Here are the exercises for the last two posts, Convolution and good kernels and Cesàro summability and Fejér’s theorem.

Exercise 1: Let $f$ be a real-valued function which is $2\pi$-periodic and integrable on any finite interval. Show that, for $a,b \in \textbf{R},$

$\displaystyle\int_a^b f(x)dx = \int_{a+2\pi}^{b+2\pi}f(x)dx = \int_{a-2\pi}^{b-2\pi}f(x)dx.$

And that

$\displaystyle\int_{-\pi}^{\pi} f(x+a)dx = \int_{-\pi}^{\pi} f(x)dx = \int_{-\pi+a}^{\pi+a} f(x)dx.$

Solution:

Let $g:\textbf{R} \rightarrow \textbf{R}$ be defined as $g(x) = x + 2\pi.$ Then $g'(x) = 1$ and by the periodicity of $f$ we have $f(x) = f(g(x)).$ So

$\displaystyle\int_a^bf(x)dx = \int_a^b f (g (x)) g'(x)dx = F \circ g(b) - F \circ g(a)$

where $F$ is a primitive for $f.$ Indeed, by the chain rule we have $\frac{d}{dx}(F \circ g (x)) = f(g(x))g'(x)$ so we only need to apply the fundamental theorem of calculus. But again by the fundamental theorem of calculus, we have

$\displaystyle\int_{a+2\pi}^{b+2\pi}f(x)dx = F(b+2\pi) - F(a+2\pi) = F \circ g(b) - F \circ g (a)$

so we may conclude.The same argument gives the second equality of the first point. What this means is that for such a function $f,$ the integral of $f$ over an integral is invariant by a translation of length equal to the period of $f.$

For the second point, note that with a similar argument as above, we get

$\displaystyle\int_{-\pi}^{\pi} f(x+a)dx = \int_{-\pi}^{\pi}f(g(x))g'(x)dx = \int_{g(-\pi)}^{g(\pi)}f(x)dx = \int_{-\pi+a}^{\pi+a}f(x)dx$

for $g(x) = x+a.$ As for the second equality, we have

$\displaystyle\int_{-\pi+a}^{\pi+a}f(x)dx = \int_{-\pi+a}^{-\pi}f(x)dx + \int_{-\pi}^{\pi}f(x)dx + \int_{\pi}^{\pi+a}f(x)dx.$

However, applying what we learned in the first set of equalities of this exercise, we can translate the domain of integration of the first term in the right side by $2\pi$ and get

$\displaystyle\int_{-\pi+a}^{-\pi}f(x)dx = \int_{\pi+a}^{\pi}f(x)dx = -\int_{\pi}^{\pi+a}f(x)dx$

hence

$\displaystyle\int_{-\pi+a}^{\pi+a}f(x)dx = \int_{-\pi}^{\pi}f(x)dx.$

QED

Exercise 2: Show that if a series of complex numbers converges to $s$, then it is Cesàro summable to $s.$

Solution:

Let $\sum c_n$ be a series converging to $s.$ Denoting the partial sums of this series by $s_n,$ we need to show that $\sigma_N := \frac{s_0 + \cdots + s_{N-1}}{N} \rightarrow s$ as $N \rightarrow \infty.$

Since $\sum c_n = s,$ for every $\epsilon > 0$ there exists $m \in \textbf{N}$ such that $|s_n - s| < \epsilon$ for all $n \geq m.$ So for large enough $N,$ we get

$|\sigma_N - s| = \left|\displaystyle\frac{s_0 + \cdots + s_{N-1} - Ns}{N} \right|$

$= \left|\displaystyle\frac{1}{N} \sum_{n=0}^{N-1} (s_n-s) \right|$

$= \left|\displaystyle\frac{1}{N}\sum_{n=0}^{m-1}(s_n-s) + \frac{1}{N}\sum_{n=m}^{N-1}(s_n-s) \right|$

$\leq \displaystyle\frac{1}{N}\sum_{n=0}^{m-1}|s_n-s| + \frac{1}{N} \sum_{n=m}^{N-1}|s_n-s|$

$< \displaystyle\frac{mM}{N} + \frac{(N-1-m)\epsilon}{N}$

where $M$ is an upper bound for $|s_n-s|.$ Since $\frac{mM}{N} \rightarrow 0$ and $\frac{N-1-m}{N} \rightarrow 1$ as $N \rightarrow \infty,$ we have

$\displaystyle\lim_{N \rightarrow \infty} |\sigma_N-s| < \epsilon$

for any $\epsilon > 0,$ hence $\sigma_N \rightarrow s$ as wanted.

QED

Exercise 3: Show that Lagrange’s trigonometric identities hold. i.e.

$\displaystyle\sum_{n=0}^{N} \cos(n\theta) = \frac{1}{2} + \frac{\sin((N+1/2)\theta)}{2\sin(\theta/2))}$

and

$\displaystyle\sum_{n=0}^N \sin(n\theta) = \frac{1}{2}\cot(\theta/2) - \frac{\cos((N+1/2)\theta)}{2\sin(\theta/2)}$

Solution: I’ll actually only post the solution to the first identity. The idea is that $1 + \cos(\theta) + \cdots + \cos(n\theta)$ is the real part of $1+e^{i\theta} + \cdots e^{in\theta}$ which can be written as

$\displaystyle\frac{1-e^{(n+1)\theta}}{1-e^{i\theta}} = \frac{1-\cos((n+1)\theta) + i\sin((n+1)\theta)}{1-\cos(\theta) + i\sin(\theta)}.$

We then multiply both the numerator and denominator by the complex conjugate of the denominator. The denominator will then be real and given by:

$(1-\cos(\theta))^2 + \sin^2(\theta) = 1-2\cos(\theta) + 1 = 2(1-\cos(\theta)) = 4\sin^2(\theta/2)$

while the real part of the numerator will be given by

$(1-\cos((n+1)\theta))(1-\cos(\theta)) + \sin((n+1)\theta)\sin(\theta)$

$=(1-\cos((n+1)\theta)(2\sin^2(\theta/2)) + \sin((n+1)\theta)2\sin(\theta/2)\cos(\theta/2)$

$=2\sin^2(\theta/2) - 2\sin(\theta/2)[\cos((n+1)\theta)\sin(\theta/2)-\sin((n+1)\theta)\cos(\theta/2)]$

$=2\sin^2(\theta/2) - 2\sin(\theta/2)[\sin(\theta/2-(n+1)\theta)].$

Now, note that the expression in brackets is equal to

$-\sin(\theta/2 + n\theta) = -\sin((N+1/2)\theta).$

Putting the numerator over the denominator, we thus get

$\displaystyle\sum_{n=0}^N \cos(n\theta) = \frac{2\sin^2(\theta/2) + 2\sin(\theta/2)\sin((N+1/2)\theta)}{4\sin^2(\theta/2)}$

$= \displaystyle\frac{1}{2} + \frac{\sin((N+1/2)\theta)}{2\sin(\theta/2)}.$

QED

I actually use the other identity too in the next exercise but I am too lazy to write it down.

Exercise 4: Show that the Fejér kernel is given by

$F_N(x) = \displaystyle\frac{1}{N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)}.$

Recall that the Fejér kernel is by definition $F_N = \frac{1}{N}\sum_{n=0}^{N-1}D_n$ where $D_n$ is the n-th Dirichlet kernel. Well, in Exercise 5, I’ll show that

$D_N(x) = \displaystyle\frac{\sin((N+1/2)x}{\sin(x/2)}.$

So we get

$NF_N(x) = \displaystyle\sum_{n=0}^{N-1} \frac{\sin(Nx+x/2)}{\sin(x/2)}$

$= \displaystyle\frac{1}{\sin(x/2)}\sum_{n=0}^{N-1}(\cos(Nx)\sin(x/2) + \sin(Nx)\cos(x/2))$

$= 1- \displaystyle\frac{1}{2} + \frac{\sin((N-1+1/2)x)}{2\sin(x/2)} + \cot(\frac{x}{2})\left(\frac{1}{2}\cot(\frac{x}{2})-\frac{\cos((N+1/2)x)}{2\sin(x/2)}\right)$

where the last equality is justified by the identities of Exercise 3. This is equal to

$\displaystyle\frac{1}{2} + \frac{\sin((N-1/2)x}{2\sin(x/2)} + \frac{\cos^2(x/2)}{2\sin^2(x/2)} - \frac{\cos(x/2)}{\sin(x/2)}\frac{\cos((N-1/2)x)}{2\sin(x/2)}$

$=\displaystyle\frac{\sin^2(\frac{x}{2})+\sin(N-\frac{1}{2})x)\sin(\frac{x}{2})+\cos^2(\frac{x}{2})-\cos(\frac{x}{2})\cos((N-\frac{1}{2})x)}{2\sin^2(x/2)}$

$= \displaystyle \frac{1-\cos((N-1/2)x + x/2)}{2\sin^2(x/2)} = \frac{1-\cos(2(Nx/2))}{2\sin^2(x/2)}$

$= \displaystyle\frac{2\sin^2(Nx/2)}{2\sin^2(x/2)}$

Which gives exactly what we wanted.

QED

Exercise 5: Show that the N-th Dirichlet kernel is given by

$D_N(x) = \displaystyle\frac{\sin((N+1/2)x}{\sin(x/2)}.$

Solution:

Recall that by definition $D_N(x) = \sum_{n=-N}^N e^{inx}.$ Let $\omega = e^{ix}.$ Then since

$(1+\omega + \omega^2 + \cdots + \omega^N)(1-\omega) = 1-\omega^{N+1}$

and

$(\omega^{-N} + \omega^{-N+1} + \cdots + \omega^{-1})(1-\omega) = \omega^{-N} - 1,$

we get

$D_N(x) = \displaystyle\sum_{n=-N}^{N} \omega^n = \sum_{n=0}^{N}\omega^n + \sum_{n=-N}^{-1} \omega^n$

$= \displaystyle\frac{1-\omega^{N+1}}{1-\omega} + \frac{\omega^{-N}-1}{1-\omega}$

$=\displaystyle\frac{\omega^{-N} - \omega^{N+1}}{1-\omega}.$

Multiplying by $\frac{\omega^{-1/2}}{\omega^{-1/2}},$ we thus get

$D_N(x) = \displaystyle\frac{\omega^{-N-1/2} - \omega^{N+1/2}}{\omega^{-1/2} - \omega^{1/2}}$

$=\displaystyle\frac{e^{-i(N+1/2)x} - e^{i(N+1/2)x}}{e^{-ix/2} - e^{ix/2}}.$

But since $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} = \frac{e^{-ix} - e^{ix}}{-2i},$ this last equality gives

$D_N(x) = \displaystyle\frac{\sin((N+1/2)x}{\sin(x/2)}$

as we wanted.

QED

From → Analysis, Exercices