Good day, we have just covered uniform continuity and polynomial approximations of continuous functions and I do not quite think I have the hang of it. We were given an example but it wasn't thoroughly explained.
$f:(0,1] \rightarrow \Bbb R \ \ $ where $f(x)=\frac{1}{x}$
Weierstrass' Approximation Theorem fails to hold here and so there does not exist a sequence of polynomials ${p_n}$ s.t $\ p_n \rightarrow f$ uniformly
Can someone explain why that is the case? We were told that it is because the interval is not closed but I wanted more explanation on this.
Best Answer
In general, if $X$ is any set whatever and $P_n$ is a sequence of bounded functions on $X$ which converge uniformly to a function $f$, then $f$ is bounded. Proof: there exists $n$ such that $|P_n(x) - f(x)| \le 1$ for all $x \in X$, and there exists $M$ such that $|P_n(x)| \le M$ for all $x \in X$. Then for all $x \in X$, $|f(x)| \le |P_n(x)| + |P_n(x) - f(x)| \le M+1$. Continuity plays no role in this part of the proof.
In the original situation, each polynomial is bounded on $(0, 1]$ since it extends continuously to $[0, 1]$. (This is the only place that continuity enters.) Since $f(x) = 1/x$ is not bounded on $(0,1]$, it cannot be uniformly approximated by polynomials.