Let $f(x)=x^2$ and let
$$
K_n(x) = \left\{
\begin{array}{ll}
\frac{n}{2} & \text{if } |x|<\frac{1}{n} \\
0 & \text{else}.
\end{array}
\right.
$$
Compute
$$\displaystyle f*K_n(x) = \int_{-\infty}^{\infty} f(y)K_n(x-y) dy$$
explicitly and show that $f*K_n(x) \rightarrow f(x)$ as $n\rightarrow\infty$.
First, we can plug the simple kernel ($K_n(x)$) into the formula for $f*K_n(x)$. We can also plug $x^2$ in the place of $f(y)$, by simply exchanging $x$ for $y$.
$$\displaystyle f*K_n(x) = \int_{-\infty}^{\infty} y^2K_n(x-y) dy$$
To successfully plug in the kernel we must divide the integral into ${-\infty}$ to $-1/n$, then $-1/n$ to $1/n$ and $1/n$ to ${\infty}$. The integral must be divided because the kernel has a different value for each section. The bounds of integration also have to be changed because we are using $y$ as our variable.
Thus the upper bound becomes $x+1/n$ and the lower bound becomes $x-1/n$. This is found by solving the inequality $-1/n < x-y < 1/n$.
Thus
$$\displaystyle f*K_n(x) = \int_{-\infty}^{x-1/n} y^2(0) dy+ \int_{x-1/n}^{x+1/n} y^2(n/2) dy\int_{x+1/n}^{\infty} y^2(0) dy$$
This means the first and last integral are negligible so the equation simplifies to:
$$\displaystyle f*K_n(x) = \int_{x-1/n}^{x+1/n} y^2(n/2) dy$$
Once integrated this becomes:
$$f*K_n(x) = (n/2)[1/3(x-1/n)^3-1/3(x+1/n)^3]$$
Which simplifies to:
$$f*K_n(x) = (n/2)[\frac{2x^2}{n}+\frac{2}{3n^3}]$$
And then simplifies more to:
$$f*K_n(x) = x^2+\frac{1}{3n^2}$$
As $n\rightarrow\infty$ the equation $f*K_n(x)$ approaches $x^2$ which is exactly $f*K_n(x) \rightarrow f(x)$