Return to Answer

That is the case because the real line is not compact.

The Pontryagin dual group of a locally compact abelian group $G$ is discrete (or has any isolated point) if and only if $G$ is compact. But the circle group giving the Fourier series is compact. It is an essential point that the characters of a non-compact lca group are not square-integrable (you can easily proof that). If there would be an isolated point $x$ in the dual space, then the characteristic function $\chi_{\left\{x\right\}}$ (in your language that is a “spike at a discrete frequency”) would be non-zero in the $L^2$-space. The inverse of the Fourier transform would have to map this non-zero $L^2$-function to a non-zero $L^2$-function, which would simply be a character $\exp(ikx)$, but this character is not square-integrable.

Actually the characterisation by square-integrable characters/representations is more fundamental than the characterisation using compactness: In the non-abelian case an irreducible unitary representation is a discrete series representation (i. e. it contributes to the inverse of the Fourier transform as a simple summand, or ”spike”) if and only if there is any square-integrable matrix coefficient. But the group might be non-compact

That is the case because the real line is not compact.

The Pontryagin dual group of a locally compact abelian group $G$ is discrete (or has any isolated point) if and only if $G$ is compact. But the circle group giving the Fourier series is compact. It is an essential point that the characters of a non-compact lca group are not square-integrable (you can easily prove that). If there would be an isolated point $x$ in the dual space, then the characteristic function $\chi_{\left\{x\right\}}$ (in your language that is a “spike at a discrete frequency”) would be non-zero in the $L^2$-space. The inverse of the Fourier transform would have to map this non-zero $L^2$-function to a non-zero $L^2$-function, which would simply be a character $\exp(ikx)$, but this character is not square-integrable.

Actually the characterisation by square-integrability of the characters/representations is more fundamental than the characterisation using compactness: In the non-abelian case an irreducible unitary representation is a discrete series representation (i. e. it contributes to the inverse of the Fourier transform as a simple summand, or ”spike”) if and only if there is any square-integrable matrix coefficient. But the group might be non-compact

That is the case because the real line is not compact.

The Pontryagin dual group of a locally compact abelian group $G$ is discrete (or has any isolated point) if and only if $G$ is compact. But the circle group giving the Fourier series is compact. It is an essential point that the characters of a non-compact lca group are not square-integrable (you can easily proof that). If there would be an isolated point $x$ in the dual space, then the characteristic function $\chi_{\left\{x\right\}}$ would be non-zero in the $L^2$-space. The inverse of the Fourier transform would have to map this non-zero $L^2$-function to a non-zero $L^2$-function, which would simply be a character $\exp(ikx)$, but this character is not square-integrable.

Actually the characterisation by square-integrable characters/representations is more fundamental than the characterisation using compactness: In the non-abelian case an irreducible unitary representation is a discrete series representation (i. e. it contributes to the inverse of the Fourier transform as a simple summand, or ”spike”) if and only if there is any square-integrable matrix coefficient. But the group might be non-compact

That is the case because the real line is not compact.

The Pontryagin dual group of a locally compact abelian group $G$ is discrete (or has any isolated point) if and only if $G$ is compact. But the circle group giving the Fourier series is compact. It is an essential point that the characters of a non-compact lca group are not square-integrable (you can easily proof that). If there would be an isolated point $x$ in the dual space, then the characteristic function $\chi_{\left\{x\right\}}$ (in your language that is a “spike at a discrete frequency”) would be non-zero in the $L^2$-space. The inverse of the Fourier transform would have to map this non-zero $L^2$-function to a non-zero $L^2$-function, which would simply be a character $\exp(ikx)$, but this character is not square-integrable.

Actually the characterisation by square-integrable characters/representations is more fundamental than the characterisation using compactness: In the non-abelian case an irreducible unitary representation is a discrete series representation (i. e. it contributes to the inverse of the Fourier transform as a simple summand, or ”spike”) if and only if there is any square-integrable matrix coefficient. But the group might be non-compact

That is the case because the real line is not compact.

The Pontryagin dual group of a locally compact abelian group $G$ is discrete (or has any isolated point) if and only if $G$ is compact. But the circle group giving the Fourier series is compact. It is an essential point that the characters of a non-compact lca group are not square-integrable (you can easily proof that). If there would be an isolated point $x$ in the dual space, then the characteristic function $\chi_{\left\{x\right\}}$ would be non-zero in the $L^2$-space. The inverse of the Fourier transform would have to map this non-zero $L^2$-function to a non-zero $L^2$-function, which would simply be a character $\exp(ikx)$, but this character is not square-integrable.

That is the case because the real line is not compact.

The Pontryagin dual group of a locally compact abelian group $G$ is discrete (or has any isolated point) if and only if $G$ is compact. But the circle group giving the Fourier series is compact. It is an essential point that the characters of a non-compact lca group are not square-integrable (you can easily proof that). If there would be an isolated point $x$ in the dual space, then the characteristic function $\chi_{\left\{x\right\}}$ would be non-zero in the $L^2$-space. The inverse of the Fourier transform would have to map this non-zero $L^2$-function to a non-zero $L^2$-function, which would simply be a character $\exp(ikx)$, but this character is not square-integrable.

Actually the characterisation by square-integrable characters/representations is more fundamental than the characterisation using compactness: In the non-abelian case an irreducible unitary representation is a discrete series representation (i. e. it contributes to the inverse of the Fourier transform as a simple summand, or ”spike”) if and only if there is any square-integrable matrix coefficient. But the group might be non-compact