# Weierstrass Approximation Theorem on $\frac{1}{x}$

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.

• For any $P$, $\sup_{(0,1]} |f(x) - P(x)| = +\infty$ – user371663 Mar 2 '18 at 15:42
• Yes, is that really it? Is it because the supremum of $f(x)-p_n(x)$ on (0,1] is $\infty$? – ʎpoqou Mar 2 '18 at 15:43
• Yes it is always $\infty$ so it will not converges to $0$. – user371663 Mar 2 '18 at 15:45
• @Lucas What about a $xsin(\frac{1}{x})$ on the same interval? This seems to satisfy the conditions unless I am missing something. – ʎpoqou Mar 2 '18 at 15:47
• Sorry, you're true. – user371663 Mar 2 '18 at 15:57

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.

• What if we consider the interval say $[1,\infty)$ ? Polynomials wont be bounded there, so the only $f(x)$ that can be approximated by polynomials is a polynomial of the same degree. How can I show that $\frac{1}{x}$ is not a polynomial? – ʎpoqou Mar 2 '18 at 18:22

Since the OP is interested about unifom convergence on interval like $[1,\infty)$ ;

If $f \mapsto_{\infty} l$ then $f$ has to be equal to $l$ on $[1,\infty)$ to allow the uniform convergence of a sequence of polynomial towards $f$. Indeed, if you take $P$ which is not constant then $\sup_{[1,\infty)}|f(x)-P(x)| = \infty$

since $|P(x)| \to_{\infty} \infty$ while $f \to l$

And, if you take $P \neq l$ then there would exists $C>0$ (which is valid for all such $P$) such that $\sup_{[1,\infty)}|f(x)-P(x)|≥ C$ since $f \to l$. The same if $P=l$ since $f$ is not contant.

Actually, if $(p_n)$ converges uniformly towards $f$ on $[1,+∞)$, $f$ is a polynomial.

Proof:

$\exists N \in \mathbb N, \forall n ≥ N, \forall x \in \mathbb R, |P_n(x)-P_N(x)|≤1$

For all $n ≥ N, P_n - P_N$ is bounded on $[1,\infty)$, hence it is constant.

$\forall n ≥ N, P_n = P_N + a_n$

Otherwise $(P_n(1))$ converges, so (a_n) converges too, towards $a$.

Finally, $\forall x \in [1, +\infty), f(x) = \lim_{n \to \infty} P_n(x) = P_N(x) + a$.

So $f = P_N + a$ is a polynomial.

As a consequence, one can prove there is no sequence of polynomial that converges toward $f$ on $[1, \infty)$ by showing $f$ is not a polynomial (just as showing $f$ is not continuous would suffice (since an uniform limit of continuous function is continuous)).

One way is to prove the derivative $f^{(n)} \neq 0, \forall n$.

• For answering the question, the above answer is what was needed. But this is amazing! Thank you for taking the time to write this up @Lucas – ʎpoqou Mar 4 '18 at 12:26