Hilbert Space Foundations & Orthonormal Systems






Hilbert Spaces — Lecture 1: Inner Products, Orthogonality, Projections (Expanded Proofs)





MATH 621 — Real Analysis I (Hilbert Spaces Unit)

Lecture 1: Inner Products, Orthogonality, Orthogonal Projections, Completeness


Purpose, Context, and Learning Outcomes

These notes establish the core geometric and analytic facts used throughout the Bases & Frames module:
how inner products induce norms, how orthogonality organizes subspaces via orthogonal complements, and why
closedness/completeness are indispensable for best approximation and projections. Every later theorem (Bessel,
Parseval, Riesz bases, frames) depends on this material.

By the end, you can:

  1. State the inner-product axioms (complex convention: linear in the first slot) and derive the induced norm.
  2. Prove Cauchy–Schwarz, triangle inequality, and the parallelogram identity with explicit step-by-step arguments.
  3. Compute orthogonal complements and orthogonal projections (both ON and non-ON settings).
  4. Prove the Projection Theorem with all details and deduce operator properties of $P_M$.
  5. Use $(\overline M)^\perp=M^\perp$ and $(M^\perp)^\perp=\overline M$; characterize $H=M\oplus M^\perp$.

Inner Products and Induced Norms

Definition 1.1 (Inner product). An inner product on a vector space $H$ over $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$ is a map $\langle \cdot,\cdot\rangle:H\times H\to\mathbb{K}$ such that for all $x,y,z\in H$, $\alpha\in\mathbb{K}$:

  1. Conjugate symmetry: $\langle x,y\rangle=\overline{\langle y,x\rangle}$.
  2. Linearity (first slot): $\langle \alpha x+y,z\rangle=\alpha\langle x,z\rangle+\langle y,z\rangle$.
  3. Positive-definiteness: $\langle x,x\rangle\ge0$ with equality iff $x=0$.

The induced norm is $\|x\|:=\sqrt{\langle x,x\rangle}$.

Examples.

  • $\mathbb{C}^n$: $\langle x,y\rangle=\sum_{k=1}^n x_k\overline{y_k}$.
  • $\ell^2$: $\langle x,y\rangle=\sum_{k\ge1} x_k\overline{y_k}$ with $\sum |x_k|^2<\infty$.
  • $L^2[a,b]$: $\langle f,g\rangle=\int_a^b f\overline g$ (identify a.e.-equal functions).

Worked examples (inner products & inequalities).

  1. Finite-dimensional, standard inner product. In $\mathbb{R}^2$ with $\langle x,y\rangle=x_1y_1+x_2y_2$ and $x=(3,-1)$, $y=(2,5)$:
    1. $\langle x,y\rangle=3\cdot2+(-1)\cdot5=6-5=1$.
    2. $\|x\|=\sqrt{3^2+(-1)^2}=\sqrt{10}$, $\|y\|=\sqrt{2^2+5^2}=\sqrt{29}$.
    3. $|\langle x,y\rangle|=1\le \sqrt{10}\sqrt{29}$ (C–S). Equality fails since $x,y$ are not colinear.
    4. $\|x+y\|^2=\|(5,4)\|^2=41$ and $(\|x\|+\|y\|)^2=(\sqrt{10}+\sqrt{29})^2$; hence $\|x+y\|\le \|x\|+\|y\|$.
    5. Parallelogram check: $\|x-y\|^2=\|(1,-6)\|^2=37$ and $41+37=2\cdot10+2\cdot29$.
  2. Weighted inner product. On $\mathbb{R}^2$, define $\langle x,y\rangle_W = x^{\top} W y$ with $W=\begin{pmatrix}2&0\\0&1\end{pmatrix}$. For $x=(1,2)$, $y=(3,1)$:
    1. $Wy=\begin{pmatrix}6\\1\end{pmatrix}$, so $\langle x,y\rangle_W=(1,2)\cdot(6,1)=8$.
    2. $\|x\|_W=\sqrt{x^{\top}Wx}=\sqrt{6}$; $\|y\|_W=\sqrt{y^{\top}Wy}=\sqrt{19}$.
    3. $|\langle x,y\rangle_W|=8\le \sqrt{6}\,\sqrt{19}$, confirming C–S in a nonstandard inner product.
  3. Function space. In $L^2[0,1]$, take $f(t)=t$, $g(t)=1$:
    1. $\langle f,g\rangle=\int_0^1 t\,dt = \frac{1}{2}$.
    2. $\|f\|^2=\int_0^1 t^2 dt=\frac{1}{3}$, $\|g\|^2=\int_0^1 1\,dt=1$.
    3. $|\langle f,g\rangle|^2=\frac{1}{4}\le \frac{1}{3}\cdot 1$; strict inequality holds.

Theorem 1.2 (Cauchy–Schwarz via completing the square). For all $x,y\in H$, $|\langle x,y\rangle|\le\|x\|\|y\|$.

Strategy. Consider the nonnegative quadratic $p(\lambda)=\|x-\lambda y\|^2$ in a complex variable $\lambda$. Minimizing $p$ forces its discriminant-like term to be nonnegative, which is precisely C–S.
  1. If $y=0$, both sides are $0$ and there is nothing to prove. Assume $y\ne 0$.
  2. Define $p(\lambda):=\|x-\lambda y\|^2=\langle x-\lambda y,\,x-\lambda y\rangle\ge 0$ for all $\lambda\in\mathbb{C}$.
  3. Expand using sesquilinearity:
    $$\begin{aligned}
    p(\lambda)&=\langle x,x\rangle-\lambda\langle y,x\rangle-\overline{\lambda}\langle x,y\rangle+|\lambda|^2\langle y,y\rangle\\
    &=\|x\|^2-\lambda\overline{\langle x,y\rangle}-\overline{\lambda}\langle x,y\rangle+|\lambda|^2\|y\|^2.
    \end{aligned}$$
  4. Complete the square by choosing $\lambda_0:=\dfrac{\langle x,y\rangle}{\|y\|^2}$. Then
    $$p(\lambda)=\|y\|^2\bigg|\lambda-\lambda_0\bigg|^2+\Big(\|x\|^2-\frac{|\langle x,y\rangle|^2}{\|y\|^2}\Big)\ge0.$$
  5. At the minimum $\lambda=\lambda_0$, the first term vanishes, so the constant term is $\ge 0$:
    $$\|x\|^2-\frac{|\langle x,y\rangle|^2}{\|y\|^2}\ge 0\quad\Rightarrow\quad |\langle x,y\rangle|\le \|x\|\,\|y\|.$$
  6. Equality case. Equality holds iff $p(\lambda_0)=0$, i.e., $x-\lambda_0 y=0$, so $x$ and $y$ are linearly dependent (or $y=0$).

Corollary 1.3 (Triangle inequality). For all $x,y\in H$, $\|x+y\|\le \|x\|+\|y\|$.

Strategy. Expand $\|x+y\|^2$ and control the cross term by C–S: $2\Re\langle x,y\rangle\le 2\|x\|\|y\|$.
  1. Compute
    $$\|x+y\|^2=\langle x+y,\,x+y\rangle=\|x\|^2+2\Re\langle x,y\rangle+\|y\|^2.$$
  2. By C–S, $|\langle x,y\rangle|\le \|x\|\|y\|$, hence $2\Re\langle x,y\rangle\le 2\|x\|\|y\|$.
  3. Therefore $\|x+y\|^2\le (\|x\|+\|y\|)^2$. Taking square roots (both sides nonnegative) gives the result.
  4. Equality case. Occurs iff $\langle x,y\rangle=\|x\|\|y\|$ with nonnegative real inner product, e.g., when $x,y$ are positively linearly dependent.

Identity 1.4 (Parallelogram law). For all $x,y\in H$, $$\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.$$

Strategy. Expand both squares and observe the cross terms cancel.
  1. $\|x+y\|^2=\|x\|^2+2\Re\langle x,y\rangle+\|y\|^2$.
  2. $\|x-y\|^2=\|x\|^2-2\Re\langle x,y\rangle+\|y\|^2$.
  3. Add and simplify to obtain the identity.

Orthogonality and Orthogonal Complements

Definition 2.1. For a linear subspace $M\subset H$, $$M^\perp:=\{u\in H:\ \langle u,m\rangle=0\ \text{for all }m\in M\}.$$

Lemma 2.2. $M^\perp$ is a closed subspace.

Strategy. Linearity is immediate from linearity of $\langle\cdot,\cdot\rangle$. Closedness uses continuity of the inner product.
  1. Subspace. If $u,v\in M^\perp$ and $\alpha,\beta\in\mathbb K$, then for any $m\in M$,
    $$\langle \alpha u+\beta v, m\rangle=\alpha\langle u,m\rangle+\beta\langle v,m\rangle=0.$$
    Hence $\alpha u+\beta v\in M^\perp$.
  2. Closedness. Suppose $u_n\in M^\perp$ and $u_n\to u$ in norm. For any fixed $m\in M$,
    $$|\langle u_n-u, m\rangle|\le\|u_n-u\|\,\|m\|\to 0$$
    (C–S). Since $\langle u_n,m\rangle=0$, we obtain $\langle u,m\rangle=\lim_n\langle u_n,m\rangle=0$. Thus $u\in M^\perp$.

Lemma 2.3. $(\overline M)^\perp=M^\perp$.

Strategy. Use density: vectors orthogonal to $M$ are orthogonal to limits of sequences from $M$.
  1. Since $M\subset\overline M$, we have $(\overline M)^\perp\subset M^\perp$.
  2. Conversely, let $x\in M^\perp$ and $y\in\overline M$. Choose $m_k\in M$ with $m_k\to y$.
  3. Continuity of the inner product yields $\langle x,y\rangle=\lim_k \langle x,m_k\rangle=0$. Hence $x\in(\overline M)^\perp$.

Lemma 2.4. $(M^\perp)^\perp=\overline M$.

Proof deferred to §3.4 after the Projection Theorem.

Concrete computations.

  • $\mathbb{R}^3$: $M=\mathrm{span}\{(1,1,0),(0,1,1)\}$ gives $M^\perp=\mathrm{span}\{(1,-1,1)\}$.
  • $\ell^2$: $M=\{x:x_1=\cdots=x_k=0\}\Rightarrow M^\perp=\mathrm{span}\{e_1,\dots,e_k\}$.
  • $L^2[-1,1]$: polynomials are dense $\Rightarrow$ $M^\perp=\{0\}$ for $M=\overline{\mathrm{span}}\{1,t,t^2,\dots\}$.

Worked examples (orthogonality & complements).

  1. $\mathbb{R}^3$. Let $M=\mathrm{span}\{(1,1,0),(0,1,1)\}$. Find $M^\perp$.
    1. Let $u=(a,b,c)$. Orthogonality gives $\langle u,(1,1,0)\rangle=a+b=0$ and $\langle u,(0,1,1)\rangle=b+c=0$.
    2. Solve: $a=-b$ and $c=-b$. Hence $u=b(-1,1,-1)=\alpha(1,-1,1)$ with $\alpha=-b$.
    3. Therefore $M^\perp=\mathrm{span}\{(1,-1,1)\}$.
  2. $\ell^2$. For $M=\{x\in \ell^2: x_1=\cdots=x_k=0\}$:
    1. If $u\perp M$, then $\langle u, e_j\rangle=0$ for all $j>k$, so $u_j=0$ for $j>k$.
    2. Conversely, any $u$ supported on $\{1,\dots,k\}$ is orthogonal to every $m\in M$. Thus $M^\perp=\mathrm{span}\{e_1,\dots,e_k\}$.
  3. $L^2[-1,1]$. Let $M=\mathrm{span}\{1,t\}$. Show $h(t)=t^2-\tfrac{1}{3}\in M^\perp$:
    1. $\langle h,1\rangle=\int_{-1}^1 (t^2-\tfrac{1}{3})\,dt=\Big[\tfrac{t^3}{3}-\tfrac{t}{3}\Big]_{-1}^1=0$.
    2. $\langle h,t\rangle=\int_{-1}^1 (t^3-\tfrac{t}{3})\,dt=0$ (odd integrand on a symmetric interval).

The Projection Theorem and Orthogonal Projections

Theorem 3.1 (Projection Theorem). If $M\subset H$ is closed, then for each $x\in H$ there exists a unique decomposition
$$x=m+u,\qquad m\in M,\ u\in M^\perp.$$
Equivalently, there is a unique $m\in M$ minimizing $\|x-y\|$ ($y\in M$), and the residual $x-m$ is orthogonal to $M$.

Strategy. (Existence) Minimize $\|x-y\|$ over $M$ using completeness/closedness: a minimizing sequence is Cauchy by the parallelogram identity and converges into $M$. (Orthogonality) Perturb the minimizer along any $z\in M$ and use calculus of a quadratic to see that the directional derivative at $0$ vanishes. (Uniqueness) If two decompositions exist, subtract and use $M\cap M^\perp=\{0\}$.
  1. Define the distance. Set $d:=\inf_{y\in M}\|x-y\|$. Choose a sequence $(y_n)\subset M$ with $\|x-y_n\|\to d$.
  2. Show $(y_n)$ is Cauchy. Apply the parallelogram law to $x-y_n$ and $x-y_m$:
    $$\|y_n-y_m\|^2=2\|x-\tfrac{y_n+y_m}{2}\|^2-\|x-y_n\|^2-\|x-y_m\|^2\le 2(d+\varepsilon)^2-2d^2,$$
    for large $n,m$. Hence $\|y_n-y_m\|\to 0$ as $n,m\to\infty$.
  3. Use completeness and closedness. Since $(y_n)$ is Cauchy in $H$ and $M$ is closed, $y_n\to m\in M$. By continuity of the norm, $\|x-m\|=d$.
  4. Orthogonality of the residual. Fix any $z\in M$ and consider $\phi(t):=\|x-(m+tz)\|^2=\|x-m\|^2-2t\,\Re\langle x-m,z\rangle+t^2\|z\|^2$ for real $t$.
    Since $t=0$ minimizes $\phi$, the derivative at $0$ vanishes: $\Re\langle x-m,z\rangle=0$. Replacing $z$ by $iz$ (complex case) yields also $\Im\langle x-m,z\rangle=0$; thus $\langle x-m,z\rangle=0$ for all $z\in M$ and $x-m\in M^\perp$.
  5. Uniqueness. If $x=m_1+u_1=m_2+u_2$ with $m_i\in M$ and $u_i\in M^\perp$, then $m_1-m_2=u_2-u_1$ lies in $M\cap M^\perp$, hence equals $0$. So $m_1=m_2$ and $u_1=u_2$.

Orthogonal projector. Define $P_Mx:=m$ in the decomposition $x=m+u$. Then
$$P_M^2=P_M,\quad P_M^*=P_M,\quad \operatorname{ran}P_M=M,\quad \ker P_M=M^\perp,\quad \|P_M\|=1\ (M\neq\{0\}).$$

Strategy. Each property follows immediately from the decomposition $x=m+u$ with orthogonality $m\perp u$ and uniqueness.
  1. Idempotence. $P_M(P_Mx)=P_M(m)=m=P_Mx$; hence $P_M^2=P_M$.
  2. Self-adjointness. For $x=m_x+u_x$ and $y=m_y+u_y$,
    $$\langle P_Mx,\,y\rangle=\langle m_x,\,m_y+u_y\rangle=\langle m_x,m_y\rangle=\langle m_x+u_x,\,m_y\rangle=\langle x,\,P_My\rangle.$$
    Thus $P_M^*=P_M$.
  3. Range and kernel. $\operatorname{ran}P_M=\{m: x=m+u\}=M$; $\ker P_M=\{x: m=0\}=M^\perp$.
  4. Norm. From Pythagoras, $\|P_Mx\|^2=\|m\|^2\le\|m\|^2+\|u\|^2=\|x\|^2$. Hence $\|P_M\|\le1$. If $M\neq\{0\}$ choose $x\in M$ with $\|x\|=1$; then $\|P_Mx\|=\|x\|=1$, giving $\|P_M\|=1$.

Computational forms.

  • ON family. If $\{e_1,\dots,e_k\}$ is orthonormal and $M=\mathrm{span}\{e_j\}$, then
    $$P_M x=\sum_{j=1}^k \langle x,e_j\rangle e_j,\qquad \|x-P_Mx\|^2=\|x\|^2-\sum_{j=1}^k|\langle x,e_j\rangle|^2.$$

    Why this formula? Orthogonality forces $x-P_Mx\perp e_j$ for each $j$; taking inner products solves for the coefficients.
  • Non-ON family (finite-dim). With $A=[v_1\ \cdots\ v_k]$ and $M=\mathrm{span}\{v_j\}$, seek $c\in\mathbb{K}^k$ with $P_Mx=Ac$ and $x-Ac\perp M$.
    Orthogonality means $A^*(x-Ac)=0$, i.e., the normal equations $(A^*A)c=A^*x$.
    If $A^*A$ is invertible (the $v_j$ are linearly independent), then $c=(A^*A)^{-1}A^*x$ and
    $$P_M= A(A^*A)^{-1}A^*.$$

    Why normal equations? We impose $\langle x-Ac, v_j\rangle=0$ for each basis vector $v_j$; stacking these conditions gives $A^*(x-Ac)=0$.

Worked examples.

  1. $\mathbb{R}^3$: $v_1=(1,1,0)$, $v_2=(0,1,1)$, $x=(1,2,3)$. Then $A=\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}$, $A^TA=\begin{bmatrix}2&1\\1&2\end{bmatrix}$, $A^Tx=(3,5)^T$, so $c=(1,2)$ and $P_Mx=(1,3,2)$; residual $(0,-1,1)\in M^\perp$.
  2. $\ell^2$: $M=\mathrm{span}\{e_1,\dots,e_k\}$ gives $P_M(x_1,x_2,\dots)=(x_1,\dots,x_k,0,\dots)$ and $\|x-P_Mx\|^2=\sum_{j>k}|x_j|^2$.
  3. $L^2[-1,1]$: With ON set $\{p_0,p_1\}=\{1/\sqrt2,\sqrt{3/2}\,t\}$ and $M=\mathrm{span}\{1,t\}$, the best affine $L^2$-approximation of $f(t)=t^2$ is $P_Mf=1/3$; residual $t^2-1/3\perp\{1,t\}$.

Additional worked examples (projections & least squares).

  1. Projection onto a line in $\mathbb{R}^2$. Let $M=\mathrm{span}\{v\}$ with $v=(2,1)$ and $x=(3,4)$.
    1. Use $P_M x= \dfrac{\langle x,v\rangle}{\langle v,v\rangle}v$ (valid in any inner product space).
    2. $\langle x,v\rangle=3\cdot2+4\cdot1=10$, $\langle v,v\rangle=2^2+1^2=5$.
    3. $P_M x= \dfrac{10}{5}(2,1)=(4,2)$; residual $r=x-P_Mx=(-1,2)$ and $\langle r,v\rangle=0$ (orthogonal).
  2. Least-squares line via normal equations. Fit $y\approx a+bx$ to data $(1,1),(2,2),(3,2),(4,3)$.
    1. Design matrix $A=\begin{bmatrix}1&1\\1&2\\1&3\\1&4\end{bmatrix}$, vector $y=\begin{bmatrix}1\\2\\2\\3\end{bmatrix}$. Coefficients $c=\begin{bmatrix}a\\b\end{bmatrix}$ solve $(A^{\!*}A)c=A^{\!*}y$ (with $\!*=$ transpose in $\mathbb{R}$).
    2. $A^{\!*}A=\begin{bmatrix}4&10\\10&30\end{bmatrix}$, $A^{\!*}y=\begin{bmatrix}8\\23\end{bmatrix}$.
    3. Solve: $\det=20$, $a=\dfrac{30\cdot8-10\cdot23}{20}=0.5$, $b=\dfrac{-10\cdot8+4\cdot23}{20}=0.6$.
    4. The fitted line is $\hat y=0.5+0.6x$; algebraically, $Ac$ is the orthogonal projection of $y$ onto $\operatorname{col}(A)$.
  3. Projection in $L^2[-1,1]$ onto $\mathrm{span}\{1,t\}$. For $f(t)=t^2$, find $P_M f=c_0\cdot 1+c_1\cdot t$.
    1. Set $\langle f-P_M f,1\rangle=0$ and $\langle f-P_M f,t\rangle=0$.
    2. Inner products: $\langle 1,1\rangle=2$, $\langle 1,t\rangle=0$, $\langle t,t\rangle=\int_{-1}^1 t^2 dt= \frac{2}{3}$; $\langle f,1\rangle= \frac{2}{3}$, $\langle f,t\rangle=\int_{-1}^1 t^3 dt=0$.
    3. Equations: $ \frac{2}{3}-2c_0=0 \Rightarrow c_0= \frac{1}{3}$; $0- \frac{2}{3}c_1=0 \Rightarrow c_1=0$.
    4. Thus $P_M f= \frac{1}{3}$ and $f-P_M f=t^2- \frac{1}{3} \perp \{1,t\}$.

3.4 Proof of $(M^\perp)^\perp=\overline M$ (re: Lemma 2.4)

Strategy. One inclusion is immediate: $M\subset(M^\perp)^\perp$ by definition, hence $\overline M\subset(M^\perp)^\perp$. For the converse, take $x\in(M^\perp)^\perp$, project $x$ onto $\overline M$ and use orthogonality of the residual.
  1. Trivial inclusion. If $m\in M$ and $u\in M^\perp$, then $\langle m,u\rangle=0$, so $m\in(M^\perp)^\perp$. Therefore $\overline M\subset(M^\perp)^\perp$.
  2. Reverse inclusion via projection. Let $x\in(M^\perp)^\perp$. Apply Theorem 3.1 to the closed subspace $\overline M$ to write $x=m_0+r$ with $m_0\in\overline M$ and $r\perp\overline M$.
  3. Because $\overline M\subset (M^\perp)^\perp$, we have $r\in (\overline M)^\perp\subset M^\perp$. Since also $x\in(M^\perp)^\perp$, orthogonality gives
    $$0=\langle x,r\rangle=\langle m_0+r,\,r\rangle=\langle m_0,r\rangle+\|r\|^2=\|r\|^2,$$
    hence $r=0$ and $x=m_0\in\overline M$.
  4. Therefore $(M^\perp)^\perp\subset\overline M$, completing the proof.

Pythagoras & Distance. For $x=m+u$ with $m\in M$, $u\in M^\perp$,
$$\|x\|^2=\|m\|^2+\|u\|^2,\qquad \mathrm{dist}(x,M)=\|u\|=\|x-P_Mx\|.$$

Completeness and the Role of Closedness

Definition 4.1 (Hilbert space). An inner-product space $H$ is a Hilbert space if it is complete: every Cauchy sequence converges in $H$.

Examples. $\ell^2$ and $L^2[a,b]$ are complete (Riesz–Fischer). By contrast, the space of polynomials $V$ with $\langle f,g\rangle=\int_0^1 f g$ is not complete; its completion is $L^2[0,1]$.

Closedness is necessary for best approximation. If $M$ is not closed, a minimizing sequence for $\inf_{y\in M}\|x-y\|$ may converge to a point outside $M$—no best approximant exists in $M$. Passing to $\overline M$ restores existence/uniqueness.

Corollary 4.2 (Decomposition iff closed). $H=M\oplus M^\perp$ iff $M$ is closed.

Strategy. (⇐) Apply Theorem 3.1 when $M$ is closed. (⇒) If a bounded projection onto $M$ exists, then $M$ is the kernel of a bounded operator and hence closed.
  1. (⇐) If $M$ is closed, Theorem 3.1 yields $x=m+u$ with $m\in M$, $u\in M^\perp$ for all $x\in H$.
  2. (⇒) Suppose $H=M\oplus M^\perp$. Define $P(x)=m$ for $x=m+u$. Then $P$ is linear, $\|P\|\le 1$ (as above), and $M=\operatorname{ran}P=\ker(I-P)$. Kernels of bounded operators are closed, so $M$ is closed.

Counterexample for nonclosed $M$. In $L^2[0,1]$, let $M$ be the (nonclosed) set of polynomials. Take $f(t)=|t-\tfrac12|$. Polynomials are dense, so $\inf_{p\in M}\|f-p\|=0$, but no polynomial equals $f$ in $L^2$. Thus the distance is not attained in $M$.

Illustrative examples (completeness & closedness).

  1. Closed subspace with explicit projection. In $L^2[-1,1]$, let $M=\{\text{even functions}\}$. For any $f$, the orthogonal projection is the even part $P_M f(t)= \dfrac{f(t)+f(-t)}{2}$.
    1. Decompose $f=f_{\text{even}}+f_{\text{odd}}$ with $f_{\text{even}}(t)= \dfrac{f(t)+f(-t)}{2}$ and $f_{\text{odd}}(t)= \dfrac{f(t)-f(-t)}{2}$.
    2. $f_{\text{even}}\in M$ and $ \langle f_{\text{even}},f_{\text{odd}} \rangle=0$ by symmetry, so $f_{\text{odd}}\in M^\perp$.
    3. Hence $f=P_M f+(I-P_M)f$ is the desired orthogonal decomposition.
  2. Nonclosed subspace with no best approximant. In $\ell^2$, let $M=c_{00}$ (finite-support sequences). Take $x=(1,\tfrac{1}{2},\tfrac{1}{4},\tfrac{1}{8},\dots)$, which lies in $\ell^2$.
    1. Let $x^{(N)}=(1,\tfrac{1}{2},\dots,\tfrac{1}{2^{N-1}},0,0,\dots)\in M$. Then \[
      \|x-x^{(N)}\|^2= \sum_{n\ge N} \frac{1}{4^{n-1}}= \frac{1}{4^{N-1}}\cdot \frac{1}{1- \frac{1}{4}} \to 0.
      \]
    2. Thus $ \inf_{m\in M} \|x-m\|=0$, but no $m\in M$ equals $x$; the infimum is not attained. This demonstrates the necessity of closedness in Theorem 3.1.

Formative Checks and Short Exercises

FC1 (norm from inner product). Prove $\|x\|=\sup_{\|y\|=1}|\langle x,y\rangle|$.
Hint: C–S gives $\le$; equality at $y=x/\|x\|$ for $x\ne0$.

FC2 (characterize $M^\perp$ in $\ell^2$). For $M=\{x\in\ell^2: x_1=\cdots=x_k=0\}$, show $M^\perp=\mathrm{span}\{e_1,\dots,e_k\}$.

FC3 (best affine $L^2$-fit). On $[-1,1]$, find the best $ax+b$ approximating $t^2$ in $L^2$ (answer: $b=1/3$, $a=0$).

© 2025 — V. Oussa. Lecture notes with fully expanded proofs and strategy boxes. Complex convention: linear in the first slot.