Parseval's Formula For Fourier Series

Fourier Coefficients as an Energy Accounting

A Fourier coefficient measures the component of a function in one oscillatory direction. Parseval’s formula says that, for a square-integrable function on the circle, the total squared size of all these components equals the squared $L^2$ norm of the function. This is more than a formal identity for finite trigonometric sums: it applies to every $L^2$ function because the trigonometric system is complete and its Fourier partial sums converge in $L^2$.

We fix the normalization carefully. Write the circle as $\mathbb T=[0,2\pi)$ with endpoints identified, and let $\mu_{\mathbb T}$ be normalized arc-length measure: $d\mu_{\mathbb T}(\theta)=d\theta/(2\pi)$. This is a finite measure space. Functions may be complex-valued and are identified when they agree $\mu_{\mathbb T}$-almost everywhere. For $f\in L^2(\mathbb T,\mu_{\mathbb T})$, define its $n$-th Fourier coefficient, for each integer $n$, by

\widehat f(n)=\int_{\mathbb T}f(\theta)e^{-in\theta}\,d\mu_{\mathbb T}(\theta) =\frac{1}{2\pi}\int_0^{2\pi}f(\theta)e^{-in\theta}\,d\theta.

The exponential functions $e_n(\theta)=e^{in\theta}$, $n\in\mathbb Z$, have norm one and are pairwise orthogonal in this $L^2$ space. Indeed, for integers $j,k$,

\langle e_j,e_k\rangle =\frac{1}{2\pi}\int_0^{2\pi}e^{i(j-k)\theta}\,d\theta =\begin{cases}1,&j=k,\\0,&j\ne k.\end{cases}

Here the inner product is $\langle f,g\rangle=\int_{\mathbb T}f\overline g\,d\mu_{\mathbb T}$, linear in its first argument. Thus $\widehat f(n)=\langle f,e_n\rangle$. The trigonometric system is complete in $L^2(\mathbb T,\mu_{\mathbb T})$, and the earlier result on $L^2$ convergence of Fourier partial sums states that, for every such $f$, the sums $S_Nf=\sum_{|n|\le N}\widehat f(n)e^{in\theta}$ converge to $f$ in $L^2$. We use that result to prove the energy formula.

Parseval’s Formula with Normalized Measure

Theorem (Parseval’s formula on the circle): If $f$ is a complex-valued member of $L^2(\mathbb T,\mu_{\mathbb T})$, then $$ \int_{\mathbb T}|f(\theta)|^2\,d\mu_{\mathbb T}(\theta) =\|f\|_2^2 =\sum_{n\in\mathbb Z}|\widehat f(n)|^2. $$ The series on the right is a sum of nonnegative terms and is finite. The identity concerns the $L^2$ equivalence class of $f$, so changing $f$ on a null set does not affect either side.

Proof. Fix $N\ge0$. By the finite orthogonality computation above, the terms in $S_Nf$ are pairwise orthogonal and each $e_n$ has norm one. Expanding the squared norm and using those inner products gives

\|S_Nf\|_2^2 =\left\langle\sum_{|n|\le N}\widehat f(n)e_n, \sum_{|k|\le N}\widehat f(k)e_k\right\rangle =\sum_{|n|\le N}|\widehat f(n)|^2.

By the $L^2$ convergence of Fourier partial sums, $\|S_Nf-f\|_2\to0$. The triangle inequality gives $\big|\|S_Nf\|_2-\|f\|_2\big|\le\|S_Nf-f\|_2$, so $\|S_Nf\|_2\to\|f\|_2$. Squaring these nonnegative real numbers shows that $\|S_Nf\|_2^2\to\|f\|_2^2$. Meanwhile, the finite sums $\sum_{|n|\le N}|\widehat f(n)|^2$ increase with $N$, since every added term is nonnegative. Their limit is, by the definition of a series of nonnegative terms, $\sum_{n\in\mathbb Z}|\widehat f(n)|^2$. Taking limits in the finite identity proves the formula. In particular, the series is finite because its limit is $\|f\|_2^2<\infty$. $\square$

If instead the integral is written with unnormalized arc length $d\theta$, the same coefficients still carry the factor $1/(2\pi)$ in their definition, but the energy formula becomes $\int_0^{2\pi}|f(\theta)|^2\,d\theta=2\pi\sum_{n\in\mathbb Z}|\widehat f(n)|^2$. The factor $2\pi$ comes from converting the normalized integral to the unnormalized one.

Real Sine-and-Cosine Form

For a real-valued $f\in L^2(\mathbb T,\mu_{\mathbb T})$, another common convention is to write $a_0=\frac1\pi\int_0^{2\pi}f(\theta)\,d\theta$, $a_n=\frac1\pi\int_0^{2\pi}f(\theta)\cos(n\theta)\,d\theta$, and $b_n=\frac1\pi\int_0^{2\pi}f(\theta)\sin(n\theta)\,d\theta$ for $n\ge1$. These coefficients satisfy $\widehat f(0)=a_0/2$, $\widehat f(n)=(a_n-i b_n)/2$, and $\widehat f(-n)=(a_n+i b_n)/2$. Substitution into Parseval’s formula gives

\int_{\mathbb T}|f|^2\,d\mu_{\mathbb T} =\frac{a_0^2}{4}+\frac12\sum_{n=1}^{\infty}(a_n^2+b_n^2).

For example, the coefficient pair at frequencies $n$ and $-n$ contributes $\frac14(a_n^2+b_n^2)+\frac14(a_n^2+b_n^2)=\frac12(a_n^2+b_n^2)$. The constant coefficient contributes $a_0^2/4$. This explains each factor in the real form; it is not a different energy principle.

Worked Example: A Trigonometric Polynomial

Let $f(\theta)=2+3\cos\theta-4\sin(2\theta)$ on $\mathbb T$. Compute its Fourier coefficients and verify Parseval’s formula both from the coefficients and by integrating $|f|^2$.

Use $\cos\theta=(e^{i\theta}+e^{-i\theta})/2$ and $\sin(2\theta)=(e^{2i\theta}-e^{-2i\theta})/(2i)$. Thus $3\cos\theta$ contributes $3/2$ at each of $n=1,-1$, while $-4\sin(2\theta)=2i e^{2i\theta}-2i e^{-2i\theta}$ contributes $2i$ at $n=2$ and $-2i$ at $n=-2$. The constant term contributes $2$ at $n=0$. All other coefficients vanish. Therefore

\sum_{n\in\mathbb Z}|\widehat f(n)|^2 =|2|^2+\left|\frac32\right|^2+\left|\frac32\right|^2+|2i|^2+|-2i|^2 =4+\frac94+\frac94+4+4=\frac{33}{2}.

For an independent integral check, the normalized integrals of $\cos^2\theta$ and $\sin^2(2\theta)$ are both $1/2$. The normalized integrals of $\cos\theta$, $\sin(2\theta)$, and $\cos\theta\sin(2\theta)$ are zero, as follows by integrating their exponential terms over a full period. Expanding the square therefore gives $\int_{\mathbb T}|f|^2\,d\mu_{\mathbb T}=4+9(1/2)+16(1/2)=4+9/2+8=33/2$, agreeing with the coefficient sum.

Check: the only nonzero coefficient magnitudes squared are $4,9/4,9/4,4,4$; their sum is $4+9/2+8=33/2$, exactly the directly integrated energy.

Worked Example: A Single Complex Frequency

Consider $g(\theta)=2e^{-2i\theta}-ie^{i\theta}$. Its only nonzero Fourier coefficients are $\widehat g(-2)=2$ and $\widehat g(1)=-i$, so the coefficient energy is $|2|^2+|-i|^2=4+1=5$. To check directly, write $|g|^2=(2e^{-2i\theta}-ie^{i\theta})(2e^{2i\theta}+ie^{-i\theta})$. Multiplication gives $4+1+2i e^{-3i\theta}-2i e^{3i\theta}$. Each nonconstant exponential has integral zero over $[0,2\pi)$, because its antiderivative has equal endpoint values. Hence $\int_{\mathbb T}|g|^2\,d\mu_{\mathbb T}=5$, as required.

Check: the cross terms have frequencies $3$ and $-3$, not frequency zero; their normalized integrals vanish, leaving $4+1=5$.

Worked Example: The Indicator of a Half-Circle

Let $h=\chi_{[0,\pi)}$ on $\mathbb T$. The endpoint convention is explicit: $h=1$ for $0\le\theta<\pi$, and $h=0$ for $\pi\le\theta<2\pi$. In particular, endpoint choices do not change its $L^2$ class because individual points have measure zero. Directly, $\int_{\mathbb T}|h|^2\,d\mu_{\mathbb T}=\pi/(2\pi)=1/2$.

For $n=0$, $\widehat h(0)=\frac1{2\pi}\int_0^\pi1\,d\theta=1/2$. For $n\ne0$, integrating the exponential gives

\widehat h(n)=\frac{1}{2\pi}\int_0^\pi e^{-in\theta}\,d\theta =\frac{1-(-1)^n}{2\pi i n}.

If $n$ is a nonzero even integer, $(-1)^n=1$, so $\widehat h(n)=0$. If $n$ is odd, $(-1)^n=-1$, so $\widehat h(n)=1/(\pi i n)$, whose squared magnitude is $1/(\pi^2n^2)$. Parseval therefore yields the exact series identity

\frac12=\frac14+\frac{1}{\pi^2}\sum_{\substack{n\in\mathbb Z\\ n\ {\rm odd}}}\frac1{n^2} =\frac14+\frac{2}{\pi^2}\sum_{\substack{n\ge1\\ n\ {\rm odd}}}\frac1{n^2}, \qquad \sum_{\substack{n\ge1\\ n\ {\rm odd}}}\frac1{n^2}=\frac{\pi^2}{8}.

The second equality uses the equality of the positive and negative odd-index contributions, since $1/(-n)^2=1/n^2$. Solving the displayed equation gives the final value: subtract $1/4$ from both sides to obtain $1/4=(2/\pi^2)\sum_{n\ge1,\ n\ {\rm odd}}1/n^2$, then multiply by $\pi^2/2$. As a check independent of summing the coefficients, the direct norm calculation above gives $1/2$; the coefficient side consists of the constant contribution $1/4$ and the nonzero odd-frequency contributions summing to $1/4$.

What Parseval Does—and Does Not—Say

Parseval’s formula is an exact accounting identity for $L^2$ energy. It also explains why many different-looking descriptions of the same $L^2$ function have the same Fourier energy: coefficients and norm depend only on the almost-everywhere equivalence class. The formula applies to all complex- or real-valued $L^2$ functions on this finite measure space; it does not require continuity or pointwise smoothness.

A common error is to switch normalizations halfway through a calculation. For the constant function $f(\theta)=1$, our convention gives $\widehat f(0)=1$ and $\widehat f(n)=0$ for $n\ne0$. Thus the coefficient energy is $1$, matching $\int_{\mathbb T}1\,d\mu_{\mathbb T}=1$. If one instead writes the unnormalized integral, it is $\int_0^{2\pi}1\,d\theta=2\pi$, so the correct formula with our coefficients is $2\pi\sum_n|\widehat f(n)|^2=2\pi$, not $\sum_n|\widehat f(n)|^2=2\pi$. The latter incorrectly inserts an extra factor.

Another pitfall is to read Parseval as a pointwise convergence statement. Its proof uses $S_Nf\to f$ in $L^2$, a norm-convergence assertion; it does not claim that $S_Nf(\theta)\to f(\theta)$ at every point. Nor does the equality of the energies alone provide pointwise values of the Fourier series. Those questions require separate convergence results. The essential hypotheses here are that $f\in L^2(\mathbb T,\mu_{\mathbb T})$ and that the complete trigonometric system is used.

Check Your Understanding

Use normalized arc-length measure and the coefficient convention fixed above in each question.

For $f(\theta)=1+2\cos(2\theta)$, list its nonzero complex Fourier coefficients and compute both sides of Parseval’s formula.
For real $f$, explain why the $n$ and $-n$ coefficients together contribute $(a_n^2+b_n^2)/2$ to the normalized energy.
For the indicator of $[0,\pi)$, why do the even nonzero frequencies have zero coefficient while the odd frequencies do not?
State the corresponding energy formula when the left-hand side is written as an integral with respect to $d\theta$, rather than $d\theta/(2\pi)$.
Does Parseval’s formula assert pointwise convergence of the Fourier partial sums? Identify the convergence used in its proof.