Here we prove a uniform convergence result regarding Fourier series. The results will be stated for an arbitrary half-period #L#. We will need an instance of Bessel's inequality.
Suppose that #f# is a real #2L#-periodic function that is piecewise continuous on #\ivcc{-L}{L}# and that the constants #a_k# #(k\ge0)#, #b_\ell# #(\ell\ge1)# are given by the Euler formulas for #f#. Then #f^2# is integrable on #\ivcc{-L}{L}# and \[\begin{array}{rcl}\displaystyle \frac{a_0^2}{2}+\sum_{k=1}^{\infty}\left(a_k^2+b_k^2\right) \leq \displaystyle{\frac{1}{L}\int_{-L}^{L} {f(x)^2}\,\dd x}\end{array}\]
In view of the comment on scaling to Uniform convergence, we can reduce the proof to the case where #L = \pi#. Let #F# be the inner product space of real #2\pi#-periodic, piecewise continuous functions with inner product
\[\dotprod{f}{g} = \frac{1}{\pi}\int_{-\pi}^{\pi} {f(x)\cdot g(x)}\,\dd x\]
For #k=0,1,2,\ldots#, the functions #\cos(kx)#, #\sin(kx)# belong to #F#. In this inner product space, we will work with the orthonormal system
\[ \basis{\dfrac{1}{\sqrt2}, \cos(x),\sin(x),\ldots,\cos(n\,x),\sin(n\,x) } \]
Assume that the function #f# belongs to #F#. The projection of #f# on the linear span in #F# of this orthonormal system coincides with #s_n=\frac{a_0^2}{2}+\sum_{k=1}^{n}\left(a_k^2+b_k^2\right) #, since, by the Euler formulas,
\[\begin{array}{rclcl}\dfrac{a_0}{\sqrt2} &=& \displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi} \frac{f(x)}{\sqrt2}\dd x &=& \dotprod{f}{\dfrac{1}{\sqrt2}}\\ a_k &=&\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(k x) \dd x &=& \dotprod{f}{\cos(kx)}\\ b_k &=&\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(k x) \dd x &=& \dotprod{f}{\sin(kx)}\end{array}\]
Recall that Bessel's inequality states that if #\basis{f_1,f_2,\ldots}# is an orthonormal system in an inner product space #V#, then, for every #f \in V#, the series #\sum_{k=1}^{n} \left| f \cdot f_k \right|^2# converges for #n\to\infty# and\[\begin{array}{rcl}\displaystyle \sum_{k=1}^{\infty}\left| f \cdot f_k \right|^2 \leq \displaystyle\left|\left| f \right|\right|^2 \end{array}\]In this context, Bessel's inequality applied to the above orthogonal system gives
\[\begin{array}{rcl}\displaystyle \left(\dfrac{a_0}{\sqrt2}\right)^2+\sum_{k=1}^{\infty}\left(a_k^2+b_k^2\right) \leq \displaystyle\left|\left| f \right|\right|^2 \end{array}\]
The left-hand side of this inequality is a real number since the infinite sum is in fact an increasing sequence bounded from above by the right-hand side. Since \[{\parallel f\parallel} ^2= {\frac{1}{\pi}\int_{-\pi}^{\pi} {f(x)^2}\,\dd x}\]this proves the statement.
Below we shall prove that, under slightly stronger conditions, equality holds.
The constants #a_k# #(k\ge0)#, #b_\ell# #(\ell\ge1)# are given by the Euler formulas. By Dirichlet's Theorem for 1-dimensional Fourier series the trigonometric series \[s_n(x)=\frac{a_0}{2} + \sum_{k=1}^n (a_k\cos(kx)+b_k\sin(kx))\]converges pointwise to #\frac12\left(f_-(x)+f_+(x)\right)# provided #f# is piecewise smooth on #\ivcc{-L}{L}#. Here, we do not need to make that assumption. The statement is phrased in greater generality, that is, using the Euler formulas without the assumption of piecewise smoothness, so it can be used in the proof below of uniform convergence of the Fourier series of a continuous function #f# without the need to invoke smoothness of the derivative #f'#.
Here is a uniform convergence result for the Fourier series.
Let #f# be a real #2L#-periodic function that is continuous and piecewise smooth on the closed interval #\ivcc{-L}{L}#. Then the Fourier series of #f# exists and converges uniformly to #f# on #\mathbb{R}#.
In view of the comment on scaling to Uniform convergence, we may assume the period of #f# to be #2\pi#. Let #a_k# and #b_k# be given by the Euler formulas for #f# and write \[f_0(x)= \frac{a_0}{2}\phantom{xxx}\text{ and }\phantom{xxx} f_k (x)= a_k\cos(kx)+b_k\sin(kx)\] Since #f# is piecewise smooth on #\ivcc{-L}{L}#, Dirichlet's Theorem for 1-dimensional Fourier series gives that the trigonometric series \[s_n(x)=\frac{a_0}{2} + \sum_{k=1}^n (a_k\cos(kx)+b_k\sin(kx))=\sum_{k=0}^nf_k(x) \]converges pointwise to #\frac12\left(f_-(x)+f_+(x)\right)#. This establishes that the Fourier series for #f# exists is uniquely determined by the Euler formulas. Moreover, the assumption that #f# is continuous implies that #\frac12\left(f_-(x)+f_+(x)\right)=f(x)# for each #x#, and hence that #s_n# converges pointwise to #f#.
We will prove the theorem by applying the Weierstrass #M#-test to the sequence #f_k# #(k=1,2,\ldots)# This will show that the series \(s_n\) converges uniformly to #f#. We will use the sequence #M_k# #(k=1,2,\ldots)# given by \[M_0 =\frac{\abs{a_0}}{2}\phantom{xxx}\text{ and }\phantom{xxx} M_k =\abs{a_k}+\abs{b_k}\]
Let #x# be a real number. Clearly, #\abs{f_0(x)} = M_0#. For #k\ge1#, the triangle inequality and the absolute upper bound #1# on the values of sin and cos give
\[\abs{f_k(x)} =\abs{a_k\cos(kx)+b_k\sin(kx)}\le \abs{a_k\cos(kx)} + \abs{b_k \sin(kx)} \le\abs{a_k} + \abs{b_k }= M_k\] Thus, for all #x\in\mathbb{R}#, we have \(\abs{f_k(x)}\le M_k\).
In order to derive that #\sum_{k=1}^{\infty}M_k# converges, we will use the fact that #f'# is piecewise continuous (since #f# is piecewise smooth) on #\ivcc{-L}{L}#. This allows us to apply Bessel's inequality for Fourier coefficients to #f'#. Thus, writing #M ={\frac{1}{\pi}\int_{-\pi}^{\pi} {f(x)^2}\,\dd x}#, we have
\[\begin{array}{rcl}\displaystyle\frac{ (a_0')^2}{2}+\sum_{k=1}^{\infty}\left((a'_k)^2+(b'_k)^2\right) \leq \displaystyle M\end{array}\]
Since #f# is continuous and piecewise smooth on #\ivcc{-\pi}{\pi}#, the relations between the Fourier coefficients of #f# and its derivative are given by the theorem Derivatives of Fourier series:
\[\begin{array}{rcl} \displaystyle a_0' = 0,\quad a_k=-\frac{b_k '}{k}, \quad b_k=\frac{a_k '}{k} \end{array}\]
For each #k\ge1#, we derive an explicit upper bound for #M_k# in terms of #(a_k')^2 # and #( b_k')^2#:
\[\begin{array}{rcl}\displaystyle M_k &=& \displaystyle |a_k| + |b_k| \\&=& \displaystyle \frac{|a_k'|}{k} + \frac{|b_k'|}{k} \\&&\phantom{xxx}\color{blue}{\text{see above}}\\&\le&\displaystyle\frac12\left(|a_k'| - \frac{1}{k} \right)^2+\frac{|a_k'|}{k}+ \frac12\left(|b_k'| - \frac{1}{k} \right)^2+\frac{|b_k'|}{k}\\&&\phantom{xxx}\color{blue}{\text{squares are non-negative}}\\&=&\displaystyle \frac12 |a_k'|^2 +\frac12 |b_k'|^2 +\frac{1}{k^2}\\&&\phantom{xxx}\color{blue}{\text{brackets expanded}}\\&=&\displaystyle \frac{1}{2} ((a_k')^2 +( b_k')^2) + \frac{1}{k^2} \end{array}\]
As a consequence the corresponding series is bounded:
\[\begin{array}{rcl}\sum_{k=1}^n M_k&=&\sum_{k=1}^n (|a_k|+|b_k|)\\&\le&\displaystyle\frac{1}{2}\sum_{k=1}^n \left( ((a_k')^2 +( b_k')^2) + \frac{1}{k^2}\right) \\&&\phantom{xxx}\color{blue}{\text{see above}}\\&=& \displaystyle\frac{1}{2}\sum_{k=1}^n ((a_k')^2 +( b_k')^2) +\frac{1}{2}\sum_{k=1}^n\frac{1}{k^2} \\&&\phantom{xxx}\color{blue}{\text{terms rearranged}}\\&\le& \displaystyle\frac{1}{2}\sum_{k=1}^n ((a_k')^2 +( b_k')^2) +\frac{\pi^2}{12} \\&&\phantom{xxx}\color{blue}{\sum_{k=1}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}}\\&\le& \displaystyle \frac{6M+\pi^2}{12} \\&&\phantom{xxx}\color{blue}{\text {Bessel's inequality}}\end{array} \]
Here, we put in the precise value of #\sum_{k=1}^\infty\frac{1}{k^2}# for the fun of it; any upper bound of this series would suffice for the proof.
We conclude that #\sum_{k=0}^\infty M_k# is convergent, and so, by the Weierstrass #M#-test, the series #s_n# #(n=1,2,\ldots)# converges uniformly to the function #f#.
Since the functions appearing in the Fourier series of #f# are continuous, the Uniform limit theorem implies that the limit of the series must be a continuous function for the convergence to be uniform. Therefore, we need to require that #f# is continuous.
Let #f# be a real #2L#-periodic function that is continuous and
piecewise smooth on the closed interval #\ivcc{-L}{L}#. Earlier, in the comment
Uniqueness to the theorem Euler formulas, we stated that the
Fourier series of #f# is uniquely determined by the
Euler formulas. Here we prove the statement by use of the Uniform convergence theorem applied to the constant function #0#.
Suppose that
\[\begin{array}{rcl} &&\displaystyle \frac{a_0}{2}+\sum_{k=1}^{\infty}\left(a_k\cos(kx)+b_k\sin(kx)\right)\\ &&\text{and}\\ &&\displaystyle \frac{c_0}{2}+\sum_{k=1}^{\infty}\left(c_k\cos(kx)+d_k\sin(kx)\right)\end{array}\] are two trigonometric series converging pointwise to #f#. Then
\[s_n (x)= \frac{a_0-c_0}{2}+\sum_{k=1}^{n}\left((a_k-c_k)\cdot\cos(kx)+(b_k-d_k) \sin(kx)\right)\]
is a series converging to the constant function #0#. Since #0# is continuous and piecewise smooth, the current theorem implies that the convergence is uniform. As a consequence of the theorem Interchange of limit and integral, the sum and integral in the proof of the Euler formulas may be interchanged, so the coefficients appearing in #s_n# are #0#. This implies #a_k = c_k# for #k=0,1,2,\ldots# and #b_k = d_k# for #k=1,2,\ldots#. Therefore, the coefficients of the Fourier series for #f# are unique.
We are now in a position to prove that, in Bessel's inequality, equality holds.
Let #f# be a real #2L#-periodic, continuous function that is piecewise smooth on #\ivcc{-L}{L}# and has Fourier coefficients #a_k# #(k\ge0)#, #b_\ell# #(\ell\ge1)#.
Then \[\frac{1}{L} \int_{-L}^L f(x)^2 \dd x =\frac{ a_0^2}{2} + \sum_{k=1}^{\infty} (a_k^2 + b_k^2) \]
Again, in view of the comment on scaling to Uniform convergence, we can reduce the proof to the case where #L = \pi#. We know that, under these assumptions, the Fourier series of #f# converges uniformly to it, so certainly \[\int_{-\pi}^{\pi} f(x) \dd x = \int_{-\pi}^{\pi} \left(\frac{a_0}{2} +\sum_{n=1}^{\infty} a_n \cos(nx) + b_n \sin(nx) \right)\dd x\]
Since uniform convergence allows us to interchange limits with integrals and an infinite sum can be written as a limit of partial sums, we find, by use of the unique convergence of Fourier series,
\[\begin{array}{rcl}\displaystyle\int_{-\pi}^{\pi} f(x)^2 \dd x &=& \displaystyle\int_{-\pi}^{\pi} f(x)\left( \frac{a_0}{2} +\sum_{n=1}^{\infty} a_n \cos(nx) + b_n \sin(nx) \right) \dd x \\&&\phantom{xxx}\color{blue}{\text{one factor }f(x)\text{ substituted by its Fourier series}}\\ &=& \displaystyle\frac{a_0}{2}\int_{-\pi}^{\pi} f(x)\dd x + \sum_{n=1}^{\infty}\left(a_n\int_{-\pi}^{\pi} f(x)\cos(nx) \dd x + b_n\int_{-\pi}^{\pi} f(x) \cos(nx) \dd x\right) \\&&\phantom{xxx}\color{blue}{\text{infinite sum and integral interchanged}}\\&=& \displaystyle \pi\cdot\frac{ a_0^2}{2} + \pi\sum_{n=1}^{\infty} (a_n^2 +b_n^2)\\&&\phantom{xxx}\color{blue}{\text{definition of }a_0, a_n, b_n} \end{array}\]
Dividing by #\pi# we get the conclusion.
We proved Parseval's identity with a direct computation but we could have also employed the notion of orthonormal basis. We know that if #\{f_k\}_{k \in \mathbb{N}}# is an orthonormal basis for an inner product space #F#, then each #f\in F# can be written as
\[f = \sum_{k=1}^{\infty} (\dotprod{f}{f_k})f_k\]
On the other hand
\[\begin{array}{rcl}\displaystyle \lvert\lvert f \rvert\rvert^2 &=& \displaystyle\lvert\lvert f - \sum_{k=1}^N (\dotprod{f}{f_k})f_k + \sum_{k=1}^N (\dotprod{f}{f_k})f_k \rvert\rvert^2 \\ &&\phantom{xxx}\color{blue}{\text{subtracting and adding }\sum_{k=1}^N (\dotprod{f}{f_k})f_k}\\ &=& \displaystyle \lvert\lvert f - \sum_{k=1}^N (\dotprod{f}{f_k})f_k) \rvert\rvert^2 + \lvert\lvert \sum_{k=1}^N (\dotprod{f}{f_k})f_k\rvert\rvert^2 \\ &&\phantom{xxx}\color{blue}{\text{Pythagorean theorem ( } f - \sum_{k=1}^N (\dotprod{f}{f_k})f_k \text{ and }\sum_{k=1}^N (\dotprod{f}{f_k})f_k \text{ are orthogonal)}} \\&=& \displaystyle \lvert\lvert f - \sum_{k=1}^N (\dotprod{f}{f_k})f_k) \rvert\rvert^2 + \sum_{k=1}^N \lvert\lvert (\dotprod{f}{f_k}) f_k \rvert\rvert^2 \\&&\phantom{xxx}\color{blue}{\text{Pythagorean theorem (}\dotprod{f_k}{f_j}=0\text{ for } k\neq j\text{)}}\\&=& \displaystyle \lvert\lvert f - \sum_{k=1}^N (\dotprod{f}{f_k})f_k) \rvert\rvert^2 + \sum_{k=1}^N \lvert\lvert (\dotprod{f}{f_k})\rvert\rvert^2 \lvert\lvert f_k \rvert\rvert^2\\ &&\phantom{xxx}\color{blue}{\lvert\lvert (\dotprod{f}{f_k})f_k \rvert\rvert^2 = \lvert\lvert (\dotprod{f}{f_k})\rvert\rvert^2\lvert\lvert f_k \rvert\rvert^2}\\&=& \displaystyle \lvert\lvert f - \sum_{k=1}^N (\dotprod{f}{f_k})f_k) \rvert\rvert^2 + \sum_{k=1}^N \lvert\lvert (\dotprod{f}{f_k})\rvert\rvert^2 \\&&\phantom{xxx}\color{blue}{\lvert\lvert f_k \rvert\rvert^2=1 \text{ for each } k}\end{array}\]
In the limit for #N\to \infty#, the sum #\sum_{k=1}^{N} (\dotprod{f}{f_k})f_k# converges to #f# and the norm #{\parallel \, \cdot \,\parallel}# is continuous. Therefore, we have
\[\displaystyle\lvert\lvert f \rvert\rvert^2 = \displaystyle\sum_{k=1}^{\infty} \lvert\lvert (\dotprod{f}{f_k})\rvert\rvert^2 \]
By applying this to the inner product space of real #2\pi#-periodic, continuous and piecewise smooth functions endowed with the inner product
\[\dotprod{f}{g} = \frac{1}{\pi}\int_{-\pi}^{\pi} {f(x)\cdot g(x)}\,\dd x\]
and the orthonormal basis #\basis{\dfrac{1}{\sqrt2}, \cos(x),\sin(x),\ldots,\cos(n\,x),\sin(n\,x)}# we obtain precisely
\[\frac{1}{L} \int_{-L}^L f(x)^2 \dd x = \frac{a_0^2}{2} + \sum_{k=1}^{\infty} (a_k^2 + b_k^2) \]
Bessel's inequality is proven under the assumption that the inner product space in question admits an orthogonal system. Parseval's identity states that if the orthonormal system is also a basis, then equality holds.
We consider piecewise continuous real functions on a closed interval #\ivcc{a}{b}#, where #a# and #b# are real numbers with #a\lt b#. Recall that these functions form a real inner product space with inner product of functions given by \[\dotprod{f}{g} = \int_{a}^{b} f(x)\cdot g(x) \dd x\]
We say that a sequence #f_n# #(n=1,2,\ldots)# of such functions converges in the norm to a real piecewise continuous function #f# on #\ivcc{a}{b}# if #\lim _{n\to\infty} \norm{f_n-f} = 0#.
Uniform convergence implies convergence in the norm induced by the inner product of functions on #\ivcc{a}{b}#. In order to prove this statement, we suppose that #f_n# #(n=1,2,\ldots)# is a sequence of functions that converges uniformly to the function #f:\ivcc{a}{b} \rightarrow \mathbb{R}#. Let #\varepsilon# be a positive number. By uniform convergence of the sequence, we can find an integer #N \gt 0# such that, for all #n \geq N#, \[\sup\left\{\left. |f_n(x)-f(x)|\,\right|\, x \in \ivcc{a}{b}\right\} \lt\sqrt{\frac{\varepsilon}{b-a}}\] Hence
\[\norm{f_n-f} = \int_{a}^{b} (f_n(x)-f(x))^2 \dd x \leq \int_a^b{ \left(\sqrt{\frac{\varepsilon}{b-a}}\right)^2 }\dd x \leq (b-a)\cdot\frac{\varepsilon}{b-a} =\varepsilon\]
Prove that the Fourier series of the periodic function #f# with period #8# given by #f(x)=\euler^{x}+\euler^ {- x }# on #\ivcc{-4}{4}# converges uniformly to #f#.
The function #f# is #8#-periodic, continuous and piecewise smooth on #\ivcc{-4}{4}#, so theorem
Uniform convergence of Fourier series can be applied, This shows that the Fourier series of #f# converges uniformly to #f#.
Uniform convergence of Fourier series
