Let´s start with a "single step" $t(x) = \chi_{[a,b]}(x)$ for some interval $[a,b]$.
Let $h_\epsilon(x) = (\chi_{[a,b]} * g_\epsilon)(x)$. By definition of the convolution we get $$ h_\epsilon(x) = \frac{1}{2\epsilon} \int_{x-\epsilon}^{x+\epsilon} \chi_{[a,b]}(y) dy $$ We have 4 cases to consider:
- Case 1: If $x+\epsilon < a$ or $x-\epsilon > b$ the intervals do not overlap, i.e. $h_\epsilon(x) = 0$.
- Case 2: If $a+\epsilon \le x \le b-\epsilon$ then $h_\epsilon(x) = \frac{1}{2\epsilon}(2\epsilon) = 1$.
- Case 3: If $a-\epsilon \le x < a+\epsilon$ then $h_\epsilon(x) = \frac{x+\epsilon-a}{2\epsilon}$.
- Case 4: If $b-\epsilon < x \le b+\epsilon$ then $h_\epsilon(x) = \frac{b-x+\epsilon}{2\epsilon}$.
Now, $$ \|h_\epsilon - \chi_{[a,b]}\|_1 = \int_{-\infty}^{\infty} |h_\epsilon(x) - \chi_{[a,b]}(x)| dx $$ The integrand has its support in $[a-\epsilon, a+\epsilon]$ and $[b-\epsilon, b+\epsilon]$.
Integral around $a$: $\int_{a-\epsilon}^{a+\epsilon} |h_\epsilon(x) - \chi_{[a,b]}(x)| dx$
- For $x \in [a-\epsilon, a)$, $\chi_{[a,b]}(x) = 0$ and $h_\epsilon(x) = \frac{x-a+\epsilon}{2\epsilon}$.
- For $x \in [a, a+\epsilon]$, $\chi_{[a,b]}(x) = 1$ and $h_\epsilon(x) = \frac{x-a+\epsilon}{2\epsilon}$. And we get
$$ \int_{a-\epsilon}^{a} \left|\frac{x-a+\epsilon}{2\epsilon} - 0\right| dx + \int_{a}^{a+\epsilon} \left|\frac{x-a+\epsilon}{2\epsilon} - 1\right| dx =\frac \epsilon 2 $$
Integral around $b$: $\int_{b-\epsilon}^{b+\epsilon} |h_\epsilon(x) - \chi_{[a,b]}(x)| dx$
By symmetry, the area will also be $\frac{\epsilon}{2}$.
Hence the total $L^1$-norm will be: $$ \|\chi_{[a,b]}*g_\epsilon - \chi_{[a,b]}\|_1 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \to 0, \quad \epsilon \to 0. $$
Now, let $t$ be a general step function, $t(x) = \sum_{i=1}^n c_i \chi_{I_i}(x)$. We use the linearity of the convolution operator and the triangle inequality:
$$ \|t*g_\epsilon - t\|_1 = \left\| \left(\sum_{i=1}^n c_i \chi_{I_i}\right)*g_\epsilon - \left(\sum_{i=1}^n c_i \chi_{I_i}\right) \right\|_1 \\ = \left\| \sum_{i=1}^n c_i \left( \chi_{I_i}*g_\epsilon - \chi_{I_i} \right) \right\|_1 \\ \le \sum_{i=1}^n |c_i| \left\| \chi_{I_i}*g_\epsilon - \chi_{I_i} \right\|_1 \\= \epsilon \left( \sum_{i=1}^n |c_i| \right) $$ The term $C = \sum_{i=1}^n |c_i|$ is a finite, positive constant that depends only on the step function $t$, not on $\epsilon$. Therefore, we have: $$ 0 \le \|t*g_\epsilon - t\|_1 \le C \cdot \epsilon \to 0.$$
