Section 2.7 Weierstrass Theorem
Theorem 2.7.1. Weierstrass Theorem.
Let \(-\infty < a < b < \infty\text{.}\) Every continuous \(f:[a,b]\to\mathbb R\) can be uniformly approximated by polynomials. In other words, for every continuous \(f:[a,b] \to \mathbb{R}\) there is a sequence of polynomials \((p_n(f))_{n \in \mathbb{N}}\) so that
\begin{equation*}
\sup_{x \in [a,b]}|p_n(f)(x)-f(x)| {\xrightarrow{n \to \infty}}\ 0.
\end{equation*}
The proof of
Theorem 2.7.1 introduces a very useful idea in analysis:
convolution and
approximation of identity.
First, a reduction is done so that we may assume \([a, b]=[0, 1]\text{,}\) and \(f(a)=f(0)=0, f(b)=f(1)=0\text{.}\) Next we extend \(f\) to be \(0\) for \(x\in \mathbb R\setminus [0, 1]\text{,}\) so that it becomes a continuous function on \(\mathbb R\text{.}\) Then we define the following convolution of \(f\)
\begin{equation*}
P_{n}(x)=\int_{\mathbb R} f(s)Q_{n}(s-x)\, ds
\end{equation*}
where \(Q_{n}(x)\) will be a polynomial , chosen to satisfy the following three properties:
-
\(Q_{n}(x) \ge 0\text{.}\)
-
\(\int_{-1}^{1} Q_{n}(x)\, dx =1\) for all \(n\text{.}\)
-
For any \(0 < \delta < 1, \int_{-\delta}^{\delta} Q_{n}(x)\, dx \to 1\) as \(n\to \infty\text{.}\)
Since \(Q_{n}(s-x)\) is a polynomial in \(x\text{,}\) it is clear that \(P_{n}(x)\) is also a polynomial in \(x\text{,}\) and using \(f=0\) for \(x\le 0\) or \(x\ge 1\text{,}\) we see that, for \(0\le x \le 1\text{,}\)
\begin{equation*}
P_{n}(x)= \int_{-1+x}^{1+x} f(s)Q_{n}(s-x)\, ds= \int_{-1}^{1} f(x+t) Q_{n}(t)\, dt.
\end{equation*}
This indicates that \(P_{n}(x)\) is a weighted average of \(f\text{,}\) with more weight near \(t=0\) due to item (3) above. We now use items (1)--(3) to see that
\begin{align*}
\amp |f(x)-P_{n}(x)|\\
= \amp\left| \int_{-1}^{1} \left(f(x)-f(x+t)\right) Q_{n}(t)\, dt\right|\\
\le \amp \left(\int_{-1}^{\delta} + \int_{\delta}^{1} \right) |f(x)-f(x+t)|Q_{n}(t) \, dt +
\int_{-\delta}^{\delta} |f(x)-f(x+t)|Q_{n}(t) \, dt\\
\le \amp 2 \max |f| \left(\int_{-1}^{\delta} + \int_{\delta}^{1} \right) Q_{n}(t) \, dt +
\int_{-\delta}^{\delta} |f(x)-f(x+t)|Q_{n}(t) \, dt.
\end{align*}
Now for any \(\epsilon>0\text{,}\) using the uniform continuity of \(f\text{,}\) we find a \(\delta >0\text{,}\) such that \(|f(x)-f(x+t)| < \epsilon \text{,}\) for all \(t,
|t| < \delta\text{.}\) Then use this \(\delta\) in the above, and item (3), there exists some \(N\) such that for \(n\ge N\) we have \(2 \max |f| \left(\int_{-1}^{\delta} + \int_{\delta}^{1} \right) Q_{n}(t) \, dt < \epsilon\text{,}\) and it follows that
\begin{equation*}
|f(x)-P_{n}(x)| < 2\epsilon.
\end{equation*}
Finally, we can check that, with \(Q_{n}(x)=c_{n}(1-x^{2})^{n}\text{,}\) and \(c_{n}\) chosen to satisfy item (2) above, then items (1) and (3) also hold---for item (3), we need to verify that for any \(0 < \delta < 1\text{,}\)
\begin{equation*}
\frac{\int_{\delta}^{1} (1-x^{2})^{n}\, dx}{ \int_{0}^{\delta} (1-x^{2})^{n}\, dx}\to 0 \text{ as }
n\to \infty.
\end{equation*}
Question.
If \(f\in C^{1}[a,b]\text{,}\) can the proof above be adapted to prove that there exists a sequence of polynomials \(P_{m}\) such that
\begin{equation*}
P_{m}\to f \text{ and } P_{m}' \to f' \text{ uniformly on } [a, b]?
\end{equation*}
