Lecture 15

More examples of tempered distributions and its Fourier transform¶

The Heviside function¶

Consider the function

called the Heviside function. It is has a jump discontinunity at 0.

This function is not Schwartz since it is a constant on positive real line, but it is a tempered distribution. It’s derivative in sense of distribution is

Proof. Let $\varphi\in \mathcal D$ be a test function. Then $(\frac{d}{dx}[H],\varphi) = -(H,\varphi') = -\int_{-\infty}^{+\infty}H(x)\varphi'(x)dx = -\int_{0}^{+\infty}\varphi'(x)dx =-( \lim_{x\to +\infty}\varphi(x) - \varphi(0) )= \varphi(0) = (\delta,\varphi)$.

The Heviside unit ramp¶

Let $u(x) = \begin{cases}0,x\leq 0 \\ x,x>0 \end{cases}$. Then $u(x)$ is a continuous function. One can verify by integration by parts that $\frac{d^2}{dx^2}[u(x)] = \delta$.

The sign function¶

Let $syn(x) = \begin{cases}1, x > 0 \\ -1, x<0 \\ 0,x = 0\end{cases}$. Then similar to the Heviside function one has $\frac{d}{dx}[syn] = 2\delta$. This can also be obtained from the relationship $syn(x) = 2H(x) - 1$.

The Cauchy principal value¶

Let $f \in \mathcal S$. Then the improper integral $\int_{-\infty}^{+\infty}\frac{f(x)}{x}dx$ may be not converge near 0. Its principal value is given by $\lim_{\epsilon\to 0}\int_{|x|>\epsilon}\frac{f(x)}{x}dx$, denoted by $p.v.\int_{-\infty}^{+\infty}\frac{f(x)}{x}dx$.

To see that the limit exists, rewrite the integral as two parts $\int_{\epsilon<|x|<1}\frac{f(x)}{x}dx + \int_{1}^{+\infty}\frac{f(x)-f(-x)}{x}dx$. For the first part, observe that $|\int_{\epsilon<|x|<1}\frac{f(x)}{x}dx |= |\int_{\epsilon}^1 \frac{f(x)-f(-x)}{x}dx|\leq \int_{\epsilon}^1 |\frac{f(x) - f(-x)}{x}|dx \leq 2\sup_{x\in \mathbb{R}}|f'(x)|\cdot (1-\epsilon)$. Where the last inequality follows from the mean value theorem. For the second part, just use the $1+x^2$-trick, $\int_1^{+\infty}|\frac{f(x) - f(-x)}{x}|dx \leq 2\sup_{x\in \mathbb{R}}|xf(x)|\int_{1}^{+\infty}\frac{1}{x^2}dx \leq C \|f\|_{1,0}$.

Then $|p.v. \int_{-\infty}^{+\infty}\frac{f(x)}{x}dx|\leq const\cdot (\|f\|_{1,0}+\|f\|_{0,1})$, which shows $p.v.\frac{1}{x}$ is a tempered distribution.

Fourier transform of the Heviside function¶

Let’s consider the Fourier tansform of $sgn$. We have, by derivative theorem, $\mathcal F\frac{d}{dx}[sgn] = 2\pi i x \mathcal F sgn$. On the other hand, we know that $\frac{d}{dx}[sgn] = 2\delta$. So $2\pi i x\cdot \mathcal Fsgn = \mathcal F 2\delta = 2$. Then it looks like $\mathcal Fsgn = \frac{1}{\pi i x}$ (to be precise, this is actually p.v. $\frac{1}{\pi i x}$). The result is correct but the argument is not, because we are dealing with distributions not functions, so the $\pi i x$ should be understood as an operator which maps a distribution to a new distribution. When we “devide” an operator it should be invertible, but the $\pi i x$ does has kernel.

Lemma.¶

Let $T$ be a distribution. if $T$ is annihilated by $x$, i.e. $xT = 0$, then $T$ must be a constant multiple of δ.

Proof. If $g(0) = 0$, then we can write $g(x) = x\psi(x)$ for smooth ψ (see problem set 6). In this case, $(T,g) = (T,x\psi) = (xT,\psi) = (0,\psi) = 0$. For general $g$, we can consider $g-g(0)\cdot \mathbf{1}$, where $\mathbf{1}$ is the constant function $\mathbf{1}(x)\equiv 1$. So $0=T(g - g(0)\mathbf{1}) = (T,g) - g(0)(T,\mathbf{1})$. So $(T,g) = (T,\mathbf{1})g(0)$, it follows that $T = (T,\mathbf{1})\delta$.

From the above lemma we see that the kernel of $\pi i x$ is given by $c\delta$. Then $\mathcal Fsgn = \frac{1}{\pi i x}+c\delta$ for some constant $c$. But since the sgn function is odd, $\mathcal F(sgn^-) = \mathcal Fsgn^- = -\frac{1}{\pi i x} + c\delta$, and on the other hand, $\mathcal F sgn^- = \mathcal F(-sgn) = -\frac{1}{\pi i x}-c\delta$. This implies $c = 0$. So we get

Also since $sgn = 2H-1$, we get

Appendix: Partition of unity¶

Let $U_1,\dots,U_n$ be open sets on $\mathbb{R}^m$, let $U = \bigcup_{j = 1}^n U_j$. Then for every $\varphi \in \mathscr C_c^{\infty}(U)$, there exists $\varphi_j \in \mathscr C_c^{\infty}(U_j)$ such that $\varphi = \sum_{j = 1}^n \varphi_j$. If $\varphi \geq 0$ then one can take all $\varphi_j \geq 0$.

Proof. Assume φ is supported on compact set $K$. Choose open cover $V_1,\dots,V_n$ of $K$ such that $V_i\subset \overline{V_i}\subset U$.

Let $\psi_j$ be smooth function such that

• $\psi_j\equiv 1$ on $K_j$.
• $supp\; \psi_j\subset U_j$.

Then let $\varphi_1:=\varphi \psi_1, \varphi_2 := \varphi\psi_2(1 - \psi_1)$, $\varphi_3:=\varphi \psi_3(1 - \psi_2)(1 - \psi_1)$, $\dots,$ $\varphi_n = \varphi \psi_n(1 - \psi_{n - 1})\cdots(1 - \psi_1)$. The $(\varphi_i)_{i=1}^n$ satisfies the desired properties.