Fourier series expansion of Dirac delta function

Question

My question is from Arfken & Weber (Ed. 7) 19.2.2:

In the first part, the question asks for Fourier series expansion of $\delta(x)$. I have found $$\delta(x)=1/2\pi + 1/\pi\sum^{\infty}_{n=1} \cos(nx)$$ Then by using the identity $$\sum^{N}_{n=1} \cos(nx)=\frac{\sin(Nx/2)}{\sin(x/2)}\cos\left[\left(N+\frac{1}{2}\right)\frac{x}{2}\right]$$, we need to find a Fourier representation which is consistent with $$\delta(x-t)=\sum^{\infty}_{n=0}\phi^{*}_{n}(t)\phi_{n}(x)$$. I have tried expanding the trigonometric functions with exponential terms countless times, however could not obtain a sufficient result. Any help is appreciated.

Cheers

Answer 1

To solve this problem you can utilize the resolution of the identity [1]. As explained in [1], take a complete orthonormal basis set of functions $\left\{\phi_n\right\}$.

For example, for the domain $x\in[0,2\pi)$, take $$\left\{\phi_n\right\} = \left\{ \frac{ a_m\cos{\left(m\,x\right)} + b_m\sin{\left(m\,x\right)} } { \sqrt{ 2\,\pi \left|a_m\right|^2 \,\delta_{m,0} + \pi \left(\left|a_m\right|^2 + \left|b_m\right|^2\right) \left(1-\delta_{m,0}\right) } } \mid m=0,1,2,\ldots \right\} .$$ Again, as explained in [1], from the resolution of the identity you have that $$ \delta(x-t) = \sum_{n=1}^\infty \varphi_n (x)\, \varphi_n^*(t). $$ Specifically, with the basis set of functions above, we have that $$ \delta(x-t) = \sum_{m=0}^\infty \frac{ \left[ a_m\cos{\left(m\,x\right)} + b_m\sin{\left(m\,x\right)} \right]\left[ a_m^*\cos{\left(m\,t\right)} + b_m^*\sin{\left(m\,t\right)} \right] } { { 2\,\pi \left|a_m\right|^2 \,\delta_{m,0} + \pi \left(\left|a_m\right|^2 + \left|b_m\right|^2\right) \left(1-\delta_{m,0}\right) } } . $$

As another example, for the domain $x\in[0,2\pi)$, take $$\left\{\phi_n\right\} = \left\{ \frac{ \exp{\left(+i\,m\,x\right)} } { \sqrt{ 2\,\pi} } \mid m=\ldots,-2,-1,0,1,2,\ldots \right\} .$$ Again, as explained in [1], from the resolution of the identity you have that $$ \delta(x-t) = \sum_{n=1}^\infty \varphi_n (x)\, \varphi_n^*(t). $$ Specifically, with the basis set of functions above, we have that $$ \delta(x-t) = \sum_{m=-\infty}^\infty \frac{ \exp{\left(+i\,m\,x\right)} \, \exp{\left(-i\,m\,t\right)} } { 2\,\pi } . $$

Bibliography

[1] https://en.wikipedia.org/wiki/Dirac_delta_function

Answer 2

It is not toooo difficult to talk precisely about convergence of Fourier series of distributions, although certainly pointwise (or $L^2$) convergence is "out the window". And we needn't go so far as to just ask for weak-*-convergence, although that's fine, too: we can easily talk about convergence in $L^2$-Sobolev spaces on products $\mathbb T^n$ of circles $\mathbb T$.

For example, on $\mathbb T^2$, for real $s$, $H^s$ is (for example) the completion of the space of finite sums $\sum_{m,n}c_{m,n}\,e^{2\pi i(mx+ny)}$ with respect to the Sobolev norm whose square is $$ \Big|\sum_{m,n}|c_{m,n}\,e^{2\pi i(mx+ny)}\Big|^2_{H^s} \;=\; \sum_{m,n} |c_{m,n}|^2\cdot (1+m^2+n^2)^s $$ The Sobolev imbedding theorem proves that $H^\infty=\bigcap_s H^s= C^\infty$, so $H^{-\infty}=\bigcup H^s$ is all distributions on $\mathbb T^2$.

The question asks for the Fourier expansion of the "integrate along the diagonal" distribution. Since it is compactly-supported on $\mathbb T^2$, we correctly compute the coefficients by applying the distribution to the smooth functions $(x,y)\to e^{2\pi i(mx+ny)}$, by integrating along the diagonal. Thus, the $m,n$-th coefficient is $0$ unless $m=n$, in which case it is $1$. The resulting Fourier series converges in the Hilbert space $H^{1+\varepsilon}$ for every $\varepsilon>0$, by direct computation.