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.

Trigonometric polynomial is defined as a function $$ f(x)=\sum_{n=1}^ka_ne^{i\lambda_nx}$$ for some positive integer $k$, complex coefficients $\{a_n\}$ and real coefficients $\{\lambda_n\}$. Let TP be the vector space of all trigonometric polynomials, we want to define an inner product on this space by $$\langle f,g\rangle=\lim_{T\to\infty} \frac{1}{2T}\int^T_{-T}\! f(x)\overline{g(x)} \:\mathrm{d}x$$ All the properties of the inner product are obvious, except for positive-definiteness. Consider the inner product of a trigonometric polynomial $f$ with itself: \begin{align*} \langle f,f\rangle&=\lim_{T\to\infty} \frac{1}{2T}\int^T_{-T}\! f(x)\overline{f(x)}\: \mathrm{d}x\\ &=\lim_{T\to\infty} \frac{1}{2T}\int^T_{-T}\!\left( \sum_{n=1}^k a_ne^{i\lambda_n x}\right)\left( \sum_{m=1}^k \overline{a_m}e^{-i\lambda_m x}\right)\: \mathrm{d}x\\ &=\sum_{n,m=1}^ka_n\overline{a_m}\lim_{T\to\infty} \frac{1}{2T}\int^T_{-T}\!e^{i(\lambda_n-\lambda_m)x}\:\mathrm{d}x \end{align*} Now, consider the limit $$ \lim_{T\to\infty} \frac{1}{2T}\int^T_{-T}\!e^{i(\lambda_n-\lambda_m)x}\:\mathrm{d}x$$ If $n=m$, the limit equals to one, but what if $n\neq m$? Do we get this limit to be $\delta_{mn}$?

share|improve this question
    
You kind of need $\lambda_n\neq \lambda_m$ when $m\neq n$. You can guarantee that by combining terms in the original polynomial. –  Thomas Andrews Jan 8 '13 at 14:19
    
If $n=m$, the integral equals one, not zero. –  Eckhard Jan 8 '13 at 14:20
    
Also, when $n=m$ the integral is $2T$, not $0$. –  Thomas Andrews Jan 8 '13 at 14:20
    
@Eckhard Depends on whether you consider the $1/2T$ part of the integral, hence our disagreement. But it is definitely not zero :) –  Thomas Andrews Jan 8 '13 at 14:21
    
sorry, I meant that the limit equals one when $m=n$. –  Jimmy R Jan 8 '13 at 14:23
add comment

2 Answers 2

up vote 1 down vote accepted

If $n=m$ and thus $\lambda_n=\lambda_m$, the integral $\frac{1}{2T}\int_{-T}^T{e^{i(\lambda_n-\lambda_m)x}dx}=\frac{1}{2T}\int_{-T}^T{1dx}$ equals one, not zero. If $\lambda_n\neq\lambda_m$, you have $$ \int_{-T}^T{e^{i(\lambda_n-\lambda_m)x}dx}=\frac{2}{\lambda_n-\lambda_m}\sin((\lambda_n-\lambda_m)T), $$ and thus $$ \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T{e^{i(\lambda_n-\lambda_m)x}dx}=0. $$

share|improve this answer
1  
Isn't it $2\sin(\lambda_n-\lambda_m)T$? –  Thomas Andrews Jan 8 '13 at 14:27
1  
@ThomasAndrews: Thanks for spotting this. I edited my answer. –  Eckhard Jan 8 '13 at 15:30
add comment

You need that $\lambda_n\neq \lambda_m$ for $m\neq n$.

If $\lambda_n\neq \lambda_n$ then the indefinite integral of $e^{i(\lambda_n-\lambda_m)x}$ is $$\frac{1}{i(\lambda_n-\lambda_m)}e^{i(\lambda_n-\lambda_m)x}$$

When evaluated from $-T$ to $T$, this is bounded by $$\frac{2}{\lambda_n-\lambda_m}$$

So multiplying by $\frac{1}{2T}$ and letting $T\to\infty$ yields 0.

On the other hand, you got the case $m=n$ wrong. In that case, the integral is $2T$, and $\frac{2T}{2T}\to 1$ as $T\to\infty$.

share|improve this answer
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.