14. Delta Function — Mass Concentrated at a Point

Calculus 6/6. The Dirac delta is not an ordinary function but a linear operation that extracts the value at one point from a test function. Keep the engineering intuition of "an infinitely narrow spike with total mass 1," but handle calculations safely only inside integrals. Background: limits and differentiation, differentiation rules, and multivariable differentiation.

Core definition — an action, not a value

The delta function $\delta$ is not a function in the ordinary sense. More precisely, it is a linear functional that returns

$$\langle \delta, \varphi \rangle = \varphi(0)$$

for a smooth and sufficiently well-decaying test function $\varphi$. Statements such as $\delta(0)=\infty$ and $\delta(x)=0$ are only intuitive pictures; the rigorous definition always lies inside an integral or pairing.

In practical notation, this action is written as

$$\int_{-\infty}^{\infty} f(x)\,\delta(x)\,dx = f(0).$$

A delta concentrated at a point $a$ only needs to be translated:

$$\int_{-\infty}^{\infty} f(x)\,\delta(x-a)\,dx = f(a).$$

This is called the sifting property: like a sieve, it filters out only the value at $a$ from the entire function.

Derivative of the Heaviside function

The step function

$$H(x) = \begin{cases} 0, & x \lt 0, \\ 1, & x \gt 0 \end{cases}$$

is discontinuous at $x=0$ and therefore has no classical derivative. From the perspective of distributions, however, we can define

$$\frac{d}{dx}H(x) = \delta(x)$$

Because the step function jumps by 1 instantaneously, its entire change appears as a derivative compressed at one point.

Delta sequences — limits of actual functions

Although the delta is not an ordinary function, it can be approached as a limit of ordinary functions. For example, the Gaussian kernel

$$\delta_\varepsilon(x) = \frac{1}{\sqrt{2\pi}\varepsilon} \exp\!\left(-\frac{x^2}{2\varepsilon^2}\right)$$

integrates to 1 for every $\varepsilon > 0$, and its mass concentrates around 0 as $\varepsilon \to 0^+$. The pointwise limit is not useful. What matters is that, for test functions,

$$\lim_{\varepsilon \to 0^+} \int_{-\infty}^{\infty} f(x)\,\delta_\varepsilon(x)\,dx = f(0)$$

holds. The limit of the delta is not a limit of function values but a limit of integral actions.

Representative delta sequences take several forms:

$$\delta(x) = \lim_{\varepsilon \to 0^+} \frac{1}{\sqrt{2\pi}\varepsilon}e^{-x^2/(2\varepsilon^2)}, \qquad \delta(x) = \lim_{\varepsilon \to 0^+} \frac{1}{\pi}\frac{\varepsilon}{x^2+\varepsilon^2}.$$

The first is a Gaussian approximation and the second a Cauchy-kernel approximation. Their shapes differ, but if their limiting actions on test functions agree, they define the same delta.

Scaling and change of variables

A delta must carry the Jacobian under a change of variables:

$$\delta(ax) = \frac{1}{\lvert a \rvert}\delta(x) \qquad (a \ne 0).$$

The sifting property verifies this identity immediately:

$$\int f(x)\delta(ax)\,dx = \frac{1}{\lvert a \rvert}f(0).$$

More generally, if the simple roots of $g(x)$ are $x_i$ and $g'(x_i) \ne 0$, then

$$\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{\lvert g'(x_i) \rvert}.$$

For example, if $g(x)=x^2-a^2$, its roots are $\pm a$, and

$$\delta(x^2-a^2) = \frac{1}{2\lvert a \rvert}\left[\delta(x-a)+\delta(x+a)\right].$$

The key is: "with several roots, sample once at each root, but the larger the slope, the less width that root occupies."

Derivatives of the delta

The derivative of the delta is also defined as an action rather than an ordinary function. Assuming the boundary term vanishes in integration by parts,

$$\int_{-\infty}^{\infty} f(x)\,\delta'(x-a)\,dx = -f'(a).$$

More generally,

$$\int_{-\infty}^{\infty} f(x)\,\delta^{(n)}(x-a)\,dx = (-1)^n f^{(n)}(a).$$

Thus the derivative of the delta is a device that extracts not "a function value at one point," but "a derivative value at one point." Derivatives of impulse responses in signal processing, fundamental solutions of partial differential equations, and Green's-function calculations stand on this language.

Convolution and the identity element

The delta is the identity element of convolution:

$$(f * \delta)(x) = \int_{-\infty}^{\infty} f(t)\delta(x-t)\,dt = f(x).$$

Convolution with a translated delta shifts the function:

$$(f * \delta_a)(x) = f(x-a), \qquad \delta_a(t)=\delta(t-a).$$

The delta is therefore a "filter that applies no filtering," while a translated delta is a "pure delay." This is why one impulse response can represent an entire linear time-invariant system.

Fourier perspective

Under the Fourier transform, the delta is paired with a constant. Constants may vary with the normalization convention, but the core structure is the same:

$$\mathcal{F}\{\delta(x)\}(k) = 1, \qquad \mathcal{F}\{\delta(x-x_0)\}(k) = e^{-2\pi i k x_0}.$$

An object completely localized at one point in space is spread uniformly over the entire frequency domain. Conversely, the Fourier transform of a constant function is a delta at the origin. This is why the delta repeatedly appears in sampling theory, impulse signals, and fundamental-solution calculations.

Multivariable delta

The delta in $\mathbb{R}^n$ follows the same principle:

$$\int_{\mathbb{R}^n} f(\mathbf{x})\,\delta(\mathbf{x}-\mathbf{a})\,d\mathbf{x} = f(\mathbf{a}).$$

A coordinate transformation introduces a Jacobian determinant. For an invertible linear transformation $A$,

$$\delta(A\mathbf{x}) = \frac{1}{\lvert \det A \rvert}\delta(\mathbf{x}).$$

The one-dimensional $1/\lvert a \rvert$ rule is a special case of this expression.

Common misconceptions

1. "The delta is a function that is infinite at 0"—useful as a picture, but not a definition. The delta is a functional acting on test functions.
2. "Anything with integral 1 is a delta"—no. Many functions integrate to 1. A limiting action is required in which the width shrinks to zero and the mass concentrates at one point.
3. "Just substitute as in $\delta(g(x)) = \delta(x)$"—no. Every root and the Jacobian $1/\lvert g'(x_i)\rvert$ must be included.
4. "The derivative of the delta is just the slope of a sharp function"—its exact meaning is the action $f \mapsto -f'(a)$.

What to remember from this page

1. The delta is not a function but a generalized function satisfying $\langle \delta, f \rangle = f(0)$.
2. Every safe calculation happens within an action such as an integral, pairing, Fourier transform, or convolution.
3. $\delta(ax)=\delta(x)/\lvert a\rvert$ and $\delta(g(x))=\sum_i \delta(x-x_i)/\lvert g'(x_i)\rvert$ are Jacobian rules for change of variables.
4. Since $\int f(x)\delta^{(n)}(x-a)dx=(-1)^n f^{(n)}(a)$, derivatives of the delta sample derivatives at one point.
5. In the Fourier domain the delta corresponds to a constant, and under convolution it acts as the identity element.

Reference

• Wolfram MathWorld, Delta Function.

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