# The Dirac Delta as a Tempered Distribution **Context**: Companion notes to the "On the Function Spaces for Fourier Transform" slides. Fills in the derivations the slides state, and connects them back to where the delta first showed up in the courses. **Topics**: Schwartz space, tempered distributions, the delta as a distributional limit, the distributional derivative, $H' = \delta$, Fourier transforms of distributions **Related**: the Spring 2026 Day 12 notes (first appearance of $\delta$ in the transform of $\cos t$), the Spring 2026 Day 14 notes (the working definition, "a tool of the devil"), MATH235 Laplace unit ($\mathcal{L}\{\delta(t-a)\} = e^{-sa}$) --- ## Symbol Conventions These are fixed for the whole document; the slides use the same ones. | Object | Symbol | |---|---| | Heaviside (unit step) function | $H(t - t_0)$ | | Schwartz space of test functions | $S(\mathbb{R})$, or $S(\mathbb{R},\mathbb{C})$ when the codomain matters | | A test function | $\phi$; a sequence of them $\phi_n$ | | A tempered distribution | $T \in S'(\mathbb{R})$ | | Pairing, **distribution first, test function second** | $\langle T, \phi \rangle$ | | The distribution induced by an ordinary function $f$ | $T_f$, with $\langle f, \phi \rangle := \int_{\mathbb{R}} f(t)\phi(t)\,dt$ | | Distributional derivative | $\langle T', \phi \rangle := -\langle T, \phi' \rangle$ | | Dirac delta at $t_0$ | $\delta_{t_0}$, with $\langle \delta_{t_0}, \phi \rangle = \phi(t_0)$; $\delta := \delta_0$ | | Fourier transform (symmetric convention) | $\hat{\phi}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \phi(t) e^{-i\omega t}\,dt$ | Two rules keep the notation honest. Letters like $f$, $T$, $H$, $\delta$ always sit in the *first* slot of a pairing and name the (possibly generalized) object; letters from the $\phi$ family always sit in the *second* slot and name a test function. Never let $f$ name a test function. --- ## 0. Why We Are Here: Two Roads to the Same Object The delta shows up in two places in the courses, and it arrives looking like an emergency each time. **The Fourier road.** We would like to get all the way back to Fourier series from the Fourier transform: take the transform of something periodic. But $\cos t$ is not integrable on $\mathbb{R}$, so its classical transform $\frac{1}{\sqrt{2\pi}} \int \cos(t)\, e^{-i\omega t}\,dt$ does not converge. Day 12 wrote down the answer anyway, $ \mathcal{F}\{\cos t\}(\omega) = \sqrt{\frac{\pi}{2}}\,\big[\delta(\omega - 1) + \delta(\omega + 1)\big], $ and Day 14 called the object that makes this work "a tool of the devil." The Fall 2023 boards did the same thing from the other direction: send the period of a square pulse to infinity and watch $\hat{f}$ pile up into $\sqrt{2\pi}\,A\,\delta(\omega)$. The physical picture was a pop on a record or a CD: an impulse in time whose spectrum is a spike. Day 12's takeaway line, "delta functions bridge discrete and continuous frequency domains," is exactly right and is what Section 5 proves. **The ODE road.** In differential equations we drive systems with switches. The Heaviside step $ H(t - t_0) = \begin{cases} 0, & t < t_0 \\ 1, & t > t_0 \end{cases} $ turns a forcing term on at $t_0$. A solution written in terms of $H$ has a corner, and it is natural to ask, "what is the derivative of the step?" Classically there is none at $t_0$. The Laplace unit of MATH235 answers with $\delta(t - t_0)$ and the transform pair $\mathcal{L}\{\delta(t-a)\} = e^{-sa}$, and uses it to model impulses: a hammer strike, a kick. Both roads run into the same wall, and the wall is the word *function*. --- ## 1. Historical Sketch The behavior had been in the air for a long time. Fourier's inversion formula (1822) already contains it in disguise, Cauchy used sharply peaked kernels, and Heaviside's operational calculus (1890s) manipulated "the derivative of the step" with a straight face and correct answers. Sobolev (1936) introduced weak derivatives for partial differential equations. The notation and the name come from P. A. M. Dirac (1902–1984), who introduced $\delta(x)$ in 1927 to handle the normalization of continuous-spectrum states in quantum mechanics: he needed $\langle x | x' \rangle = \delta(x - x')$ to play the role that $\delta_{ij}$ plays for a discrete basis. He was open about it not being a function. The theory that makes it rigorous is Laurent Schwartz's theory of distributions, announced in a 1945 paper and laid out in full in the two-volume *Théorie des distributions* (1950–51), for which he received the Fields Medal in 1950. Schwartz's idea, which is the whole content of these notes: stop asking what $\delta$ *is* at each point and ask instead what it *does* to nice functions. A distribution is a linear map on a vector space of test functions. --- ## 2. The Working Definition, and Why It Cannot Be a Function Every course states the same three properties. For $t_0 \in \mathbb{R}$, the delta at $t_0$ is "the function" $\delta(t - t_0)$ with 1. **Localization**: $\delta(t - t_0) = 0$ for $t \neq t_0$. 2. **Normalization**: $\displaystyle\int_{\mathbb{R}} \delta(t - t_0)\,dt = 1$. 3. **Sifting**: $\displaystyle\int_{\mathbb{R}} \delta(t - t_0)\, f(t)\,dt = f(t_0)$ for every continuous $f$. Properties (1) and (2) are already incompatible for a function. If a function vanishes everywhere except at one point, its integral (Riemann or Lebesgue) is zero: a single point has no length. So no function satisfies both, and "$\delta(t_0) = \inftyquot; does not repair this; $\infty \cdot 0$ is not $1$. What is actually true, and what the three properties are shorthand for, is this: *there are sequences of honest functions $\phi_n$ for which the integrals $\int \phi_n(t) f(t)\,dt$ converge to $f(t_0)$.* The delta is the name we give to the limit of those integrals, viewed as a rule $f \mapsto f(t_0)$. In the language of the handwritten notes: the three properties "should be thought of as the value of a sequence of functionals." Making that sentence precise takes three steps. --- ## 3. Step 1: Schwartz Space, the "Really Nice" Functions We need a supply of functions so well behaved that anything reasonable can be tested against them. The right class for Fourier analysis is the **Schwartz space** $ S(\mathbb{R}, \mathbb{C}) = \left\{ \phi \in C^\infty(\mathbb{R}, \mathbb{C}) \;\middle|\; \forall \alpha, \beta \in \mathbb{N}_0,\; \sup_{x \in \mathbb{R}} \left| x^\alpha \frac{d^\beta \phi}{dx^\beta} \right| < \infty \right\}. $ In words: smooth functions which, together with every derivative, decay faster than any power grows. The quantities $ \|\phi\|_{\alpha,\beta} := \sup_{x \in \mathbb{R}} \left| x^\alpha \phi^{(\beta)}(x) \right| $ are the **seminorms** of $S(\mathbb{R})$, and they define what convergence means there: $\phi_n \to \phi$ in $S(\mathbb{R})$ if and only if $\|\phi_n - \phi\|_{\alpha,\beta} \to 0$ for every pair $(\alpha, \beta)$. That is a strong requirement (uniform convergence of every derivative, with polynomial weights), which is why continuity with respect to it will be easy to satisfy. **Examples.** - $\phi(x) = x^a e^{-bx^2}$ with $b > 0$ and $a \in \mathbb{N}_0$. The integer restriction on $a$ matters: for fractional $a$ the function is not smooth at $0$, and for negative $x$ may not even be defined. These are the integrands in the moments of the normal distribution, $\mathbf{E}[X^a]$ for $X \sim \mathcal{N}(0, \sigma^2)$. - Any $C^\infty$ function with compact support. - $e^{-|x|}$ is **not** in $S(\mathbb{R})$ (not smooth at $0$); $(1 + x^2)^{-1}$ is not (decays only like $x^{-2}$); polynomials, $\cos x$, and $H$ are not (no decay at all). **Why this space.** Schwartz space is stable under everything the Fourier transform does. Differentiation and multiplication by polynomials map $S(\mathbb{R})$ to itself, and the transform swaps them: $ \widehat{\phi'}(\omega) = i\omega\, \hat{\phi}(\omega), \qquad \widehat{x\phi}(\omega) = i \frac{d}{d\omega} \hat{\phi}(\omega). $ Decay in $x$ buys smoothness in $\omega$ and vice versa, so a function that has both stays a function that has both. Consequently $\mathcal{F} : S(\mathbb{R}) \to S(\mathbb{R})$ is a bijection, continuous in the seminorms, with inverse given by the usual inversion formula. Schwartz space is exactly the space on which the Fourier transform can do no harm. That is the fact Section 5 leans on. --- ## 4. Step 2: Distributions as Continuous Linear Functionals Now generalize the notion of function. A **tempered distribution** is a map $ T : S(\mathbb{R}) \to \mathbb{C}, \qquad \phi \mapsto \langle T, \phi \rangle, $ that is - **linear**: $\langle T, c_1 \phi_1 + c_2 \phi_2 \rangle = c_1 \langle T, \phi_1 \rangle + c_2 \langle T, \phi_2 \rangle$, and - **continuous**: if $\phi_n \to \phi$ in $S(\mathbb{R})$ then $\langle T, \phi_n \rangle \to \langle T, \phi \rangle$. A distribution takes a function in and returns a number. The set of all tempered distributions is the **dual space** of $S(\mathbb{R})$, written $S'(\mathbb{R})$. **Ordinary functions are distributions.** Let $f$ be locally integrable with at most polynomial growth, $|f(t)| \le C(1 + |t|)^N$. Then $ T_f(\phi) := \langle f, \phi \rangle := \int_{\mathbb{R}} f(t)\, \phi(t)\, dt $ converges for every $\phi \in S(\mathbb{R})$ (the test function's decay beats $fs growth), is linear in $\phi$, and is continuous: $|\langle f, \phi_n - \phi \rangle| \le \int |f|\,|\phi_n - \phi| \to 0$ because $\phi_n \to \phi$ in $S(\mathbb{R})$ controls $|\phi_n - \phi|$ against any polynomial weight. So $T_f \in S'(\mathbb{R})$. Distributions of this form are called **regular**. Everything we have met so far ($H$, $\cos$, polynomials, Gaussians) is a regular tempered distribution. This is the sense in which the generalization "naturally includes the functions we have met up until now." Two remarks that save confusion later. *On terminology.* We will say "the distribution $Tquot; and also "the function $fquot; for $T_f$, and by abuse write $\langle f, \phi \rangle$ for $\langle T_f, \phi \rangle$. When someone says "the Dirac delta function," they mean the value of its associated distribution, $\langle \delta, \phi \rangle$. The slides' rule, distribution first and test function second in every pairing, is what keeps this abuse from becoming an error. *On the word "tempered."* The general theory uses compactly supported test functions and gets a larger space of distributions, $\mathcal{D}'(\mathbb{R})$. Tempered distributions, $S'(\mathbb{R}) \subset \mathcal{D}'(\mathbb{R})$, are the ones with at most polynomial growth. They are precisely the distributions that have Fourier transforms; $e^{t^2}$ is a distribution but not a tempered one. Since Fourier is the goal, we work in $S'$ throughout. --- ## 5. Step 3: The Delta Is a Tempered Distribution Define, for each $t_0 \in \mathbb{R}$, $ \langle \delta_{t_0}, \phi \rangle := \phi(t_0), \qquad \phi \in S(\mathbb{R}), $ and write $\delta := \delta_0$. This is linear in $\phi$ trivially. It is continuous because evaluation at a point is controlled by the crudest seminorm: $|\langle \delta_{t_0}, \phi_n - \phi \rangle| = |\phi_n(t_0) - \phi(t_0)| \le \|\phi_n - \phi\|_{0,0} \to 0$. So $\delta_{t_0} \in S'(\mathbb{R})$. The **sifting property is now the definition**, and nothing is being integrated. The delta is *not* regular: there is no function $f$ with $\int f \phi = \phi(0)$ for all $\phi$ (take $\phi$ supported away from $0$ to force $f = 0$ almost everywhere, then take $\phi$ with $\phi(0) = 1$ for a contradiction). So $\delta$ lives in $S'(\mathbb{R})$ but not in the image of the functions. That is the precise sense in which it is "not a function." ### 5.1 The delta as a limit of Gaussians Here is the picture behind the three properties. Take the sequence of test functions (this choice is not unique) $ \phi_n(t) = \frac{n}{\sqrt{\pi}}\, e^{-n^2 t^2}, \qquad n = 1, 2, 3, \ldots $ Each $\phi_n$ is a Gaussian with $\int_{\mathbb{R}} \phi_n(t)\,dt = 1$ (substitute $u = nt$ and use $\int e^{-u^2}du = \sqrt{\pi}$), height $n/\sqrt{\pi}$, and width of order $1/n$: taller and narrower as $n$ grows, with fixed unit area. Each is in $S(\mathbb{R})$, so each defines a regular distribution $T_{\phi_n}$. **Claim.** For every test function $\phi$, $ \langle \delta, \phi \rangle = \lim_{n \to \infty} \langle \phi_n, \phi \rangle = \lim_{n \to \infty} \int_{-\infty}^{\infty} \phi_n(t)\, \phi(t)\, dt = \phi(0). $ **Proof.** Using $\int \phi_n = 1$, $ \int_{\mathbb{R}} \phi_n(t)\phi(t)\,dt - \phi(0) = \int_{\mathbb{R}} \phi_n(t)\,\big[\phi(t) - \phi(0)\big]\,dt. $ Fix $\varepsilon > 0$. By continuity of $\phi$ at $0$ choose $\eta > 0$ with $|\phi(t) - \phi(0)| < \varepsilon$ for $|t| < \eta$. Split the integral at $|t| = \eta$: $ \left| \int_{\mathbb{R}} \phi_n\,[\phi - \phi(0)] \right| \le \varepsilon \int_{|t| < \eta} \phi_n + 2\|\phi\|_\infty \int_{|t| > \eta} \phi_n \le \varepsilon + 2\|\phi\|_\infty \cdot \frac{2}{\sqrt{\pi}} \int_{n\eta}^{\infty} e^{-u^2}\,du. $ The last integral tends to $0$ as $n \to \infty$. Since $\varepsilon$ was arbitrary, the claim follows. $\blacksquare$ Notice the proof used only that $\phi$ is continuous at $0$ and bounded. That is why the physicists' version, with $f$ merely continuous, is correct as far as it goes; Schwartz space is needed for differentiation and Fourier transforms, not for this limit. ### 5.2 What kind of limit this is This is the one place where the slides compress two ideas into one line, and it is worth having the honest answer ready. The continuity axiom of Section 4 says: for a *fixed* distribution $T$, if $\phi_n \to \phi$ *in $S(\mathbb{R})$* then $\langle T, \phi_n \rangle \to \langle T, \phi \rangle$. That is continuity of one functional along sequences that converge inside the test space. The Gaussian sequence does **not** converge in $S(\mathbb{R})$. Its sup norm $\|\phi_n\|_{0,0} = n/\sqrt{\pi}$ blows up, and there is no Schwartz function it could converge to, because the limit would have to be $\delta$, which is not a function at all. What converges is the sequence of *distributions* $T_{\phi_n}$: for every fixed test function $\phi$, the numbers $\langle \phi_n, \phi \rangle$ converge to $\langle \delta, \phi \rangle$. This is convergence in $S'(\mathbb{R})$, sometimes called weak-$*$ convergence, and it is a different notion sitting next to the continuity axiom, not a consequence of it. We write $\phi_n \to \delta$ in $S'(\mathbb{R})$, or $T_{\phi_n} \to \delta$. If someone asks "does $\phi_n$ converge in $S(\mathbb{R})$?", the answer is no, and that is the point: the delta is a limit that exists only after you have enlarged the space. ### 5.3 The three properties, recovered Each of the working properties from Section 2 is now a true statement about $\delta \in S'(\mathbb{R})$. 1. **Localization.** If $\phi$ vanishes on a neighborhood of $t_0$ then $\langle \delta_{t_0}, \phi \rangle = 0$. The delta cannot see test functions that avoid $t_0$; its *support* is $\{t_0\}$. 2. **Normalization.** $\int \phi_n = 1$ for every $n$; and if $\phi$ equals $1$ on a neighborhood of $t_0$ then $\langle \delta_{t_0}, \phi \rangle = 1$. (The constant function $1$ is not itself a test function, which is why the property has to be phrased this way.) 3. **Sifting.** $\langle \delta_{t_0}, \phi \rangle = \phi(t_0)$, by definition. And the shift is what it should be: writing $\delta(t - t_0)$ for $\delta_{t_0}$ agrees with the substitution rule for regular distributions, $\langle f(\cdot - t_0), \phi \rangle = \langle f, \phi(\cdot + t_0) \rangle$. --- ## 6. Step 4: Differentiation, and $H' = \delta$ Now the ODE road. We want to differentiate things that have no classical derivative. The idea is to move the derivative onto the test function, where it is always allowed. **Motivation.** For smooth $f$ of polynomial growth and $\phi \in S(\mathbb{R})$, integrate by parts: $ \langle f', \phi \rangle = \int_{\mathbb{R}} f'(t)\phi(t)\,dt = \Big[ f(t)\phi(t) \Big]_{-\infty}^{\infty} - \int_{\mathbb{R}} f(t)\phi'(t)\,dt = -\langle f, \phi' \rangle. $ The boundary term vanishes because $\phi$ decays faster than $f$ grows. The right-hand side makes sense for *any* distribution, so we take it as the definition. **Definition.** For $T \in S'(\mathbb{R})$, the **distributional derivative** $T'$ is $ \langle T', \phi \rangle := -\langle T, \phi' \rangle, \qquad \phi \in S(\mathbb{R}). $ Since $\phi \mapsto \phi'$ is linear and continuous on $S(\mathbb{R})$, $T'$ is again in $S'(\mathbb{R})$. Every tempered distribution is therefore infinitely differentiable, and for smooth $f$ the new derivative agrees with the old one. Continuity in $S'$ is automatic too: if $T_n \to T$ in $S'$ then $T_n' \to T'$. Differentiation, which is delicate for functions, is painless for distributions. **The step function.** With $t_0 = 0$ for simplicity, $H(t) = 0$ for $t < 0$ and $1$ for $t > 0$. Its value at $t = 0$ is immaterial: $T_H$ is an integral, and a single point does not change an integral. (Distributions do not see points. That is also why the delta had to be built as a functional rather than a function.) Since $H$ is bounded, $T_H \in S'(\mathbb{R})$ with $\langle H, \phi \rangle = \int_0^\infty \phi(t)\,dt$. **Theorem.** $H' = \delta$ in $S'(\mathbb{R})$. **Proof.** For $\phi \in S(\mathbb{R})$, $ \langle H', \phi \rangle = -\langle H, \phi' \rangle = -\int_{-\infty}^{\infty} H(t)\,\phi'(t)\,dt = -\int_0^\infty \phi'(t)\,dt = -\Big[\phi(t)\Big]_0^\infty = -\big(0 - \phi(0)\big) = \phi(0) = \langle \delta, \phi \rangle, $ using $\phi(t) \to 0$ as $t \to \infty$. Since this holds for every $\phi$, $H' = \delta$. $\blacksquare$ Shifting gives $\frac{d}{dt} H(t - t_0) = \delta_{t_0}$. The derivative of the switch is the impulse. The question from the ODE road, "what is the derivative of the step?", has an answer, and the answer is a theorem rather than a convention. **Two consequences worth seeing once.** - The delta has a derivative too: $\langle \delta', \phi \rangle = -\langle \delta, \phi' \rangle = -\phi'(0)$. It is not a limit of anything that looks like a bump; it is a "dipole." - $t\,\delta(t) = 0$: multiplying a distribution by a smooth function $g$ is defined by $\langle gT, \phi \rangle := \langle T, g\phi \rangle$, and $\langle t\delta, \phi \rangle = \langle \delta, t\phi \rangle = 0 \cdot \phi(0) = 0$. This is the identity that lets one "cancel a $tquot; against a delta in the sifting integrals. --- ## 7. Step 5: Closing the Loop, the Fourier Transform of a Distribution Finally the Fourier road, and the reason the slides are titled the way they are. **The trick is the same.** For $f, \phi \in S(\mathbb{R})$, Fubini gives the *multiplication formula* $ \int_{\mathbb{R}} \hat{f}(\omega)\,\phi(\omega)\,d\omega = \int_{\mathbb{R}} f(t)\,\hat{\phi}(t)\,dt, \qquad \text{i.e.} \qquad \langle \hat{f}, \phi \rangle = \langle f, \hat{\phi} \rangle. $ (Both sides equal $\frac{1}{\sqrt{2\pi}} \iint f(t)\phi(\omega)e^{-i\omega t}\,dt\,d\omega$.) Again the right-hand side makes sense for any distribution, because $\hat{\phi}$ is again a test function by Section 3. **Definition.** For $T \in S'(\mathbb{R})$, $ \langle \hat{T}, \phi \rangle := \langle T, \hat{\phi} \rangle, \qquad \phi \in S(\mathbb{R}). $ Since $\mathcal{F}$ is linear and continuous on $S(\mathbb{R})$, $\hat{T} \in S'(\mathbb{R})$; since $\mathcal{F}$ is a bijection on $S(\mathbb{R})$, it is a bijection on $S'(\mathbb{R})$ too, with inverse $\langle \check{T}, \phi \rangle = \langle T, \check{\phi} \rangle$. *Every tempered distribution has a Fourier transform, and every ordinary function of polynomial growth is a tempered distribution.* That single sentence is what licenses $\mathcal{F}\{\cos t\}$. **The delta and the constant.** $ \langle \hat{\delta}, \phi \rangle = \langle \delta, \hat{\phi} \rangle = \hat{\phi}(0) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \phi(t)\,dt = \Big\langle \tfrac{1}{\sqrt{2\pi}}, \phi \Big\rangle, \qquad \text{so} \qquad \hat{\delta} = \frac{1}{\sqrt{2\pi}}. $ An impulse in time has a flat spectrum. Conversely, using the inversion formula $\phi(0) = \frac{1}{\sqrt{2\pi}} \int \hat{\phi}(\omega)\,d\omega$, $ \langle \hat{1}, \phi \rangle = \langle 1, \hat{\phi} \rangle = \int_{\mathbb{R}} \hat{\phi}(\omega)\,d\omega = \sqrt{2\pi}\,\phi(0), \qquad \text{so} \qquad \hat{1} = \sqrt{2\pi}\,\delta. $ This is the Fall 2023 board computation, $\lim_{L \to \infty} \hat{f} = \sqrt{2\pi}\,A\,\delta(\omega)$ for a pulse of height $A$ whose width $L \to \infty$, now as a theorem rather than a limit taken on faith. (The limit statement is also true, and it is a statement of convergence in $S'(\mathbb{R})$, exactly the kind in Section 5.2.) **A pure tone.** For $\omega_0 \in \mathbb{R}$, the function $e^{i\omega_0 t}$ is bounded, hence a tempered distribution, and $ \langle \widehat{e^{i\omega_0 t}}, \phi \rangle = \int_{\mathbb{R}} e^{i\omega_0 t}\,\hat{\phi}(t)\,dt = \sqrt{2\pi}\,\phi(\omega_0) = \langle \sqrt{2\pi}\,\delta_{\omega_0}, \phi \rangle, $ by the inversion formula evaluated at $\omega_0$. Hence $ \mathcal{F}\{e^{i\omega_0 t}\} = \sqrt{2\pi}\,\delta(\omega - \omega_0). $ **The Day 12 formula, proved.** Since $\cos(\omega_0 t) = \frac{1}{2}(e^{i\omega_0 t} + e^{-i\omega_0 t})$ and $\mathcal{F}$ is linear on $S'(\mathbb{R})$, $ \mathcal{F}\{\cos \omega_0 t\} = \frac{\sqrt{2\pi}}{2}\,\big[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)\big] = \sqrt{\frac{\pi}{2}}\,\big[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)\big]. $ With $\omega_0 = 1$ this is exactly the Day 12 result. The only frequencies present are $\pm\omega_0$, each carrying weight $\sqrt{\pi/2}$, and the classical integral that refused to converge was never the right object; the pairing $\langle \cos, \hat{\phi} \rangle$ was. **All the way back to Fourier series.** If $f$ is periodic with Fourier series $f(t) = \sum_n c_n e^{in\omega_0 t}$ (convergent in $S'$, which is weaker than pointwise and always holds for reasonable $f$), then by linearity and continuity of $\mathcal{F}$ on $S'(\mathbb{R})$, $ \hat{f}(\omega) = \sqrt{2\pi} \sum_{n} c_n\, \delta(\omega - n\omega_0). $ The Fourier transform of a periodic function is a comb of deltas at the series frequencies, weighted by the series coefficients. Fourier series are the special case of the Fourier transform in which the spectrum is supported on a lattice. This is the sentence the first slide promised, "we would like to get all the way back, i.e., Fourier series," and it is the precise content of "delta functions bridge discrete and continuous frequency domains." --- ## 8. The Same Object in MATH235 The Laplace pair $\mathcal{L}\{\delta(t - a)\} = e^{-sa}$ for $a > 0$ is the sifting property with the test function $e^{-st}$: $ \mathcal{L}\{\delta_a\}(s) = \langle \delta_a, e^{-st} \rangle = e^{-sa}. $ Strictly, $e^{-st}$ is not in $S(\mathbb{R})$ (it grows as $t \to -\infty$), but $\delta_a$ has support $\{a\}$ and so extends to any function smooth near $a$. The step pair $\mathcal{L}\{H(t - a)\} = e^{-sa}/s$ and the derivative rule $\mathcal{L}\{y'\} = sY - y(0)$ then make $H' = \delta$ visible in the transform domain: $s \cdot \frac{e^{-sa}}{s} = e^{-sa}$. The ODE road and the Fourier road are using the same distribution with different test functions. --- ## 9. Summary | Statement in the course | What it means in $S'(\mathbb{R})$ | Where proved | |---|---|---| | $\delta(t - t_0) = 0$ for $t \ne t_0$ | $\operatorname{supp} \delta_{t_0} = \{t_0\}$ | §5.3 | | $\int \delta(t - t_0)\,dt = 1$ | $\int \phi_n = 1$; $\langle \delta_{t_0}, \phi \rangle = 1$ if $\phi \equiv 1$ near $t_0$ | §5.3 | | $\int \delta(t - t_0) f(t)\,dt = f(t_0)$ | $\langle \delta_{t_0}, \phi \rangle := \phi(t_0)$ (definition) | §5 | | "$\delta$ is a limit of narrow Gaussians" | $T_{\phi_n} \to \delta$ in $S'(\mathbb{R})$, not in $S(\mathbb{R})$ | §5.1, §5.2 | | "$\delta$ is the derivative of the step" | $\langle H', \phi \rangle = -\langle H, \phi' \rangle = \phi(0)$ | §6 | | $\mathcal{F}\{\cos t\} = \sqrt{\pi/2}\,[\delta(\omega-1) + \delta(\omega+1)]$ | $\langle \hat{T}, \phi \rangle := \langle T, \hat{\phi} \rangle$ applied to $T = \cos$ | §7 | | $\mathcal{L}\{\delta(t-a)\} = e^{-sa}$ | sifting against $e^{-st}$ | §8 | The one idea underneath all of it: **a distribution is known by what it does to test functions, and every operation on distributions (shift, derivative, Fourier transform) is defined by moving the operation onto the test function, where it is harmless.** Schwartz space is chosen so that all those moves land back inside it. --- ## 10. Exercises 1. Show directly from the definition that $\langle \delta', \phi \rangle = -\phi'(0)$, and that $\delta'$ is not a limit in $S'$ of any sequence of nonnegative functions. 2. Compute $\mathcal{F}\{\sin \omega_0 t\}$ and check that it is purely imaginary and odd in $\omega$, as it must be for a real odd function. 3. The Cauchy (Lorentzian) kernel $\psi_\varepsilon(t) = \dfrac{1}{\pi}\dfrac{\varepsilon}{t^2 + \varepsilon^2}$ has unit integral. Show $\psi_\varepsilon \to \delta$ in $S'(\mathbb{R})$ as $\varepsilon \to 0^+$. Note that $\psi_\varepsilon \notin S(\mathbb{R})$ (it decays only like $t^{-2}$), yet it is a perfectly good regular tempered distribution: the approximating sequence need not consist of test functions. Compare with the heavy-tailed Cauchy density from the Fall 2024 probability boards. 4. Show that $\frac{d}{dt}|t| = \operatorname{sgn}(t)$ and $\frac{d^2}{dt^2}|t| = 2\delta$ in $S'(\mathbb{R})$. 5. Using $H' = \delta$ and the derivative rule, show that the Fourier transform of $H$ is not a function: it must satisfy $i\omega\,\hat{H}(\omega) = \frac{1}{\sqrt{2\pi}}$, which no locally integrable $\hat{H}$ can do at $\omega = 0$. (The actual answer involves a principal value, $\hat{H} = \frac{1}{\sqrt{2\pi}}\big[\pi\delta(\omega) + \frac{1}{i}\,\mathrm{p.v.}\frac{1}{\omega}\big]$, which is where these notes stop.) --- ## References - L. Schwartz, *Théorie des distributions*, Hermann, 1950–51. - P. A. M. Dirac, "The physical interpretation of the quantum dynamics," *Proc. R. Soc. A* 113 (1927). - R. Strichartz, *A Guide to Distribution Theory and Fourier Transforms*, World Scientific, 2003. (The friendliest full treatment at this level.) - [Dirac delta function](https://en.wikipedia.org/wiki/Dirac_delta_function), [Distribution (mathematics)](https://en.wikipedia.org/wiki/Distribution_(mathematics)), [Schwartz space](https://en.wikipedia.org/wiki/Schwartz_space)