1
$\begingroup$

Prove or disprove that there does not exist a real valued continuous function $g$ on $[0,1]$ with $g (x ) \neq x$ for all $x \in(0,1)$ such that given any $\varepsilon > 0$ and any real valued continuous function $f$ on $[0,1]$ there exist real numbers $a_1, a_2,...,a_n$ (depending only on $f$,$g$ and $\varepsilon$) such that $$|f(x)-\sum_{k=0}^na_k(g(x))^k|<\varepsilon$$for all $x\in[0,1]$.

I believe the statement is false, but I cannot really prove it. The reason I believe it is false is that I should not be able to find a polynomial in a continuous function which approximates EVERY $f$. I feel this is a consequence of Weierstrass Approximation Theorem. Any hint is appreciated.

The first question that came to my mind is: Should $g$ be a polynomial to start with? But even then, how one can proceed is eluding me. I tried to take $f(x)=x^d$ for natural numbers $d$ but nothing comes to my mind.

$\endgroup$
1
$\begingroup$

Well, without the assumption "$g (x ) \neq x$ for all $x \in(0,1)$", putting $g(x)=x$ fits the bill (Weierstrass killin' it).

You can tweak that a little bit and set $g(x)=\frac{x}{2}$. This satisfies all the requirements.

$\endgroup$
  • $\begingroup$ Thanks, so such a $g$ indeed exists, in fact $g(x)=\dfrac{x}{2}$. I was getting confused with $n$ as I could not make sure whether $n$ is fixed for every $f$ or not. $\endgroup$ – Landon Carter Apr 13 '15 at 20:01
  • $\begingroup$ At this point, I would like to ask a question. Is there any theory on approximation of continuous functions by polynomials in continuous functions? My professor had once said that it is possible if injectivity is given, but I do not seem to get it. $\endgroup$ – Landon Carter Apr 13 '15 at 20:03
  • $\begingroup$ I don't know. Fourier series basically tells you you can approximate even discontinuous functions with infinite sums of cosines and sines, but that's not really the answer to your question. $\endgroup$ – DetroitRapper Apr 13 '15 at 20:06
  • $\begingroup$ Yes ok. Anyway, at least this question has its answer!! Thank you and I am going to accept it. $\endgroup$ – Landon Carter Apr 13 '15 at 20:09

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.