# Tuesday November 12 ## Closed Subspaces and Orthogonal Projections **Definition:** Let $H$ be a Hilbert space, then a subspace $M \subseteq H$ is *closed* if $x_n \mapsvia{H} x$ with $\theset{x_n} \subset M$ implies that $x\in M$. > Note that finite-dimensional subspaces are *always* closed, so this is a purely infinite-dimensional phenomenon. **Proposition:** Given any *set* $M$, then $$ M^\perp \definedas \theset{x\in H \suchthat \inner{x}{y} = 0 ~\forall y\in M} $$ is always a closed subspace. *Proof:* Homework problem. **Lemma:** Let $M$ be a closed subspace of $H$ and $x\in H$. Then 1. There exists a unique $y \in M$ that is *closest* to $y$, i.e. $$ \exists y\in M \suchthat \norm{x - y} = \inf_{y' \in M}\norm{x - y'} .$$ 2. Defining $z \definedas x-y$, then $z\in M^\perp$. **Consequence 1:** If $M \subseteq H$ is a closed subspace, then $(M^\perp)^\perp = M$. > Note that $M \subseteq M^{\perp \perp}$ by definition. (Easy to check) To show that $M^{\perp \perp} \subseteq M$, let $x\in M^{\perp \perp}$, then $x = y + z$ where $y\in M$ and $z\in M^\perp$. Then $$ \inner{x}{z} = \inner{y}{z} + \inner{z}{z} \implies \norm{z}^2 = 0 \implies z = 0 \implies x=y .$$ **Consequence 2:** **Theorem:** If $M \subseteq H$ is a closed subspace, then $H = M \oplus M^\perp$, i.e. $$ x\in H \implies x = y + z, \quad y\in M, ~z\in M^\perp ,$$ and $y,z$ are the unique elements in $M, M^\perp$ that are closest to $x$. *Proof of Lemma (Part 1):* Let $\delta \definedas \displaystyle\inf_{y' \in M} \norm{x - y'}$, which is a sequence of real numbers that is bounded below, and thus this infimum is attained. Then there is a sequence $\theset{y_n} \subseteq M$ such that $\norm{x - y_n} \to \delta$. Consider the following parallelogram: ![Image](figures/2019-11-12-11:25.png)\ Then by the parallelogram theorem, we have $$ 2(\norm{y_n - x}^2 + \norm{y_m - x}^2) = \norm{y_n - y_m}^2 + \norm{y_n + y_m - 2x}^2. $$ which yields \begin{align*} \norm{y_n - y_m}^2 &= 2 \norm{y_n - x}^2 + 2\norm{y_m - x}^2 - 4\norm{\frac 1 2 (y_n + y_m) - x}^2 \\ &\leq 2 \norm{y_n - x}^2 + 2\norm{y_m - x}^2 - 4\delta^2 \to 0, \end{align*} since $\norm{y_n - x}_H \to 0$ since $y_n \to_H x$. It follows that $\theset{y_n}$ is Cauchy in $H$, so $y_n \mapsvia{H} y \in H$. But since the $y_n$ were in $M$ and $M$ is closed, we in fact have $y\in M$. Since $\norm{x - y_n} \to \norm{x - y} = \delta$, we have the existence of $x$. > We'll establish uniqueness after part 2. *Proof of Lemma (Part 2):* Let $u\in M$, we want to show that $$ \inner{z}{u} = \inner{x-y}{u} = 0 .$$ Without loss of generality, we can assume that $\inner{z}{u} \in \RR$, since $u$ satisfies this property iff any complex scalar multiple does. Let $f(t) = \norm{z + tu}^2$ where $t\in \RR$. Then $f(t) = \norm{z}^2 + zt\inner{z}{y} = t^2 \norm{u}^2$. We know that $t$ attains a minimum at $t=0$, since $z + tu = x - (y + u)$, but $y$ was the closest element to $x$ and thus the norm is minimized exactly when $z + tu = x - y \implies t=0$. Because of this fact, we know that $f'(0) = 0$. But by using Calculus, we can compute that $f'(0) = 2 \inner{z}{u}$, so $\inner{z}{u}$ must equal zero. Now to show uniqueness, let $y' \in M$ and suppose $y' \neq u$ but $\norm{x-y'} = \delta$. Then $x- y' = (x-y) + (y-y')$. But these are two orthogonal terms, so we can apply Pythagoras to obtain \begin{align*} \norm{x-y'}^2 &= \norm{x-y}^2 + \norm{y-y'}^2 \\ \implies \delta &= \delta + c \\ \implies c &= 0 \\ \implies \norm{y-y'} &= 0 \\ \implies y &= y' .\end{align*} > Note: the statement is the important thing here, less so this particular proof. ## Trigonometric Series **Theorem:** Let $e_n(x) \definedas e^{2\pi i n x}$ for all $x\in [0, 1]$ and $n\in \ZZ$. Then $\theset{e_n}_{n\in \ZZ}$ is an *orthonormal basis* for $L^2([0, 1])$. > Note: Orthonormality is easily check, so the crux of the proof is showing it's a basis. > Note: Elements in $\mathrm{span}\theset{e_n}$ are referred to as *trigonometric polynomials*. Goal: We'll show that the span of the trigonometric polynomials are dense in $L^2([0, 1])$. This will be a consequence of the following theorem: ### Trigonometric Polynomials are Dense in $C^0([0, 1])$ **Theorem (Periodic Analogue of the Weierstrass Approximation Theorem):** If $f\in C(\Pi)$ (where $\Pi$ is a torus) and $\varepsilon > 0$, then there exists a trigonometric polynomial $P$ such that $\abs{f(x) - P(x)} < \varepsilon$ uniformly for all $x\in \Pi$. > Note that this measures closes in the *uniform* norm. We can relate these by $$ \norm{f(x) -P(x)}_{L^2} \leq \norm{f(x) - P(x)}_{\infty} \text{, i.e. } \int_0^1 \abs{f(x) - P(x)}^2 \leq \sup_x \abs{f(x) - P(x)}^2 .$$ *Proof:* Identify $\Pi = [- \frac 1 2, \frac 1 2)$. Suppose there exists a sequence $\theset{Q_k}$ of trigonometric polynomials such that \begin{align*} Q_k(x) &\geq 0 \quad \text{ for all } x, k \\ \int_{-1/2}^{1/2} Q_k(x) ~dx &= 1 \quad \forall k \\ \forall \delta>0, \quad Q_k(x) &\mapsvia{u} 0 \text{ uniformly on } \Pi\setminus[-\delta, \delta] .\end{align*} > Note that these properties are similar to what we wanted from approximations to the identity. Define $$ P_k(x) = \int_{-1/2}^{1/2} f(y) Q_k(x - y) ~dy $$ by convolving over the circle, then $P_k$ is also a trigonometric polynomial. We then have \begin{align*} I = \abs{ P_k(x) - f(x) } \leq \int_{-1/2}^{1/2} \abs{ f(x-y) - f(x) } Q_k(y) ~dy \quad \text{by Property 2} .\end{align*} We can now note that $f$ is continuous on a compact set, so it is uniformly continuous, and thus for $y$ small enough, we can find a $\delta$ such that $$ \abs{f(x-y) - f(x)} < \varepsilon/2 \text{ for all } x \in B(\delta, x) .$$ But this lets us break the integral into two pieces, \begin{align*} I &\leq \int_{y \in B_\delta} \abs{ f(x-y) - f(x) } Q_k(y) ~dy + \int_{y \in B_\delta^c} \abs{ f(x-y) - f(x) } Q_k(y) ~dy \\ &< \int_{y \in B_\delta} \frac \varepsilon 2~ Q_k(y) ~dy + \int_{y \in B_\delta^c} \abs{ f(x-y) - f(x) } Q_k(y) ~dy \\ &\leq \int_{y \in B_\delta} \frac \varepsilon 2~ Q_k(y) ~dy + \frac \varepsilon 2~ \quad \text{(for $k$ large enough)} \\ &\to 0 \quad \quad \text{since } Q_k \mapsvia{u} 0 .\end{align*} $\qed$ *Constructing $Q_k$:* Define $$ Q_k(x) = c_k \left( \frac {1 + \cos(2\pi x)}{2} \right)^k, $$ where $c_k$ is chosen to normalize the integral to 1 to satisfy property 2. Property 1 is clear, so we just need to show 3, Since $\cos(x)$ is decreasing on $[\delta, \frac 1 2]$, $$ Q_k(x) \leq Q_k(\delta) = c_k \left( \frac{1 + \cos(2\pi \delta)}{2} \right)^k .$$ Note that the numerator is less than 2, so the entire term is a constant that is less than 1 being raised to the $k$ power. So this goes to zero exponentially, the question now depends on the growth of $c_k$. It turns out that $c_k \leq (k+1)\pi$, so it only grows linearly. So the whole quantity indeed goes to zero. We can now write \begin{align*} 1 &= 2c_k \int_0^{1/2} \left( \frac{1 + \cos(2\pi x)}{2} \right)^k dx \\ &= 2c_k \int_0^{1/2} \left( \frac{1 + \cos(2\pi x)}{2} \right)^k \sin(2\pi x) dx \\ &= \frac{2c_k}{\pi} \int_0^1 u^k ~du = \frac{2c_k}{\pi(k+1)} .\end{align*} $\qed$ > Note: this is a nice proof! Question: when is a function equal to its Fourier series? We have $L^2$ convergence, but when do we get pointwise? **Theorem from the 1960s:** any $L^2$ function (in particular continuous functions) converges to its Fourier series *almost everywhere*.