Convolution | Real Analysis

Flip one function, slide it, and measure how it overlaps the other. Overlapping two indicator blocks gives an overlap length that traces a tent curve. Convolving with a smooth kernel softens even a sharp step function. The narrower the kernel, the closer the softened function sits to the original. For example, it is the blur effect in a 📷 photo, replacing each pixel by a weighted average of its neighbors. The center pixel gets the largest weight.

Examples & Exercises

Prove that convolution commutes: for integrable $f, g$ , $(f * g)(x) = (g * f)(x)$ for every $x$ . Use the substitution $u = x - t$ , making the limit-flip step and the sign cancellation explicit.

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

f, g \in L^1(\mathbb{R}), \qquad \int_{-\infty}^{\infty} |f(t)\, g(x - t)|\, dt < \infty

Tent mode,

x = 0.7

(f * g)(0.7) = 0.7

-- here

f = g

makes commutativity look trivial, but the proof below works for every integrable pair.

Show proof

Hypothesis: $f, g \in L^1(\mathbb{R})$ , and for the fixed $x$ the product integral converges absolutely (second display line). Under this condition the substitution below is legitimate. Start from the definition: $(f * g)(x) = \int_{-\infty}^{\infty} f(t)\, g(x - t)\, dt$ .
Apply the substitution $u = x - t$ . Then $t = x - u$ and $dt = -du$ . The endpoints map as $u \to +\infty$ when $t \to -\infty$ and $u \to -\infty$ when $t \to +\infty$ -- that is, $t: -\infty \to \infty$ corresponds to $u: \infty \to -\infty$ . Substituting gives $(f * g)(x) = \int_{+\infty}^{-\infty} f(x - u)\, g(u)\, (-du)$ .
Now the two signs cancel: the factor $-1$ in $-du$ pairs exactly with flipping the limits ( $\int_{+\infty}^{-\infty} = -\int_{-\infty}^{+\infty}$ ). Hence $(f * g)(x) = \int_{-\infty}^{\infty} f(x - u)\, g(u)\, du$ .
The right side is, by definition, $(g * f)(x) = \int_{-\infty}^{\infty} g(u)\, f(x - u)\, du$ (the dummy variable is just named $u$ ). Hence $(f * g)(x) = (g * f)(x)$ , and $x$ was arbitrary, so $f * g = g * f$ .

For $f = g = \mathbf{1}_{[0,1]}$ , compute $(f * g)(x)$ for every real $x$ . The answer must be the tent function: $x$ on $[0,1]$ , $2 - x$ on $[1,2]$ , and $0$ elsewhere -- matching the trace in the scene.

(f * g)(x) = \int_{-\infty}^{\infty} \mathbf{1}_{[0,1]}(t)\, \mathbf{1}_{[0,1]}(x - t)\, dt = \text{length of } [0,1] \cap [x-1, x]

Tent mode,

x = 1.4

: the overlap is

[0.4, 1]

and

(f * g)(1.4) = 2 - 1.4 = 0.6

-- the falling case of the case split below.

Show proof

The product $\mathbf{1}_{[0,1]}(t)\, \mathbf{1}_{[0,1]}(x - t)$ equals $1$ exactly when both conditions hold: $0 \le t \le 1$ and $0 \le x - t \le 1$ . The second condition is equivalent to $x - 1 \le t \le x$ . Hence the integral is the length of $\{t : 0 \le t \le 1,\ x - 1 \le t \le x\} = [0,1] \cap [x-1, x]$ .
The endpoints of the intersection are $\max(0, x-1)$ and $\min(1, x)$ . The comparisons split at $x - 1 \le 0 \iff x \le 1$ and at $x \le 1$ itself: if $x \le 1$ then $\max(0, x-1) = 0$ and $\min(1, x) = x$ (for $x \ge 0$ ), while if $x \ge 1$ then $\max(0, x-1) = x-1$ and $\min(1, x) = 1$ . These comparisons justify the case split below.
Case 1, $x < 0$ or $x > 2$ : if $x < 0$ then $\min(1, x) = x < 0 \le \max(0, x-1)$ , so the intersection is empty; if $x > 2$ then $\max(0, x-1) = x - 1 > 1 \ge \min(1, x)$ , again empty. In both cases $(f * g)(x) = 0$ . Case 2, $0 \le x \le 1$ : the intersection is $[0, x]$ , of length $x$ .
Case 3, $1 \le x \le 2$ : the intersection is $[x-1, 1]$ , of length $1 - (x - 1) = 2 - x$ . In summary $(f * g)(x)$ is the tent function $0$ (outside), $x$ (on $[0,1]$ ), $2 - x$ (on $[1,2]$ ), with peak value $1$ at $x = 1$ -- exactly the trace drawn in the scene.

Let $\varphi \ge 0$ , $\operatorname{supp} \varphi \subseteq [-1,1]$ , $\int \varphi = 1$ , and $\varphi_{\varepsilon}(t) = \frac{1}{\varepsilon} \varphi(t/\varepsilon)$ . If $f$ is bounded ( $\|f\|_{\infty} = M$ ) and continuous at $x_0$ , prove $(f * \varphi_{\varepsilon})(x_0) \to f(x_0)$ as $\varepsilon \to 0^+$ . Start by checking that the scene's tent kernel $\varphi(t) = \max(0, 1 - |t|)$ meets the hypotheses.

\varphi_{\varepsilon}(t) = \frac{1}{\varepsilon} \varphi(t/\varepsilon), \qquad \int_{-\infty}^{\infty} \varphi_{\varepsilon} = 1, \qquad \operatorname{supp} \varphi_{\varepsilon} \subseteq [-\varepsilon, \varepsilon]

(f * \varphi_{\varepsilon})(x_0) - f(x_0) = \int_{-\infty}^{\infty} [f(x_0 - t) - f(x_0)]\, \varphi_{\varepsilon}(t)\, dt

Smooth mode,

x = 0.05

\varepsilon = 0.1

v = x/\varepsilon = 0.5

, so

(f * \varphi_{\varepsilon})(0.05) = 1 - (1 - 0.5)^2/2 = 0.875

-- approaching

f(0.05) = 1

\varepsilon \to 0

Show proof

Checking the tent kernel: $\varphi(t) = \max(0, 1 - |t|)$ is nonnegative, vanishes outside $[-1,1]$ , and $\int \varphi$ is the area of a triangle of base $2$ and height $1$ , namely $\frac{1}{2} \cdot 2 \cdot 1 = 1$ . So the tent kernel satisfies $\varphi \ge 0$ , $\operatorname{supp} \varphi \subseteq [-1,1]$ , and $\int \varphi = 1$ .
Scaling keeps the mass: $\int \varphi_{\varepsilon} = 1$ . With the substitution $s = t/\varepsilon$ ( $dt = \varepsilon\, ds$ ), $\int \frac{1}{\varepsilon} \varphi(t/\varepsilon)\, dt = \int \varphi(s)\, ds = 1$ . The same substitution gives $\operatorname{supp} \varphi_{\varepsilon} \subseteq [-\varepsilon, \varepsilon]$ : $\varphi_{\varepsilon}(t) \ne 0$ requires $|t/\varepsilon| \le 1$ , i.e. $|t| \le \varepsilon$ . Also $|f| \le M$ with $\varphi_{\varepsilon} \in L^1$ makes the convolution well defined.
The error identity: since $f(x_0) = \int f(x_0)\, \varphi_{\varepsilon}(t)\, dt$ (the mass is $1$ , so this is $f(x_0) \cdot 1$ ), subtracting gives $(f * \varphi_{\varepsilon})(x_0) - f(x_0) = \int [f(x_0 - t) - f(x_0)]\, \varphi_{\varepsilon}(t)\, dt$ . Moving the absolute value inside and using $\varphi_{\varepsilon} \ge 0$ gives $|(f * \varphi_{\varepsilon})(x_0) - f(x_0)| \le \int |f(x_0 - t) - f(x_0)|\, \varphi_{\varepsilon}(t)\, dt$ .
Let $\eta > 0$ be given. By continuity of $f$ at $x_0$ there is $\delta > 0$ with $|f(x_0 - t) - f(x_0)| < \eta/2$ whenever $|t| < \delta$ . Now take any $\varepsilon < \delta$ .
Finish: since $\operatorname{supp} \varphi_{\varepsilon} \subseteq [-\varepsilon, \varepsilon] \subseteq [-\delta, \delta]$ , the whole error integral lives on $|t| < \delta$ , where the factor $|f(x_0 - t) - f(x_0)|$ is below $\eta/2$ . Hence the error is at most $(\eta/2) \int \varphi_{\varepsilon} = \eta/2 < \eta$ . As $\eta$ was arbitrary, $(f * \varphi_{\varepsilon})(x_0) \to f(x_0)$ as $\varepsilon \to 0^+$ . Unlike a general approximate identity there is no tail term, because the support shrinks to $[-\varepsilon, \varepsilon]$ .