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Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution).

But then once you know that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".

Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. The same kinds of mollifications or similar calculations are used in proofs of Stone Weierstrass theorem; by a similar reasoning, once you have enough characters to separate points, you know you have a complete set.

But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution).

But then once you have established that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".

If your purpose of calculating the dual of the torus is to have the Fourier inversion formula, I just want to add the remark that you don't need to go through all the machinery of locally compact topological groups in order to prove the basic results of Fourier analysis.

For the circle, one needs only note that the distribution $u(x) = \sum_{n \in {\mathbb Z}} e^{2 \pi i n x}$ on the torus ${\mathbb R}/{\mathbb Z}$ is invariant under multiplication by $e^{2 \pi i x}$, and is therefore a multiple of the delta-function $\delta(x)$ -- integrating against the smooth function $1$, you see that, in fact, $u(x) = \delta(x)$. (This move is simply how you usually calculate the "sum of a geometric series".) Substituting into the general formula $f(x) = \int_{{\mathbb R}/{\mathbb Z}} f(y) \delta(x - y) dy$ gives the Fourier inversion formula.

The above is really just for a conceptual point of view, since it's unnecessary to quote distribution theory for this purpose. To give a direct proof, suppose $f(x)$ is smooth; after translating, it is enough to show that $f(0) = \sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_n \int f(x) e^{- 2 \pi i n x} dx$ (which converges absolutely because you can integrate by parts enough times to prove decay of $\hat{f}(n)$). We can assume without loss of generality that $f(0) = 0$ by subtracting the constant $f(0)$ (this is how we know "the constant" in the inversion formula is correct and corresponds to testing against $1$ in the previous argument).

In the case $f(0) = 0$, we can write $f(x) = (e^{2 \pi i x} - 1) g(x)$ for some smooth function $g(x)$ (which proves the previous claim that if $e^{2 \pi i x} u = u$ then $u(x) = C \delta(x)$). But then,

$\sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_{n \in {\mathbb Z}} (\hat{g}(n - 1) - \hat{g}(n))$

but this is $0$ because $\hat{g}$ is integrable. (In particular, assuming $f$ to be $C^\infty$ was certainly unnecessary, and the Fourier inversion formula holds pointwise for functions in a certain class.)

Another remark about distribution theory here is that for the argument of KConrad above, the equation $\gamma'(x + s) = \gamma'(s) \gamma(x)$'' already makes sense when $\gamma$ is a distribution (in particular, if $\gamma$ is continuous). The way it makes sense is by integrating by parts against a test function, which is related to the integration performed the accepted proof. One can use a very similar argument to KConrad's and show that the only distributions solving $\gamma(x + y) = \gamma(x)\gamma(y)$ are exponentials as I'll (almost) show below.

Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution).

But then once you know that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".

If your purpose of calculating the dual of the torus is to have the Fourier inversion formula, I just want to add the remark that you don't need to go through all the machinery of locally compact topological groups in order to prove the basic results of Fourier analysis.

For the circle, one needs only note that the distribution $u(x) = \sum_{n \in {\mathbb Z}} e^{2 \pi i n x}$ on the torus ${\mathbb R}/{\mathbb Z}$ is invariant under multiplication by $e^{2 \pi i x}$, and is therefore a multiple of the delta-function $\delta(x)$ -- integrating against the smooth function $1$, you see that, in fact, $u(x) = \delta(x)$. (This move is simply how you usually calculate the "sum of a geometric series".) Substituting into the general formula $f(x) = \int_{{\mathbb R}/{\mathbb Z}} f(y) \delta(x - y) dy$ gives the Fourier inversion formula.

The above is really just for a conceptual point of view, since it's unnecessary to quote distribution theory for this purpose. To give a direct proof, suppose $f(x)$ is smooth; after translating, it is enough to show that $f(0) = \sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_n \int f(x) e^{- 2 \pi i n x} dx$ (which converges absolutely because you can integrate by parts enough times to prove decay of $\hat{f}(n)$). We can assume without loss of generality that $f(0) = 0$ by subtracting the constant $f(0)$ (this is how we know "the constant" in the inversion formula is correct and corresponds to testing against $1$ in the previous argument).

In the case $f(0) = 0$, we can write $f(x) = (e^{2 \pi i x} - 1) g(x)$ for some smooth function $g(x)$ (which proves the previous claim that if $e^{2 \pi i x} u = u$ then $u(x) = C \delta(x)$). But then,

$\sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_{n \in {\mathbb Z}} (\hat{g}(n - 1) - \hat{g}(n))$

but this is $0$ because $\hat{g}$ is integrable. (In particular, assuming $f$ to be $C^\infty$ was certainly unnecessary, and the Fourier inversion formula holds pointwise for functions in a certain class.)

If you like to think in terms of Stone Weierstrass, you can view this argument is saying that the map $Tf$ which takes a Fourier transform and then performs the inverse Fourier transform is the identity. It relies on the fact that $f(x) = (e^{2 \pi i x} - e^{2 \pi i x_0}) g(x)$ whenever $f$ vanishes at $x_0$ which is a strong sense in which trigonometric polynomials separate points. The conclusion that $T$ is a multiple of the identity then follows from $T$ being linear not only over ${\mathbb C}$ but also over the algebra of trigonometric polynomials.

Another remark about distribution theory here is that for the argument of KConrad above, the equation $\gamma'(x + s) = \gamma'(s) \gamma(x)$'' already makes sense when $\gamma$ is a distribution (in particular, if $\gamma$ is continuous). The way it makes sense is by integrating by parts against a test function, which is related to the integration performed the accepted proof. One can use a very similar argument to KConrad's and show that the only distributions solving $\gamma(x + y) = \gamma(x)\gamma(y)$ are exponentials as I'll (almost) show below.

Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution).

But then once you know that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".

If your purpose of calculating the dual of the torus is to have the Fourier inversion formula, I just want to add the remark that you don't need to go through all the machinery of locally compact topological groups in order to prove the basic results of Fourier analysis.

For the circle, one needs only note that the distribution $u(x) = \sum_{n \in {\mathbb Z}} e^{2 \pi i n x}$ on the torus ${\mathbb R}/{\mathbb Z}$ is invariant under multiplication by $e^{2 \pi i x}$, and is therefore a multiple of the delta-function $\delta(x)$ -- integrating against the smooth function $1$, you see that, in fact, $u(x) = \delta(x)$. Substituting into the general formula $f(x) = \int_{{\mathbb R}/{\mathbb Z}} f(y) \delta(x - y) dy$ gives the Fourier inversion formula.

The above is really just for a conceptual point of view, since it's unnecessary to quote distribution theory for this purpose. To give a direct proof, suppose $f(x)$ is smooth; after translating, it is enough to show that $f(0) = \sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_n \int f(x) e^{- 2 \pi i n x} dx$ (which converges absolutely because you can integrate by parts enough times to prove decay of $\hat{f}(n)$). We can assume without loss of generality that $f(0) = 0$ by subtracting the constant $f(0)$ (this is how we know "the constant" in the inversion formula is correct and corresponds to testing against $1$ in the previous argument).

In the case $f(0) = 0$, we can write $f(x) = (e^{2 \pi i x} - 1) g(x)$ for some smooth function $g(x)$ (which proves the previous claim that if $e^{2 \pi i x} u = u$ then $u(x) = C \delta(x)$). But then,

$\sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_{n \in {\mathbb Z}} (\hat{g}(n - 1) - \hat{g}(n))$

but this is $0$ because $\hat{g}$ is integrable. (In particular, assuming $f$ to be $C^\infty$ was certainly unnecessary, and the Fourier inversion formula holds pointwise for functions in a certain class.)

Another remark about distribution theory here is that for the argument of KConrad above, the equation $\gamma'(x + s) = \gamma'(s) \gamma(x)$'' already makes sense when $\gamma$ is a distribution (in particular, if $\gamma$ is continuous). The way it makes sense is by integrating by parts against a test function, which is related to the integration performed the accepted proof. One can use a very similar argument to KConrad's and show that the only distributions solving $\gamma(x + y) = \gamma(x)\gamma(y)$ are exponentials as I'll (almost) show below.

Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution).

But then once you know that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".

If your purpose of calculating the dual of the torus is to have the Fourier inversion formula, I just want to add the remark that you don't need to go through all the machinery of locally compact topological groups in order to prove the basic results of Fourier analysis.

For the circle, one needs only note that the distribution $u(x) = \sum_{n \in {\mathbb Z}} e^{2 \pi i n x}$ on the torus ${\mathbb R}/{\mathbb Z}$ is invariant under multiplication by $e^{2 \pi i x}$, and is therefore a multiple of the delta-function $\delta(x)$ -- integrating against the smooth function $1$, you see that, in fact, $u(x) = \delta(x)$. (This move is simply how you usually calculate the "sum of a geometric series".) Substituting into the general formula $f(x) = \int_{{\mathbb R}/{\mathbb Z}} f(y) \delta(x - y) dy$ gives the Fourier inversion formula.

The above is really just for a conceptual point of view, since it's unnecessary to quote distribution theory for this purpose. To give a direct proof, suppose $f(x)$ is smooth; after translating, it is enough to show that $f(0) = \sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_n \int f(x) e^{- 2 \pi i n x} dx$ (which converges absolutely because you can integrate by parts enough times to prove decay of $\hat{f}(n)$). We can assume without loss of generality that $f(0) = 0$ by subtracting the constant $f(0)$ (this is how we know "the constant" in the inversion formula is correct and corresponds to testing against $1$ in the previous argument).

In the case $f(0) = 0$, we can write $f(x) = (e^{2 \pi i x} - 1) g(x)$ for some smooth function $g(x)$ (which proves the previous claim that if $e^{2 \pi i x} u = u$ then $u(x) = C \delta(x)$). But then,

$\sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_{n \in {\mathbb Z}} (\hat{g}(n - 1) - \hat{g}(n))$

but this is $0$ because $\hat{g}$ is integrable. (In particular, assuming $f$ to be $C^\infty$ was certainly unnecessary, and the Fourier inversion formula holds pointwise for functions in a certain class.)

Another remark about distribution theory here is that for the argument of KConrad above, the equation $\gamma'(x + s) = \gamma'(s) \gamma(x)$'' already makes sense when $\gamma$ is a distribution (in particular, if $\gamma$ is continuous). The way it makes sense is by integrating by parts against a test function, which is related to the integration performed the accepted proof. One can use a very similar argument to KConrad's and show that the only distributions solving $\gamma(x + y) = \gamma(x)\gamma(y)$ are exponentials as I'll (almost) show below.

Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution).

But then once you know that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".