Weierstrass approximation theorem

Theorem

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function on the interval $[a,b]$. Then, for any positive $\epsilon$, there exists $n\in \mathbb{N}$ where for some $n$-degree polynomial $p_n$, $\Vert f-p_n{\Vert }_{C^0}\le \epsilon$ holds.

Proof

Without loss of generality, let $a=0$ and $b=1$. Since $f$ is continuous on $[0,1]$, let $M$ as:

$$M=\underset{x\in [0,1]}{\sup }|f(x)|:=\Vert f{\Vert }_0.$$

Consequently, for any $\xi \in [0,1]$, the following holds:

$$|f(x)-f(\xi )|\le \{\begin{array}{llc}\frac{\epsilon }{2}& & \text{if }|x-\xi |<\delta \\ |f(x)|+|f(\xi )|& \le 2M& \\ & \le 2M\cdot 1& \\ & \le 2M{\left(\frac{|x-\xi |}{\delta }\right)}^2& \text{otherwise.}\end{array}$$

Therefore we can conclude that

$$|f(x)-f(\xi )|\le 2M{\left(\frac{|x-\xi |}{\delta }\right)}^2+\frac{\epsilon }{2}.$$

Now, we will prove that the polynomial of interest is actually the Bernstein polynomial.

$$\begin{array}{rl}|B_n(x;f)-f(\xi )|& =|B_n(x;f)-B_n\left(x;f(\xi )\right)|=|B_n\left(x;f-f(\xi )\right)|\\ & =\left|\sum _{\nu =0}^n\left(f\left(\frac{\nu }{n}\right)-f(\xi )\right)\left(\genfrac{}{}{0ex}{}{n}{\nu }\right)x^{\nu }(1-x{)}^{n-\nu }\right|\\ & \le \sum _{\nu =0}^n\left|f\left(\frac{\nu }{n}\right)-f(\xi )\right|\left(\genfrac{}{}{0ex}{}{n}{\nu }\right)x^{\nu }(1-x{)}^{n-\nu }\\ & =B_n\left(x;|f-f(\xi )|\right)\\ & \le B_n\left(x;2M{\left(\frac{|x-\xi |}{\delta }\right)}^2+\frac{\epsilon }{2}\right)\\ & =\frac{2M}{{\delta }^2}\left(x^2+\frac{1}{n}(x-x^2)\right)-\frac{4M\xi }{{\delta }^2}+\frac{2M{\xi }^2}{{\delta }^2}+\frac{\epsilon }{2}\end{array}$$

The last line is due to the properties of Bernstein polynomials.

Next, let $\xi =x$. Then consequently,

$$\begin{array}{rl}|B_n(x;f)-f(x)|& \le \frac{2M}{{\delta }^2}\cdot \frac{\xi -{\xi }^2}{n}+\frac{\epsilon }{2}\\ & \le \frac{M}{2{\delta }^2}\cdot \frac{1}{n}+\frac{\epsilon }{2}\end{array}$$

The last line is due to the fact that the maximum value of the function $x-x^2$ on the interval $[0,1]$ is $\frac{1}{4}$.

Finally, choose $n$ such that it satisfies $n\ge \frac{M}{{\delta }^2\epsilon }$. Then for any $x\in [0,1]$,

$$|B_n(x;f)-f(x)|\le \epsilon$$

and equivalently,

$$\Vert B_n(\cdot ;f)-f{\Vert }_{C^0}\le \epsilon .$$

This completes the proof. \tag*{$||$}