Week 4 Tutorial Solutions

Integrating

1. For every $n \in \N$ we have $|\phi_n(x)| \le 1$ for all $x \in [0,1)$ so \[ \int |\phi_n|^2 \intd \mu \le \int 1 \intd \mu \le 2 \pi \] and $\phi_n$ therefore belongs to $\mathsf{L}^2(X,\mathscr{B},\mu)$.
2. First we check that $\langle f, g \rangle$ is well-defined. This follows from Young's inequality \[ \left| \int f \cdot g \intd \mu \right| \le \int | f \cdot g | \intd \mu \le \int \dfrac{|f|^2}{2} + \dfrac{|g|^2}{2} \intd \mu \l \infty \] because $f$ and $g$ belong to $\mathsf{L}^2(X,\mathscr{B},\mu)$. It is bilinear because the integral is linear and certainly symmetric. Lastly we must check that $\langle f,f \rangle = 0$ implies $f = 0$. This is indeed the case because \[ 0 = \int |f|^2 \intd \mu = \langle f, f \rangle \] means exactly that $f = 0$ in $L^2(X,\mathscr{B},\mu)$.
3. We need to verify that $\langle \phi_n, \phi_j \rangle = 0$ for all $n \ne j$ in $\N$ and that $\langle \phi_n, \phi_n \rangle = 1$ for all $n \in \N$. As all functions are continuous we can calculate the integral by calculating the Riemann integral of the product $\phi_n \phi_k$. This is then a calculus exercise.
4. We calculate that \[ \begin{aligned} 0 \le {} & \left\| f - \sum_{n=0}^N \langle f, \phi_n \rangle \phi_n \right\|_\mathsf{2}^\mathsf{2} \\ = {} & \left\langle f - \sum_{n=0}^N \langle f, \phi_n \rangle \phi_n, f - \sum_{k=0}^N \langle f, \phi_k \rangle \phi_k \right\rangle \\ = {} & \langle f,f \rangle - \left\langle f, \sum_{k=0}^N \langle f, \phi_k \rangle \phi_k \right\rangle - \left\langle \sum_{n=0}^N \langle f, \phi_n \rangle \phi_n , f \right\rangle\\ & \qquad + \sum_{n=0}^N \sum_{k=0}^N \langle f, \phi_n \rangle \langle f, \phi_k \rangle \langle \phi_k, \phi_n \rangle \\ = {} & \| f \|_\mathsf{2}^2 - \sum_{n=0}^N |\langle f, \phi_n \rangle|^2 \end{aligned} \] giving the desired inequality. Since \[ \sum_{n=0}^N |(\mathcal{F} f)(n) |^2 \le \| f \|_\mathsf{2}^2 \] for all $N \in \N$ it must be the case that \[ \sum_{n=0}^N |(\mathcal{F} f)(n) |^2 \] converges to a finite value, which means $\mathcal{F} f$ belongs to $\ell^\mathsf{2}(\N \cup \{0\})$.