Convolution

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 $f * g = g * f$.

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.

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 $[-\varepsilon, \varepsilon]$.