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Let $C\subset \mathbb{R}$ be compact and let $M=\Vert f{\Vert }_{\infty }$. Since every $k_n$ is supported in $[-R,R]$, the function $f$ is uniformly continuous on the compact setGiven $\epsilon >0$, choose $\delta >0$ so thatwhenever $x\in C$ and $|t|<\delta$. Since $\int k_n=1$ and $k_n\ge 0$,Property 3 makes the final term smaller than $\epsilon /2$ for all sufficiently large $n$, uniformly in $x\in C$. Hence $f_n\to f$ uniformly on every compact set. The sequence $(k_n)$ is an approximate identity.

For the second part, extend $g$ to a continuous function $\tilde{g}$ on $\mathbb{R}$ by setting it equal to zero outside $[0,1]$; continuity at the endpoints uses $g(0)=g(1)=0$. DefineThese nonnegative kernels have integral one. For every $\delta >0$, their mass outside $[-\delta ,\delta ]$ tends to zero exponentially relative to the mass near zero, so they satisfy property 3.

For $x\in [0,1]$, the convolution isBecause $|x-y|\le 1$ on the square $[0,1{]}^2$, no cutoff remains in this formula. Expanding the $n$th power shows that $p_n$ is a polynomial in $x$. The first part, applied to the compact interval $[0,1]$, givesThis proves the Weierstrass approximation theorem for functions with the stated endpoint values.

For $1\le p<\infty$ and $f\in L^p({\mathbb{R}}^d)$, translations converge to $f$ in the Lp space norm. First prove this for continuous compactly supported functions, using uniform continuity and a common bounded support. Their density in Lp space, together with the norm-preserving translation map and the triangle inequality, proves the assertion for arbitrary $f$. It implies convergence of averages over shrinking intervals, a basic approximate identity argument. The corresponding assertion fails for general $L^{\infty }$ functions, such as an indicator function with a jump.