I am studying for my introductory real analysis final exam, and here is a problem I am somewhat stuck on. It is Question 2, in page 3 of the following past exam (no answer key unfortunately!):


Give an example of each of the following, together with a brief explanation of your example. If an example does not exist, explain why not.

(c) A continuous function $f : (−1,1) → \mathbb{R}$ that cannot be uniformly approximated by a polynomial.

By Weierstrass Approximation Theorem, every continuous real-valued function on closed interval can be uniformly approximated by a sequence of polynomials. Since in this question the domain of the function is an open interval $(-1, 1)$, I have a feeling that such example must exist.

My attempts: The proof of Weierstrass approximation theory uses the fact that a continuous function a compact set (a closed interval by Heine-Borel Theorem) achieves a maximum, so we can guess that the example we are looking after will not achieve a maximum on $(-1, 1)$. Such example of continuous function is $$ f(x)=\frac{1}{x+1} $$ So now my question: is it true $f$ cannot be uniformly approximated by a sequence of polynomials? And if so, how do proceed to prove such a statement?



Your feeling seems to be right. :-D Hint: each polynomial should be bounded on $(0,1)$.

  • $\begingroup$ Thanks! :) It makes sense! $\endgroup$ – Prism Apr 18 '13 at 10:46
  • $\begingroup$ I have written my answer based on your hint below. $\endgroup$ – Prism Apr 18 '13 at 10:52

Based on Alex Ravsky's hint, I have found the solution. I will type it up for the sake of reference.

We claim that the function $$ f(x)=\frac{1}{x+1} $$ is a continuous function on $(-1, 1)$ that cannot be approximated by a polynomial. Assume not. Then, for $\epsilon=1$ in the definition of uniform convergence, there exists a polynomial $p(x)$ such that $$ |f(x)-p(x)|\le 1 $$ for $all$ $x\in (-1, 1)$. Since the polynomial $p(x)$ is bounded on $(-1, 1)$, it follows that, there exists a constant $M$ such that $|p(x)|\le M$ for all $x\in (-1, 1)$. But then, $$ |f(x)|\le |p(x)| + |f(x)-p(x)| \le M+1 $$ for all $x\in (-1,1)$ which contradicts the fact that $f(x)$ is unbounded on $(-1, 1)$.

  • $\begingroup$ This solution is an exact realization of my idea, so it seems to be OK and I vote up it. :-) $\endgroup$ – Alex Ravsky Apr 18 '13 at 12:18
  • $\begingroup$ This is a really old post, but I wanted to thank you for writing this up! $\endgroup$ – Alex Mar 14 '17 at 14:40
  • $\begingroup$ @Alex I am happy to hear that it helped :) $\endgroup$ – Prism Mar 14 '17 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.