Let \(f:[0,1]\to\RR\) be continuous and let \(\varepsilon>0\). Show that there exist \(n\in\ZZ_{\ge0}\) and rational numbers \(C_0,\dots,C_n\) such that \[ g(x)=\sum_{k=0}^n C_kx^{4k} \] satisfies \[ |g(x)-f(x)|<\varepsilon \] for every \(x\in[0,1]\).
Solution
1. The function\[ h:[0,1]\longrightarrow\RR, \qquad h(t)\coloneqq f(t^{1/4}), \]is continuous.
The map \(t\mapsto t^{1/4}\) is continuous from \([0,1]\) to \([0,1]\), and \(f\) is continuous by hypothesis. Their composition is therefore continuous.
2. There is a real polynomial\[ p(t)=\sum_{k=0}^n a_kt^k \]such that\[ |p(t)-h(t)|<\frac{\varepsilon}{2} \]for every \(t\in[0,1]\).
Apply the Weierstrass approximation theorem to the continuous function \(h\) from step 1, with tolerance \(\varepsilon/2\).
3. There are rational numbers\[ C_0,\ldots,C_n\in\QQ \]such that the polynomial\[ q(t)\coloneqq\sum_{k=0}^n C_kt^k \]satisfies\[ |q(t)-p(t)|<\frac{\varepsilon}{2} \]for every \(t\in[0,1]\).
Since \(\QQ\) is dense in \(\RR\), for each \(k\) choose \(C_k\in\QQ\) with \[ |C_k-a_k|<\frac{\varepsilon}{2(n+1)}. \] For \(0\le t\le1\), \[ \begin{aligned} |q(t)-p(t)| &\le \sum_{k=0}^n |C_k-a_k|t^k\\ &\le \sum_{k=0}^n |C_k-a_k|\\ &< (n+1)\frac{\varepsilon}{2(n+1)} = \frac{\varepsilon}{2}. \end{aligned} \]
4. Define\[ g(x)\coloneqq q(x^4) = \sum_{k=0}^n C_kx^{4k}. \]Then\[ |g(x)-f(x)|<\varepsilon \]for every \(x\in[0,1]\).
Fix \(x\in[0,1]\) and put \(t=x^4\). Then \(t\in[0,1]\) and \[ h(t)=f(t^{1/4})=f(x). \] By steps 2 and 3, \[ \begin{aligned} |g(x)-f(x)| &= |q(t)-h(t)|\\ &\le |q(t)-p(t)|+|p(t)-h(t)|\\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \end{aligned} \]
5. Q.E.D.
Step 4 gives the required polynomial with rational coefficients and only powers \(x^{4k}\).