Recall that in our space $E[-\pi,\pi]$ we identify between two function $f,g$ if $f-g$ is zero almost everywhere (namely $f\equiv g$ for all but finitely many points).

In particular, if both $f$ and $g$ have the same Fourier series, then $f\sim g$ are the same in $E[-\pi, \pi]$ (both equivalent to the same Fourier series) and therefore are equal almost everywhere.

The Fourier coefficients of $f\in E[-\pi,\pi]$ are $\angles{f,\frac{1}{\sqrt{2}}}$ , $\angles{f,\sin(nx)}$ , and $\angles{f,\cos(nx)}$ for all $n\geq 1$ . In particular we have the classic Fourier series expansion:

$$f \sim \sum \angles{f,\frac{1}{\sqrt{2}}}\frac{1}{\sqrt{2}} + \sum_{n=1}^\infty \left( \angles{f,\cos(nx)}\cos(nx) + \angles{f,\sin(nx)}\sin(nx)\right).$$

1. $f(x)=x$ : In this case, we already saw that $b_n=2\frac{(-1)^{n+1}}{n}$ and $a_n=0$ for all $n$ (since $x\cos(nx)$ is an odd function).

2. $f(x)=\chi_{[0,\pi]}-\chi_{[-\pi,0]}$ : This is also an odd function, so that $a_n=0$ for all $n$ . As for the $b_n$ , the product $f(x)\sin(nx)$ is even, so we have that

   $$\align{b_n & = \angles{\chi_{[0,\pi]}-\chi_{[-\pi,0]},\sin(nx)} = \frac{1}{\pi} \int_0^\pi \sin(nx) \dx - \frac{1}{\pi} \int_{-\pi}^0 \sin(nx) \dx \\ & = \frac{2}{\pi} \int_0^\pi \sin(nx) \dx = \frac{2}{n\pi} \cos(nx)\mid_0^\pi=\frac{2}{n\pi}\left((-1)^n-1\right).}$$

   It follows that $b_{2n}=0$ while $b_{2n+1}=\frac{4}{(2n+1)\pi}$ . All in all, we get that

   $$\chi_{[0,\pi]}-\chi_{[-\pi,0]}\sim\sum_0^\infty \frac{4}{(2n+1)\pi} \sin((2n+1)x).$$

3. $f(x)=|x|$ : Now we have an even function, so that $b_n=0$ for all $n$ . Similarly, since $f(x)\cos(nx)$ is even we get:

   $$\align{a_n&=\angles{|x|,\cos(nx)}=\frac{2}{\pi} \int_0^\pi x\cos(nx)\dx \\ &= \frac{2}{n\pi} \left( x\sin(nx)\mid_0^\pi+ \int_0^\pi \sin(nx)\dx\right)=\frac{2}{n^2\pi}\left((-1)^n-1\right)}.$$

   It follows that $a_{2n}=0$ while $a_{2n+1}=\frac{-4}{(2n+1)^2\pi}$ .

   However, if $n=0$ then the computation above is not well defined as we divided by zero! So for this case we need a separate computation, we produces

   $$\angles{|x|,1}=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\dx = \pi.$$

   Finally, we get that

   $$|x|\sim \frac{\pi}{2} - \sum_1^\infty \frac{4}{(2n+1)^2\pi} \cos((2n+1)x).$$

4. $f(x)=\sin^2(x)$ : While we can compute all the coefficients using the integrals, here we can use another method. Recall that $\sin(x)=Im(e^{ix})=\frac{e^{ix}-e^{-ix}}{2i}$ . Using this identity, we see that

   $$\align{\sin^2(x) &= \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2 = \frac{e^{2ix}-2+e^{-2ix}}{-4} \\ & = \frac{\cos(2x)-1}{-2}= \frac{1}{2}-\frac{\cos(2x)}{2}.}$$

   This is already Fourier series expression of $f(x)$ , and since there is always a unique such combination, this is the Fourier series of $f(x)$ .

Let $f\in E[-\pi,\pi]$ and denote the right limit at a point by $f(a^+)=\displaystyle{\lim_{x\to a^+}}f(x)$

We define the right half derivative as the limit, if it exists, of

$$\displaystyle{\lim_{x\to a^+}}\frac{f(x)-f(a^+)}{x-a}.$$

Similarly we define the left half derivative.

For simplicity of notation, we identify between the edge points $\pm \pi$ and consider its left and right half derivatives as the left half derivative at $\pi$ and right half derivative at $-\pi$ .

We denote by $E'[-\pi,\pi]\subseteq E[-\pi,\pi]$ all the piecewise continuous function on $[-\pi,\pi]$ that have left and right half derivative at every point in $[-\pi,\pi]$

Let $f\in E'[-\pi,\pi]$ . Then at a given point $\lambda \in(-\pi,\pi)$ the Fourier series converges pointwise to the average of the left and right limits, namely

$$\frac{f(\lambda^+)+f(\lambda^-)}{2} = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n\cos(n\lambda) + b_n\sin(n\lambda)\right).$$

At the edge points $\lambda =\pm \pi$ , it converges to $\frac{f(-\pi^+)+f(\pi^-)}{2}$ .

In particular, the series converges pointwise at each point where $f$ is continuous.

As the last theorem suggest, we should actually think about the edge points $\pm \pi$ as a single point.

1. We already saw that this theorem holds for the discontinuities of $f=\chi_{[0,\pi]}-\chi_{[-\pi,0]}$ , namely the pointwise convergence there are to the average of the left and right limits.

   Let’s consider the pointwise convergence at $x=\frac{\pi}{2}$ . Since $f(x)$ is continuous there and $f(\frac{\pi}{2})=1$ , this is going to be the limit. Writing the pointwise convergence of the Fourier series, we get:

   $$1=\sum_0^\infty \frac{4}{(2n+1)\pi} \sin((2n+1)\frac{\pi}{2})=\frac{4}{\pi}\sum_0^\infty \frac{(-1)^n}{2n+1} .$$

   Try to prove this using other methods (clue: $\arctan(1)=\frac{\pi}{4}$ ).

2. As another example, recall that the approximation for $f(x)=x$ looks like:

   True[[invalid wikilink: Pasted image 20240131120746.png]]

   Here the function is “continuous” except for the edges, where it jumps from $f(\pi^-)=\pi$ to $f(-\pi^+)=-\pi$ . The average is of course $\frac{f(-\pi^+)+f(\pi^-)}{2}=0$ , and the pointwise convergence at these points are

   $$2\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\overbrace{\frac{\sin\left(\pm k\pi\right)}{k}}^{=0}=0.$$

Compute the Fourier series for $f(x)=\chi_{[0,\pi]}$ . Where are its discontinuity points, and what should be the pointwise convergence there for the Fourier series?

Let $f_n:I\to \CC$ be a sequence of functions. If $\sum_1^\infty M_n< \infty$ for some sequence of numbers satisfying $\sup_x |f_n(x)|\leq M_n$ , then $\sum_1^\infty f_n(x)$ converges uniformly and in absolute value.

Suppose that $f\in C[-\pi, \pi]$ is continuous and $f(\pi)=f(-\pi)$ and that $f'(x)\in E[-\pi,\pi]$ (and in particular $f'(x)$ exists except for finitely many points).

Then writing the Fourier series of $f$ as:

$$f(x)\sim \frac{a_0}{2} + \sum_1^\infty (a_n \cos(nx)+ b_n\sin(nx)),$$

we have that the Fourier series of $f'$ is the element wise derivative:

$$f'(x)\sim \sum_1^\infty (-na_n \sin(nx)+ nb_n\cos(nx)).$$

Under the conditions of the Theorem above, the Fourier series of $f$ converges uniformly to $f$ .

Denote by $a_n,b_n$ , and $\tilde{a}_n,\tilde{b}_n$ the Fourier coefficients of $f(x)$ and $f'(x)$ respectively. By the theorem we know that $|a_n|=\frac{|\tilde{n}_n|}{n}, |b_n|=\frac{|\tilde{a}_n|}{n}$ for $n\geq 1$ . Using this fact and the Cauchy Shwarz inequality, we get

$$\sum_1^N (|a_n|+|b_n|)=\sum_1^N \frac{1}{n}(|\tilde{a}_n|+|\tilde{b}_n|)\leq \sqrt{\sum_1^N \frac{2}{n^2}}\sqrt{\sum_1^N\left( |\tilde{a}_n|^2+|\tilde{b}_n|^2\right)}.$$

If we can show that the last expression is bounded (and therefore converge) we could use Weierstrass theorem to prove uniform convergence. We already know that $\sum_1^\infty \frac{1}{n^2}$ is finite (and even know how to compute it), and for the other term we can use Bessel’s inequality to obtain:

$$\sum_1^N \left(|\tilde{a}_n^2| + |\tilde{b}_n^2|\right) \leq \norm{f'},$$

which completes the proof.

This function is periodic continuous, and its derivative is $f'=\chi_{[0,\pi]}-\chi_{[-\pi,0]}$ which is in $E[-\pi,\pi]$ so we can apply the last lemma. We already computed its Fourier series:

$$|x|\sim \frac{\pi}{2} - \sum_1^\infty \frac{4}{(2n+1)^2\pi} \cos((2n+1)x).$$

Since

$$\sum_1^\infty\sup_x|\frac{4}{(2n+1)^2\pi} \cos((2n+1)x)|\leq \sum_1^\infty \frac{4}{(2n+1)^2\pi}< \infty,$$

we can use Weierstrass’ theorem to conclude that its convergence is indeed uniform.

Furthermore, taking the term by term derivatives, we obtain another Fourier series that we already saw, namely:

$$\chi_{[0,\pi]}-\chi_{[-\pi,0]} \sim \sum_1^\infty \frac{4}{(2n+1)\pi} \sin((2n+1)x).$$

Note that the Fourier coefficients of $\chi_{[0,\pi]}-\chi_{[-\pi,0]}$ (which is not continuous) behave like $\frac{1}{n}$ while the coefficients of the continuous $|x|$ behave like $\frac{1}{n^2}$ .

The main idea of the proof is to move between the Fourier coefficients of $f$ and $f'$ via integration by parts. More specifically, write the Fourier series of both $f$ and $f'$ :

$$\align{f(x)& \sim \frac{a_0}{2} + \sum_1^\infty (a_n \cos(nx)+ b_n\sin(nx))\\ f'(x) &\sim \frac{\tilde{a}_0}{2} + \sum_1^\infty (\tilde{a}_n \cos(nx)+ \tilde{b}_n\sin(nx))}$$

Using integration by parts we have that

$$\align{\tilde{a}_n &= \angles{f',\cos(nx)}=\frac{1}{\pi}\int_{-\pi}^{\pi}f'(x)\cos(x)\dx\\ &=\frac{1}{\pi}\left(f(x)\cos(nx)\mid_{-\pi}^{\pi} - \int_{-\pi}^{\pi}nf(x)\sin(nx)\dx \right)\\&=\overbrace{(f(\pi)-f(-\pi))}^{=0}(-1)^n+n\angles{f,\sin(nx)} = nb_n.}$$

A similar computation shows that $\tilde{b}_n = -na_n$ . This shows that we can take the derivative element wise.

Suppose that we are given $f$ as in the theorem, however, we don’t know if $f(\pi)=f(-\pi)$ . Find the Fourier coefficients of $f'(x)$ .

Try to use the last theorem, instead of reproving it.

We usually study derivatives before integrals, and we sometimes get the feeling that they are easier. While this is usually true when trying to actually compute these functions, when speaking about the results, the integral, namely the antiderivative of a function is usually much “better” than the derivative of a function. Indeed, if we measure a function by how smooth it is, namely how many derivatives it has, then we always get that the anti derivative has at least (!) one derivative.

Suppose that $f\in E[-\pi, \pi]$ and denote its Fourier series by

$$f(x)\sim \frac{a_0}{2} + \sum_1^\infty (a_n \cos(nx)+ b_n\sin(nx)).$$

Then for any $[c,d]\subseteq [-\pi,\pi]$ we have the term by term integral:

$$\int_c^d f(t)\dt = \frac{a_0(d-c)}{2} + \sum_1^\infty (\frac{a_n}{n} (\sin(nd)-\sin(nc)) - \frac{b_n}{n}(\cos(nd)-\cos(nc)).$$

The main step here is to note that we can write

$$\align{\int_c^d f(t) \dt &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\pi\chi_{[c,d]}\dt = \angles{f,\pi\chi_{[c,d]}} \\ & =\angles{\frac{a_0}{2} + \sum_1^\infty (a_n \cos(nx)+ b_n\sin(nx)),\pi\chi_{[c,d]}}.}$$

Using the fact that the inner product is continuous (with respect to the $\norm{\cdot}_2$ norm), we can take the summation outside the integral to obtain:

$$\align{\int_c^d f(t) \dt &=\angles{\frac{a_0}{2}, \pi\chi_{[c,d]}} + \sum_1^\infty\angles{a_n \cos(nx)+ b_n\sin(nx),\pi\chi_{[c,d]}} \\ & = \int_c^d\frac{a_0}{2} \dt + \sum_1^\infty\int_c^d(a_n \cos(nx)+ b_n\sin(nx))\dt \\ & = \frac{a_0(d-c)}{2} \dt + \sum_1^\infty(\frac{a_n}{n} (\sin(nd)-\sin(bc))- \frac{b_n}{n}(\cos(nd)-\cos(nc))\dt.}$$

Let $f(x)=|x|$ which is continuous, and define

$$F(x)=\int_{-\pi}^{x} f(t)\dt-\frac{\pi^2}{2},$$

so that $F'(x)=f(x)$ .

What can you say about the convergence of the Fourier series of $F(x)$ ? What is it $\cl_2$ , $\cl_\infty$ and pointwise limit? Check to see if your guess is correct.

It is not hard to show that

$$F(x)=\cases{-\frac{x^2}{2}&x\leq 0\\ \frac{x^2}{2} & x\geq 0}.$$

Write it Fourier series as

$$F(x) \sim \frac{a_0}{2} + \sum_1^\infty (a_n \cos(nx)+ b_n\sin(nx)).$$

Since $F(x)$ is an odd function, we immediately get that $a_n=0$ for all $n$ . As for $b_n$ , using symmetry and integration by parts we get

$$\align{b_n&=\angles{F(x),\sin(nx)}=\frac{2}{\pi}\int_0^\pi \frac{x^2}{2} \sin(nx) \dx \\ & = -\frac{2}{\pi n} \left[ \frac{x^2\cos(nx)}{2}\mid_0^\pi -\int_0^\pi x\cos(nx)\dx \right] = -\frac{2}{\pi n} \left[ \frac{\pi^2(-1)^n}{2} - \int_0^\pi x\cos(nx)\dx \right].}$$

A second integration by parts gives:

$$\int_0^\pi x\cos(nx)\dx = x\frac{\sin(nx)}{n}\mid_0^\pi - \int_0^\pi \frac{\sin(nx)}{n}\dx=\frac{\cos(nx)}{n^2}\mid_0^\pi=\frac{(-1)^n-1}{n^2}.$$

All together we have that $b_n=-\frac{2}{\pi n} \left[ \frac{\pi^2(-1)^n}{2} + \frac{(-1)^n-1}{n^2} \right]$ .

Considering the converge of the Fourier series, since $F\in E[-\pi, \pi]$ we automatically get the $\cl_2$ -norm convergence. Also, the function is continuous “everywhere” (except the edge points) and has half derivatives, so Dirichlet’s theorem applies and we have pointwise convergence in $(-\pi,\pi)$ . On the edges, the pointwise convergence is to the average:

$$\frac{\displaystyle{\lim_{x\to-\pi+}F(x)+\lim_{x\to\pi-}F(x)}}{2}=\frac{\pi^2/2-\pi^2/2}{2}=0.$$

The function $F(x)$ is not continuous at the edges, since $F(-\pi)\neq F(\pi)$ so we can’t use the theorem about uniform convergence. In general the fact that the conditions of the theorem don’t hold doesn’t mean that its result doesn’t hold. However, in this case it is true, since if there is uniform convergence in $[-\pi,\pi]$ , then considering the functions as $2\pi$ periodic there will be uniform convergence everywhere, so the limit (of continuous function) is going to be continuous. As this is not true for $F$ , the convergence is not uniform.

Let $f_0(x)=sign(x)$ which is not continuous. Show that there are $f_1(x),f_2(x)$ such that $f_2'(x)=f_1(x)$ and $f_1'(x)=f_0(x)$ in almost every $x$ and both $f_1,f_2$ are continuous (including edge points).

The main idea is to start with $|x|$ as before, and to change it by a scalar to $f_1(x)=|x|+c$ . This keeps $f_1(x)$ as a continuous function with $f_1'(x)=f_0(x)$ . Let $F$ be as in the previous exercise, and define

$$F_C(x)=\int_{-\pi}^{x} (f_1(t)+C)\dt-\frac{\pi^2}{2} = F(x)+Cx.$$

The function $F_C(x)$ is still continuous in $(-\pi,\pi)$ and $F_C'(x)=f_1(x)$ . Also, it limit at the edge points are

$$\lim_{x\to-\pi^+}F_C(x)=-\frac{\pi^2}{2}-C\pi\;\;;\;\;\lim_{x\to\pi^-}F_C(x)=\frac{\pi^2}{2}+C\pi.$$

Choosing $C=-\frac{\pi}{2}$ will cause them to be equal and therefore make $F_C(x)$ continuous.

Let’s try to find continuous $f$ solving the equation

$$ \int_0^x f(t)\dt = x - f'(x) .$$

If the equation above holds, then in particular $f'(x)$ is by itself defined and continuous. Let’s write the Fourier expansion of $f$ as

$$f(x)\sim \frac{a_0}{2} + \sum_1^\infty (a_n \cos(nx)+ b_n\sin(nx)).$$

We would like to take the term by term derivative, but not all the conditions of the theorem hold - we don’t know that $f(-\pi)=f(\pi)$ . However, we can assume that it is true and continue on, but we need to check that our final solution does satisfy this condition (or alternatively, it satisfies the equation we started with). Under this assumption, we can write:

$$f'(x)\sim \sum_1^\infty (-n a_n \sin(nx)+ n b_n\cos(nx)).$$

For any function in $E[-\pi,\pi]$ we can do term by term integration, so putting everything together we obtain the equation:

$$\frac{a_0x}{2} + \sum_1^\infty (\frac{a_n}{n} \sin(nx)- \frac{b_n}{n}(\cos(nx)-1)) = x - \sum_1^\infty (-n a_n \sin(nx)+ n b_n\cos(nx)).$$

Moving everything to the same side, we get

$$\left(\frac{a_0}{2}-1\right)x + \sum_1^\infty \frac{b_n}{n} + \sum_2^\infty \left( a_n(\frac{1}{n}-n) \sin(nx) - b_n(\frac{1}{n}-n\right)\cos(nx))=0.$$

If all the “coefficients” are zero, then of course it solves the equation. However, note that the last equation is not a Fourier expansion (namely, it is not a linear combination of a basis), so we don’t know automatically that this is the only solution.

In any case, one way to make all the coefficient zero is by taking:

$$a_0=2,\; a_n =0\; \forall n\geq 2,\; b_n =0 \; \forall n\geq 1.$$

We are now left with $f(x)=1+a_1\cdot\cos(x)$ , and note that the functions in this family do satisfy $f(-\pi)=f(\pi)$ . Finally, just to be on the safe side, let’s see that they do satisfy our original equation:

$$\align{\int_0^x f(t)\dx & = \int_0^x (1+a_1\cos(t))\dx = x + a_1 \sin(t)\mid_0^x = x+a_1\sin(x)\\ x-f'(x) & = x-(1+a_1\cos(x))' = x + a_1\sin(x)}.$$