[problem: Injectivity Of The Fourier Transform On Tempered Distributions | 1] Let $u \in \mathcal{S}'(\mathbb{R}^n)$ and suppose $\hat{u} = 0$ in $\mathcal{S}'(\mathbb{R}^n)$. Show that $u = 0$. [/problem]

[solution] **Step 1: Translate the hypothesis.** The condition $\hat{u} = 0$ means that $\hat{u}(\phi) = 0$ for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. By the definition of the distributional Fourier transform, $\hat{u}(\phi) = u(\hat{\phi})$, so the hypothesis says $u(\hat{\phi}) = 0$ for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. **Step 2: Use surjectivity.** The Fourier transform $\mathcal{F}: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ is a bijection by the [topological automorphism theorem](/theorems/228). In particular, $\mathcal{F}$ is surjective: every $\psi \in \mathcal{S}(\mathbb{R}^n)$ can be written as $\psi = \hat{\phi}$ for some $\phi \in \mathcal{S}(\mathbb{R}^n)$. Therefore $u(\psi) = u(\hat{\phi}) = 0$ for every $\psi \in \mathcal{S}(\mathbb{R}^n)$. **Step 3: Conclude.** Since $u(\psi) = 0$ for every $\psi \in \mathcal{S}(\mathbb{R}^n)$, we have $u = 0$ in $\mathcal{S}'(\mathbb{R}^n)$ by definition of the zero distribution. [/solution]

[problem: Failure Of Temperedness For Gaussian Growth | 2] Show that the distribution $T_g$ with $g(x) = e^{x^2}$ on $\mathbb{R}$ is *not* a tempered distribution. That is, $T_g \in \mathcal{D}'(\mathbb{R}) \setminus \mathcal{S}'(\mathbb{R})$. [/problem]

[solution] **Step 1: Verify $T_g \in \mathcal{D}'(\mathbb{R})$.** For any $\phi \in \mathcal{D}(\mathbb{R})$ supported in $[-R, R]$, we have $|T_g(\phi)| \le e^{R^2}\|\phi\|_{L^1} \le 2R\,e^{R^2}\sup|\phi|$. This is a bound of the form $C_K \sup_K |\phi|$ for the compact set $K = [-R, R]$, which establishes that $T_g$ is a distribution of order $0$ on every compact set. **Step 2: Construct a test sequence.** Suppose for contradiction that $T_g$ satisfies the [semi-norm bound](/theorems/456): $|T_g(\phi)| \le C\sum_{j \le N, k \le M}\|\phi\|_{j,k}$ for some $C, N, M$ and all $\phi \in \mathcal{S}(\mathbb{R})$. Fix a non-negative bump $\eta \in C_c^\infty(\mathbb{R})$ with $\eta(0) = 1$ and $\mathrm{supp}(\eta) \subseteq [-1, 1]$. For $R > 0$, define $\phi_R(x) := \eta(x - R)$. Then $\phi_R \in \mathcal{D}(\mathbb{R}) \subseteq \mathcal{S}(\mathbb{R})$, $\phi_R$ is supported in $[R-1, R+1]$, and $\phi_R(R) = 1$. **Step 3: Estimate the semi-norms.** The semi-norms of the translated bump satisfy $\|\phi_R\|_{j,k} = \sup_x |x^j \partial^k\phi_R(x)| \le (R+1)^j \sup_x |\eta^{(k)}(x-R)| = (R+1)^j \|\eta^{(k)}\|_\infty$. Therefore the right-hand side of the semi-norm bound grows at most polynomially in $R$: there exists $C' > 0$ such that $C\sum_{j,k}\|\phi_R\|_{j,k} \le C'(R+1)^N$. **Step 4: Derive the contradiction.** On the other hand, since $e^{x^2} \ge e^{(R-1/2)^2}$ on $[R-1/2, R+1/2]$ and $\eta \ge \min_{|t| \le 1/2}\eta(t) > 0$ on this interval (by continuity and $\eta(0) = 1$), \begin{align*} T_g(\phi_R) &= \int e^{x^2}\eta(x-R)\,d\mathcal{L}^1(x) \ge c\, e^{(R-1/2)^2} \end{align*} for some constant $c > 0$ independent of $R$. This grows faster than any polynomial in $R$, contradicting the polynomial bound $C'(R+1)^N$ for $R$ sufficiently large. [/solution]

[problem: Fourier Transform Of The Heaviside Function | 3] Compute the Fourier transform in $\mathcal{S}'(\mathbb{R})$ of the Heaviside function $H(x) = \mathbb{1}_{[0,\infty)}(x)$. [/problem]

[solution] **Step 1: Decompose.** Write $H = \frac{1}{2} + \frac{1}{2}\mathrm{sgn}$, where $\mathrm{sgn}(x) = \mathbb{1}_{(0,\infty)}(x) - \mathbb{1}_{(-\infty,0)}(x)$. Both $H$ and $\mathrm{sgn}$ are bounded, hence tempered. By linearity of the Fourier transform, $\hat{H} = \frac{1}{2}\hat{1} + \frac{1}{2}\widehat{\mathrm{sgn}}$. From the examples on the page, $\hat{1} = 2\pi\delta_0$ (in dimension $n = 1$). **Step 2: Compute $\widehat{\mathrm{sgn}}$ via the distributional derivative.** The distributional derivative of $\mathrm{sgn}$ is $\mathrm{sgn}'(x) = 2\delta_0$ (the sign function has a jump of size $2$ at the origin). Taking the Fourier transform of both sides and applying the exchange identity $\widehat{u'} = i\xi\hat{u}$: \begin{align*} i\xi \cdot \widehat{\mathrm{sgn}}(\xi) &= 2\hat{\delta}_0 = 2. \end{align*} This gives $\widehat{\mathrm{sgn}}(\xi) = 2/(i\xi)$ away from $\xi = 0$, but the equation $i\xi \cdot v = 2$ in $\mathcal{S}'(\mathbb{R})$ has general solution $v = 2/(i\xi) + c\delta_0$ for some constant $c$ (since $\xi \cdot \delta_0 = 0$ by the multiplication example on the page). Here $1/(i\xi)$ denotes the tempered distribution defined by the Cauchy principal value: $(1/(i\xi))(\phi) := \mathrm{p.v.}\int \phi(\xi)/(i\xi)\,d\mathcal{L}^1(\xi) = -i\,\mathrm{p.v.}\int \phi(\xi)/\xi\,d\mathcal{L}^1(\xi)$. **Step 3: Determine the constant by parity.** Since $\mathrm{sgn}$ is an odd function, $\widehat{\mathrm{sgn}}$ is also odd (the Fourier transform of an odd function is odd). The distribution $1/(i\xi)$ is odd (since $\mathrm{p.v.}\,1/\xi$ is odd), and $\delta_0$ is even, so the requirement that $\widehat{\mathrm{sgn}}$ be odd forces $c = 0$. Therefore $\widehat{\mathrm{sgn}} = 2/(i\xi)$. **Step 4: Assemble.** Combining the results: \begin{align*} \hat{H} &= \frac{1}{2}(2\pi\delta_0) + \frac{1}{2}\cdot\frac{2}{i\xi} = \pi\delta_0 + \frac{1}{i\xi}. \end{align*} [/solution]

[problem: Sobolev Regularity Of The Dirac Delta | 2] Let $s \in \mathbb{R}$ and $n \ge 1$. Show that $\delta_0 \in H^s(\mathbb{R}^n)$ if and only if $s < -n/2$. [/problem]

[solution] **Step 1: Reduce to an integrability condition.** By the Fourier characterisation, $\delta_0 \in H^s(\mathbb{R}^n)$ if and only if $(1+|\xi|^2)^{s/2}\hat{\delta}_0 \in L^2(\mathbb{R}^n)$. Since $\hat{\delta}_0 = 1$ (as computed on the page), this reduces to $(1+|\xi|^2)^{s/2} \in L^2(\mathbb{R}^n)$, which means \begin{align*} \int_{\mathbb{R}^n}(1+|\xi|^2)^s\,d\mathcal{L}^n(\xi) &< \infty. \end{align*} **Step 2: Pass to polar coordinates.** Writing $\xi = r\omega$ with $r \ge 0$ and $\omega \in \mathbb{S}^{n-1}$: \begin{align*} \int_{\mathbb{R}^n}(1+|\xi|^2)^s\,d\mathcal{L}^n(\xi) &= \omega_{n-1}\int_0^\infty (1+r^2)^s \, r^{n-1}\,d\mathcal{L}^1(r), \end{align*} where $\omega_{n-1} = \mathcal{H}^{n-1}(\mathbb{S}^{n-1})$ is the surface area of the unit sphere. **Step 3: Analyse convergence at infinity.** For $r \ge 1$, the integrand behaves like $r^{2s} \cdot r^{n-1} = r^{2s+n-1}$. By the comparison test for improper integrals, $\int_1^\infty r^{2s+n-1}\,d\mathcal{L}^1(r) < \infty$ if and only if $2s + n - 1 < -1$, which simplifies to $s < -n/2$. **Step 4: Check convergence near the origin.** Near $r = 0$ the integrand is bounded: $(1+r^2)^s \le \max(1, 2^{|s|})$ for $r \le 1$, and $r^{n-1}$ is integrable on $[0,1]$ since $n \ge 1$. So there is no issue at the origin. **Step 5: Conclude.** Combining Steps 3 and 4, the integral converges if and only if $s < -n/2$, so $\delta_0 \in H^s(\mathbb{R}^n)$ if and only if $s < -n/2$. [/solution]