mathbb{R})$, then on any interval $(a, b)$, the function $G(x) := \int_a^x g(t) \, dt$ satisfies $T_G' = T_g = u'$ in $\mathcal{D}'((a,b))$. Since $(u - T_G)' = 0$ on $(a,b)$, the distribution $u - T_G$ is a constant $c_{a,b}$ on $(a,b)$ (distributions with zero derivative on a connected open set are constants). Thus $u = T_{G + c_{a,b}}$ on $(a,b)$, meaning $u$ is regular on every bounded interval, and therefore on all of $\mathbb{R}$. So: a distribution on $\mathbb{R}$ whose derivative is regular is itself regular. [/solution]