Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is one of the problem in our past comprehensive exams. I don't mind getting full solution.

Suppose $f$ is a bounded, measurable function on $[0,1]$, $\epsilon>0,$ and for all $x>\epsilon\,$ one has

$$0=\int_0^1 f(s )\exp(-xs)ds$$

Show that $f=0$ almost everywhere.

Someone gave me a hint to solve the problem using Urysohn's lemma. I am not totally comfortable with that lemma. I have a hunch that we can prove this along the line of Fourier analysis. I am not that sure on this approach either. I don't even know how to get started.

share|improve this question
    
Integral is from $0$ to $1$, I don't know how to incorporate that in the integral sign. I would appreciate if somebody do that for me. –  Deepak Dec 20 '12 at 20:46
    
is this true for all epsilon or simply a fixed epsilon? If it was a given epsilon, the result may not be true, although I haven't fully thought it through. –  toypajme Dec 20 '12 at 21:22
    
@toypajme, For me it looks like for all epsilon. This is all the information I have. –  Deepak Dec 20 '12 at 21:24
add comment

1 Answer

up vote 4 down vote accepted

Some hints, hoping they will be useful. Expand the exponential as a power series to deduce that $\int_{[0,1]}f(s)s^nds=0$ for all $n$. This gives, by Stone-Weierstrass theorem that $\int_{[0,1]}f(s)g(s)ds=0$ for all $g$ continuous on $[0,1]$. We conclude from this answer.

share|improve this answer
1  
To finish: this implies all Fourier coefficients are zero, so the function is zero a.e. See also math.stackexchange.com/questions/17026/…. –  Potato Dec 20 '12 at 21:40
    
@Davide Giraudo Seems, I still can not follow the proof from those problems mentioned in the link. I have difficulty understanding the use of Stone-Weierstrass theorem. Can somebody explain more explicitly. Sorry for being so dumb here. –  Deepak Dec 21 '12 at 19:36
    
@Deepak Stone-Weierstrass gives that polynomials are dense in the space of continuous functions on $[0,1]$ endowed with the supremum norm. The equality $\int_{[0,1]}f(s)P(s)ds=0$ holds for any polynomial $P$. Using boundedness of $f$ on $[0,1]$, we get for all sequence of polynomials $\{P_n\}$ and all continuous function $g$ that $|\int_{[0,1]}f(s)g(s)ds|\leqslant \sup_{[0,1]}|f|+\lVert g-P_n\rVert_{\infty}$. Now choosing a good sequence we get what we want. –  Davide Giraudo Dec 23 '12 at 11:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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