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I am currently writing a small paper regarding the analogy between Fourier analysis and linear algebra for my mathematics class. Unfortunately, I am stuck at proving that $\mathrm{e}^{2\pi\mathrm{t}\omega t}$ forms an orthonormal basis in the $L^2(\mathbb{R})$ for all $\omega\in\mathbb{R}$.

In my paper I already showed that $\mathrm{e}^{2\pi\mathrm{i} n/T}\in L^2([-T/2;T/2])$ forms an orthonormal basis for all $n\in\mathbb{Z}$ such that $$ \langle \mathrm{e}^{2\pi\mathrm{i} n/T},\mathrm{e}^{2\pi\mathrm{i} m/T}\rangle=\frac{1}{T}\int_{-T/2}^{T/2}\mathrm{e}^{2\pi\mathrm{i} n/T}\overline{\mathrm{e}^{2\pi\mathrm{i} m/T}}\mathrm{d}t=\delta_{nm}. $$ Therefore, the Fourier series of a function $f(t)\in L^2([-T/2;T/2])$ can be interpreted as an orthogonal projection of $f$ onto the individual basis functions with $$ f(t)=\sum_{k=-\infty}^{\infty}\underbrace{\left\langle f(t),\mathrm{e}^{2\pi\mathrm{i} kt/T}\right\rangle}_{=c_k} \mathrm{e}^{2\pi\mathrm{i} kt/T} $$ When we extend the Fourier series to a non-periodic function $f(t)\in L^2(\mathbb{R})$, we get the Fourier transform and the inverse Fourier transform give by $$ f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(\omega)\mathrm{e}^{2\pi\mathrm{i} \omega t}\mathrm{d}\omega, \quad \hat{f}(\omega)=\int_{-\infty}^{\infty}f(t)\mathrm{e}^{-2\pi\mathrm{i}\omega t}\mathrm{d}t $$ In order for the Fourier transform to work I would assume that $\mathrm{e}^{2\pi\mathrm{t}\omega t}$ also has to form an orthonormal basis for each $\omega\in\mathbb{R}$ in the $L^2(\mathbb{R})$. But can we prove something like $$ \forall \omega_1,\omega_2\in\mathbb{R}\colon \langle \mathrm{e}^{2\pi\mathrm{i} \omega_1},\mathrm{e}^{2\pi\mathrm{i} \omega_2}\rangle=\delta_{\omega_1\omega_2} $$ And if so, how could I prove it? Thanks in advance.

Fynn Zentner
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  • See Eq. 1.17.12 here. – John Barber Mar 26 '24 at 01:47
  • @JohnBarber Would you expand on how that equation answers the question? Is the idea basically that the uncountable set of Dirac delta "functions" form an orthonormal "basis" for $L^2(\mathbb{R})$, and because the Dirac deltas can be represented in terms of an integral "orthonormal basis expansion" of complex exponentials, the complex exponentials must also have the same "basis" property as the Dirac deltas? But the OP's question was about whether the Kronecker delta (not Dirac delta) property of orthonormal bases. – Chill2Macht Jan 05 '25 at 15:56
  • OP your question, about whether the uncountable set of complex exponentials can be interpreted as an "orthonormal basis" for L^2 is subtle. (And also something I also struggle with, full disclosure.) On one hand, "morally" if we are to accept the Fourier transform as the "morally" "true" extension of the Fourier series idea from L^2 spaces of compact intervals [-T/2,T/2], then the statement should be true. On the other hand, the complex exponential functions themselves technically are not even square integrable -- only their truncations to compact intervals are. – Chill2Macht Jan 05 '25 at 16:01
  • Moreover, although $L^2(\mathbb{R})$ seems like it should be a "larger" Hilbert space than any of the $L^2[-T/2,T/2]$, given the different topologies of $\mathbb{R}$ and compact intervals, nevertheless $L^2(\mathbb{R})$ is actually still a separable Hilbert space, and so should still admit a countable orthonormal basis (i.e. a series expansion, not an integral expansion). Cf. this question https://math.stackexchange.com/questions/2602736/proof-of-separability-of-l2-mathbbr-without-stone-weierstrass-theorem . – Chill2Macht Jan 05 '25 at 16:06
  • (Not to be confused with $\ell^2(\mathbb{R})$, which is non-separable, see here: https://math.stackexchange.com/questions/2448229/is-every-hilbert-space-separable In fact it seems that $\ell^2([-T/2,T/2])$ is non-separable for any $T$ as well, because the compact intervals are also compact sets, hence the determining factor for the topology of $L^2(S)$ vs. $\ell^2(S)$ for these limited examples of sets S seems to be the cardinality of S rather than the topology of S, with $\ell^2(\mathbb{N})$ being the prototypical example of a separable Hilbert space.) – Chill2Macht Jan 05 '25 at 16:12
  • Regarding the orthonormality property expressed in Kronecker deltas, that has to be false. First, for $\omega_1 = \omega_2$, the integral will be $+\infty$. (This is qualitatively similar to the property involving Dirac deltas that John Barber mentioned.) For $\omega_1 \not= \omega_2$, even when truncated to compact intervals, the inner product would be expected to be non-zero for $\omega_2$ close enough to $\omega_1$ (and/or periodic shifts away from a value close enough to $\omega_1$), e.g. $\sin(1x)$ and $\sin(1.0000000001x)$ are not orthogonal functions. – Chill2Macht Jan 05 '25 at 16:27
  • (Actually for $\omega_1 \not = \omega_2$ the inner product actually might be non-zero only when truncated to compact intervals, and zero when integrated over $\mathbb{R}$ -- that is the claim I've found on the internet, but without any clear proofs / justifications. So even though the "normality" property is definitely false and has to be replaced with a Dirac delta, the orthognoality property might still be valid. I.e. a "Dirac delta version" of the Kronecker delta might hold.) – Chill2Macht Jan 05 '25 at 17:26
  • Note that although the complex exponentials themselves are not square-integrable, they are bounded, hence for any square integrable function $f$, $|| e^{2 \pi i \omega x} f(x)||^2 \le || |e^{2 \pi i \omega x}| f(x)||^2 \le (1) ||f(x)||^2 < \infty$, basically the same reason $L^2$ is closed under scalar multiplication even though constant functions are not square integrable.

    For something that's actually a countable basis of L^2(R), possibly the Fourier series basises for L^2[-T/2, T/2] for all T rational might work, although they won't be orthogonal for the reasons given above.

     – Chill2Macht Jan 05 '25 at 17:29
  • Maybe the Fourier series bases for a countable sequence of disjoint intervals (rather than nested) might also work as a countable basis, ... [-1, 0], [0,1], [1,2], ... etc. but probably you actually have to take the limit as the interval width goes to zero, in which case again you'd lose orthogonality for the reasons given above. – Chill2Macht Jan 05 '25 at 17:31
  • Related questions: (1) https://math.stackexchange.com/questions/1303974/fourier-transform-understanding-change-of-basis-property-with-ideas-from-linear (2) https://math.stackexchange.com/questions/1927614/basis-of-l2-mathbb-r-and-fourier-transform?rq=1 (3) https://math.stackexchange.com/questions/3375359/why-is-the-infinite-period-fourier-series-the-fourier-transform – Chill2Macht Jan 05 '25 at 17:31

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