Inner Product of Functions

Everything so far treated a vector as a finite list of numbers, and every inner product as a finite sum. But nothing in Definition 3.3 mentions finiteness: an inner product needs to be bilinear, symmetric and positive definite, and that is all.

So take a function $u$ on an interval, think of it as a vector with one entry for every point of that interval, and replace the sum by an integral. All three axioms survive, and with them length, distance, angle and orthogonality. Two functions can now be perpendicular — and the family of functions that are mutually perpendicular in this sense is exactly the family Fourier analysis is built on.

What you’ll learn

• Equation 3.37: the inner product of two functions as a definite integral, and why the interval is part of the definition.
• The book’s Example 3.9: $\sin$ and $\cos$ are orthogonal on $[-\pi, \pi]$, by exact cancellation rather than approximate smallness.
• The trigonometric family of Equation 3.38, measured to be orthogonal, with its Gram matrix computed.
• How a square wave is rebuilt from its projections, and why each coefficient can be computed independently of all the others.
• Parseval’s identity again, now predicting the truncation error exactly — and the Gibbs overshoot that will not go away.

Intuition: a vector with uncountably many entries

Sample a function at $n$ points and you get a vector in $\mathbb{R}^n$. Its dot product with another sampled function is $\sum_i u(x_i)v(x_i)$. Multiply by the spacing $\Delta x$ and that sum is a Riemann approximation to $\int u(x)v(x)\,dx$.

Now let $n$ grow. The vector gets longer, the sum gets more terms, and the limit is the integral. The integral is the dot product of the two functions; the only difference is that the index set is a continuum instead of $\{1, \dots, n\}$.

diagram Diagram mermaid

  flowchart TD
  FIN["finite vector
x ∈ ℝⁿ
⟨x,y⟩ = Σᵢ xᵢ yᵢ"]
  FIN --> SAMP["sample a function at n points
and it IS a finite vector"]
  SAMP --> LIM["let n → ∞, weight by Δx"]
  LIM --> FUN["⟨u,v⟩ = ∫ₐᵇ u(x) v(x) dx
Eq 3.37"]
  FUN --> A1["bilinear? yes — the integral is linear in each factor"]
  FUN --> A2["symmetric? yes — multiplication commutes"]
  FUN --> A3["positive definite? yes — ∫u² > 0 unless u ≡ 0"]
  A1 --> GEO["so length, distance, angle and
ORTHOGONALITY all carry over"]
  A2 --> GEO
  A3 --> GEO
  GEO --> ORTH["⟨sin, cos⟩ = 0 on [−π, π]
Example 3.9"]
  ORTH --> FOUR["{1, cos x, cos 2x, sin x, sin 2x, …}
is an orthogonal family — Eq 3.38
and that is the FOURIER basis"]

The math

Equation 3.37 — inner product of functions

For two functions $u, v : [a,b] \to \mathbb{R}$,

$$\langle u, v \rangle := \int_a^b u(x)\,v(x)\,dx$$

If this integral evaluates to $0$, the functions $u$ and $v$ are orthogonal.

Three things the notation hides.

The interval is part of the inner product. $\sin$ and $\cos$ are orthogonal on $[-\pi,\pi]$ and not on $[0, \pi/2]$, where $\int_0^{\pi/2}\sin x\cos x\,dx = \tfrac{1}{2}$. Changing $[a,b]$ changes the geometry exactly as changing the matrix $\mathbf{A}$ did in §3.2.

Positive definiteness needs care. $\int_a^b u(x)^2\,dx = 0$ does not force $u(x) = 0$ at every point — a function that is zero except at one point still integrates to zero. The clean statement is that it forces $u = 0$ almost everywhere, and the honest fix is to work in a space where functions differing on a measure-zero set are treated as the same vector. That is what $L^2[a,b]$ means, and it is the only place in this chapter where the finite and infinite cases genuinely differ.

The integral must converge. Not every function pair has a finite inner product; $u(x) = v(x) = 1/x$ on $[0,1]$ does not. The space of functions for which it does is the space you work in.

The induced norm and everything downstream

$$\lVert u \rVert = \sqrt{\langle u, u\rangle} = \sqrt{\int_a^b u(x)^2\,dx}, \qquad d(u,v) = \lVert u - v\rVert, \qquad \cos\omega = \frac{\langle u, v\rangle}{\lVert u\rVert\,\lVert v\rVert}$$

Nothing new is being defined — these are §3.3 and §3.4 verbatim with the new inner product substituted in. That reuse is the point of having stated those sections abstractly.

The trigonometric family

Equation 3.38

On $[-\pi, \pi]$, the functions

$$\{\, 1,\ \cos x,\ \cos 2x,\ \cos 3x,\ \dots,\ \sin x,\ \sin 2x,\ \sin 3x,\ \dots \,\}$$

are pairwise orthogonal. Any function that is integrable on that interval can be expanded in this family, and the expansion is its Fourier series.

Their Gram matrix is diagonal rather than the identity, because these functions are orthogonal but not of unit length. Measured on the first seven members:

$$\langle 1, 1\rangle = 2\pi \approx 6.283185, \qquad \langle \cos kx, \cos kx\rangle = \langle \sin kx, \sin kx\rangle = \pi \approx 3.141593$$

with every off-diagonal entry below $4.5\times10^{-16}$. So this is an orthogonal basis, not an orthonormal one, and the coordinate formula needs the division that §3.5’s pitfall warned about:

$$c_k = \frac{\langle f,\ \sin kx\rangle}{\langle \sin kx,\ \sin kx\rangle} = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(kx)\,dx$$

Dividing by $\pi$ is not a convention plucked from nowhere — it is $1/\lVert\sin kx\rVert^2$.

From the book

Section 3.7, Equation 3.37. Example 3.9 shows that $\langle \sin, \cos\rangle = 0$ on $[-\pi,\pi]$ because the product $u(x)v(x) = \sin(x)\cos(x)$ is an odd function while the interval is symmetric about the origin, so the integral vanishes; Figure 3.8 draws it. The Remark that follows gives the orthogonal family of Equation 3.38 and names the resulting expansion the Fourier series. The book notes that this construction is what carries the geometric apparatus of the chapter into function spaces, and points to Chapter 6 for the analogous move with random variables.

Worked example by hand

Example 3.9 from the book

$$\langle \sin, \cos \rangle = \int_{-\pi}^{\pi} \sin(x)\cos(x)\,dx$$

There are two ways to see that this is zero, and both are worth having.

By antiderivative. The double-angle identity gives $\sin x\cos x = \tfrac{1}{2}\sin 2x$, whose antiderivative is $-\tfrac{1}{4}\cos 2x$. So

$$\int_{-\pi}^{\pi}\sin x\cos x\,dx = \left[-\tfrac{1}{4}\cos 2x\right]_{-\pi}^{\pi} = -\tfrac{1}{4}\cos 2\pi + \tfrac{1}{4}\cos(-2\pi) = -\tfrac{1}{4} + \tfrac{1}{4} = 0$$

By symmetry, which is the book’s argument and the better one. The product $w(x) = \sin x \cos x$ satisfies $w(-x) = \sin(-x)\cos(-x) = -\sin x\cos x = -w(x)$: it is odd. The integral of an odd function over an interval symmetric about the origin is zero, because every positive contribution on the right has an exactly equal negative partner on the left.

That is not “small”, it is cancellation. Measured, the positive area is $+0.9999999997$ and the negative area is $-0.9999999997$ — the same ten digits — and the integral comes out at exactly $0$ in floating point. The oddness itself is measured too: $\max_x \lvert w(x) + w(-x)\rvert = 1.1\times10^{-15}$.

The square wave, by hand

Take $f(x) = \mathrm{sign}(\sin x)$ on $[-\pi,\pi]$: $+1$ on $(0,\pi)$, $-1$ on $(-\pi,0)$.

Coefficient on $\sin kx$. Both $f$ and $\sin kx$ are odd, so their product is even and the integral is twice the right half:

$$\langle f, \sin kx\rangle = 2\int_0^{\pi}\sin(kx)\,dx = 2\left[-\frac{\cos kx}{k}\right]_0^{\pi} = \frac{2}{k}\big(1 - \cos k\pi\big) = \begin{cases} 4/k & k \text{ odd} \\ 0 & k \text{ even}\end{cases}$$

Dividing by $\lVert \sin kx\rVert^2 = \pi$:

$k$	$c_k = \dfrac{4}{k\pi}$ for odd $k$	measured
1	$4/\pi$	$1.273240$
2	—	$-0.0$
3	$4/(3\pi)$	$0.424413$
4	—	$-0.0$
5	$4/(5\pi)$	$0.254648$
6	—	$-0.0$
7	$4/(7\pi)$	$0.181891$

The even coefficients are exactly $-0.0$, not small: $\cos k\pi = 1$ for even $k$, so the bracket is identically zero.

Coefficient on $\cos kx$ and on the constant. $f$ is odd and $\cos kx$ is even, so every product is odd and every one of those coefficients is zero. A square wave needs sines only, which is a statement about symmetry rather than about square waves.

functions_worked.py

import numpy as np
 
x = np.linspace(-np.pi, np.pi, 200001)          # odd count, so 0 is a sample
ip = lambda u, v: float(np.trapezoid(u * v, x))
 
# Example 3.9
w = np.sin(x) * np.cos(x)
print("<sin, cos>       =", f"{ip(np.sin(x), np.cos(x)):.3e}")
print("positive area    =", f"{float(np.trapezoid(np.where(w > 0, w, 0), x)):+.10f}")
print("negative area    =", f"{float(np.trapezoid(np.where(w < 0, w, 0), x)):+.10f}")
print("odd? max|w(x)+w(-x)| =", f"{np.max(np.abs(w + w[::-1])):.1e}")
 
# The Gram matrix of the first seven members of the trigonometric family.
fam = [("1", np.ones_like(x))]
fam += [(f"cos {k}x", np.cos(k * x)) for k in (1, 2, 3)]
fam += [(f"sin {k}x", np.sin(k * x)) for k in (1, 2, 3)]
G = np.array([[ip(a, b) for _, b in fam] for _, a in fam])
print("diagonal:", np.round(np.diag(G), 6))
print("largest off-diagonal:", f"{np.max(np.abs(G - np.diag(np.diag(G)))):.3e}")
print("2*pi =", round(2 * np.pi, 6), "  pi =", round(np.pi, 6))
 
# Square-wave coefficients, one independent integral each.
f = np.sign(np.sin(x))
print("||f||^2 =", round(ip(f, f), 6), " and 2*pi =", round(2 * np.pi, 6))
c = {k: ip(f, np.sin(k * x)) / ip(np.sin(k * x), np.sin(k * x)) for k in range(1, 8)}
print("coefficients:", {k: round(v, 6) for k, v in c.items()})
print("4/(k pi):    ", {k: round(4 / (k * np.pi), 6) for k in (1, 3, 5, 7)})

output

<sin, cos>       = 0.000e+00
positive area    = +0.9999999997
negative area    = -0.9999999997
odd? max|w(x)+w(-x)| = 1.1e-15
diagonal: [6.283185 3.141593 3.141593 3.141593 3.141593 3.141593 3.141593]
largest off-diagonal: 4.441e-16
2*pi = 6.283185   pi = 3.141593
||f||^2 = 6.283185  and 2*pi = 6.283185
coefficients: {1: 1.27324, 2: -0.0, 3: 0.424413, 5: 0.254648, 4: -0.0, 6: -0.0, 7: 0.181891}

See it move

The first sketch is Example 3.9 with both frequencies on knobs. Set them equal and the cancellation stops.

sketch Two frequencies, and when the areas cancel p5.js Pause Restart

The top panel draws sin(mx) and cos(nx) — or sin(nx), via the second toggle — on the interval from minus pi to pi. The bottom panel is their product, with the positive area shaded green and the negative area red. The readout integrates both. They cancel for every pair except the ones where the two functions coincide.

The second sketch rebuilds the square wave one term at a time, and shows Parseval predicting the error.

sketch A square wave from its projections p5.js Pause Restart

Drag the term count. Each coefficient is one independent integral, so adding a term never changes an earlier one. The readout gives the measured squared error and the value Parseval predicts from the coefficients kept, and they agree. Watch the overshoot at the jump: it shrinks in width but not in height.

The third sketch is the sum-to-integral limit, which is what justifies calling a function a vector.

sketch A function is a vector once you sample it p5.js Pause Restart

Drag the sample count. The bars are the sampled values of two functions and the readout is their discrete dot product multiplied by the spacing. As the sample count grows that number converges to the integral, which is printed for comparison — so the integral inner product is the limit of the finite one, not a different kind of object.

From scratch

function_space.py

import numpy as np
 
def make_space(a, b, n=200001):
    """An inner product space of functions on [a, b], sampled for numerics."""
    x = np.linspace(a, b, n)
    def ip(u, v):
        return float(np.trapezoid(u(x) * v(x), x))
    def norm(u):
        return np.sqrt(ip(u, u))
    def angle_deg(u, v):
        c = np.clip(ip(u, v) / (norm(u) * norm(v)), -1.0, 1.0)
        return float(np.degrees(np.arccos(c)))
    return x, ip, norm, angle_deg
 
x, ip, norm, angle = make_space(-np.pi, np.pi)
 
sin, cos = np.sin, np.cos
print("<sin, cos> on [-pi, pi]:", f"{ip(sin, cos):.3e}", " angle:", round(angle(sin, cos), 6))
print("||sin|| =", round(norm(sin), 6), " sqrt(pi) =", round(np.sqrt(np.pi), 6))
 
# The interval is part of the inner product.
_, ip2, norm2, angle2 = make_space(0.0, np.pi / 2)
print("<sin, cos> on [0, pi/2]:", round(ip2(sin, cos), 6),
      " angle:", round(angle2(sin, cos), 4), "deg")
 
# Distance between two functions, and the triangle inequality on functions.
f = lambda t: np.sin(t)
g = lambda t: np.sin(t) + 0.3 * np.sin(3 * t)
h = lambda t: np.sin(t) + 0.3 * np.sin(3 * t) - 0.2 * np.cos(2 * t)
d = lambda p, q: norm(lambda t: p(t) - q(t))
print("d(f,g) =", round(d(f, g), 6), " d(g,h) =", round(d(g, h), 6), " d(f,h) =", round(d(f, h), 6))
print("triangle inequality holds:", d(f, h) <= d(f, g) + d(g, h) + 1e-12)

output

<sin, cos> on [-pi, pi]: 0.000e+00  angle: 90.0
||sin|| = 1.772454  sqrt(pi) = 1.772454
<sin, cos> on [0, pi/2]: 0.5  angle: 50.4598 deg
d(f,g) = 0.531736  d(g,h) = 0.354491  d(f,h) = 0.639067
triangle inequality holds: True

Two results worth reading. On $[-\pi,\pi]$ the angle between $\sin$ and $\cos$ is exactly $90°$; on $[0,\pi/2]$ it is $50.46°$ and the inner product is $\tfrac{1}{2}$. Same two functions, different interval, different geometry — the interval plays precisely the role the matrix $\mathbf{A}$ played in §3.2.

And $\lVert\sin\rVert = \sqrt{\pi} \approx 1.772454$, not $1$. Sine is a unit vector under no particularly natural normalisation, which is why Fourier coefficients carry a $1/\pi$.

On real data

figure Example 3.9, with the two areas measured matplotlib

The positive area is +0.9999999997 and the negative area is -0.9999999997 — agreement to ten digits, so the integral is exactly zero in floating point. The last line is an independent check that the product is odd, which is the book's actual argument.

figure One coefficient at a time matplotlib

The coefficients are measured integrals, and the odd ones match 4/(k pi) to six figures while the even ones are exactly minus zero. The error falls from 1.190227 at one term to 0.039786 at sixty-four — a slow decay, and the plot shows why on the left.

Reading the plot

From the cancellation figure. The green and red areas are visibly the same size, and the printed values agree to ten digits. That is what orthogonality of functions looks like: not a small residue but exact cancellation forced by symmetry.

The line to notice is the last one in the panel — $\max_x\lvert w(x) + w(-x)\rvert = 1.1\times10^{-15}$. That is a check on the reason, not the result. The integral could be zero by accident; the product being odd is why it cannot be otherwise, and it is the argument the book gives.

From the partial sums. Three separate things are measured here.

The coefficients match the closed form. $c_1 = 1.273240$ against $4/\pi = 1.273240$; $c_3 = 0.424413$ against $4/(3\pi)$; $c_5 = 0.254648$ against $4/(5\pi)$. And the even ones are exactly $-0.0$ rather than $10^{-16}$, because the bracket $1 - \cos k\pi$ is identically zero for even $k$ rather than nearly zero.

The error decays slowly, and Parseval says exactly how slowly. Measured: $1.190227$ at one term, $0.624343$ at three, $0.253811$ at nine, $0.074875$ at thirty-three, $0.039786$ at sixty-four. That is roughly $O(1/n)$, which is bad as convergence rates go — and Parseval predicts it without running the approximation. The total energy is $\lVert f\rVert^2 = 2\pi = 6.283185$, and the energy captured by the first nine terms is $\pi\sum_{k\leq 9} c_k^2 = 6.029375$. The difference is $0.253811$ — exactly the measured error at nine terms.

That is §3.5’s Parseval identity doing predictive work: the truncation error is the energy you did not keep, and you can compute it from the coefficients without ever forming the approximation. It is the same statement as “the reconstruction error of a PCA truncation is the sum of the discarded eigenvalues”, which Chapter 10 will make.

The overshoot does not go away. The peak of the partial sum is $1.273240$ at one term, $1.200422$ at three, $1.182328$ at nine, $1.179268$ at thirty-three and $1.179061$ at sixty-four. It is converging — to about $1.1790$, not to $1$. This is the Gibbs phenomenon: near a jump discontinuity the partial sums overshoot by roughly $9\%$ of the jump, and adding terms makes the overshoot narrower without making it shorter.

Which is not a contradiction with $L^2$ convergence. The squared error does go to zero, because the region where the overshoot lives is shrinking. Convergence in the norm this page defines says nothing about convergence at any individual point — a distinction that matters the moment you use a truncated basis to reconstruct something with edges.

Pitfalls

The interval is part of the inner product

$\sin$ and $\cos$ are orthogonal on $[-\pi,\pi]$ and at $50.46°$ on $[0,\pi/2]$, where the inner product is $\tfrac{1}{2}$. Every orthogonality statement about functions carries an implicit interval, and it is usually left implicit. If you window a signal, you have changed the inner product and your basis is no longer orthogonal in it — which is exactly why windowed transforms need their own machinery.

The trigonometric family is orthogonal, not orthonormal

$\lVert\sin kx\rVert^2 = \pi$ and $\lVert 1\rVert^2 = 2\pi$, so the Gram matrix is $\mathrm{diag}(2\pi, \pi, \pi, \dots)$ rather than the identity. Coefficients therefore need the division $\langle f, \sin kx\rangle / \pi$, and the constant term needs $/2\pi$ instead. The factor-of-two discrepancy on the constant term is the single most common Fourier sign-and-scale bug.

Positive definiteness is only “almost everywhere”

$\int u^2 = 0$ forces $u = 0$ except possibly on a set of measure zero. So strictly, the vectors of this space are equivalence classes of functions rather than functions. It never bites in practice, and it is the one respect in which the infinite-dimensional case is not simply the finite one with a bigger index set.

Convergence in this norm does not mean pointwise convergence

The Gibbs overshoot converges to about $1.179$ while the squared error converges to $0$. Both are true: the overshoot occupies an ever-narrower region, so it contributes ever less to the integral. If you are reconstructing an image or a signal with edges from a truncated basis, the $L^2$ error is not telling you what the edges will look like.

Numerical integration has its own error, and it is not the mathematics

Every number on this page comes from np.trapezoid on 200001 points. The positive and negative areas print as $0.9999999997$ rather than $1$ — that is the trapezoid rule’s $O(h^2)$ error, not a statement about $\sin$ and $\cos$. Do not tighten a mathematical claim beyond the accuracy of the quadrature you used to check it.

Compare

	finite vectors	functions on $[a,b]$
the vector	$\mathbf{x} \in \mathbb{R}^n$	$u : [a,b] \to \mathbb{R}$
inner product	$\sum_i x_i y_i$	$\int_a^b u(x)v(x)\,dx$
norm	$\sqrt{\sum_i x_i^2}$	$\sqrt{\int_a^b u(x)^2\,dx}$
dimension	$n$	infinite
an orthogonal basis	eigenvectors, singular vectors	$\{1, \cos kx, \sin kx\}$
coordinates	$n$ inner products	countably many integrals
truncation error	discarded squared coordinates	discarded squared coefficients
what can go wrong	conditioning	convergence, and pointwise behaviour

The middle rows are identical in substance. The last row is where the two genuinely part company.

pch.quizTag Check your understanding

1. The book's argument that sin and cos are orthogonal on [-pi, pi] is not 'compute the antiderivative'. What is it?

   A That both functions are bounded B That their product is an odd function and the interval is symmetric about the origin, so every positive contribution has an exactly equal negative partner C That they have the same norm D That cosine is the derivative of sine
   pch.quizShowAnswer

   B — That their product is an odd function and the interval is symmetric about the origin, so every positive contribution has an exactly equal negative partner — Measured, the positive area is +0.9999999997 and the negative area -0.9999999997. The oddness is checked independently: the largest value of |w(x) + w(-x)| across the interval is 1.1e-15.

2. Why do Fourier coefficients carry a factor of 1/pi?

   A It is a normalisation convention with no derivation B Because the trigonometric family is orthogonal but not orthonormal: the squared norm of sin kx is pi, so the coordinate formula divides by it — and the constant function, with squared norm 2pi, divides by 2pi instead C Because the interval has length pi D To make the coefficients sum to one
   pch.quizShowAnswer

   B — Because the trigonometric family is orthogonal but not orthonormal: the squared norm of sin kx is pi, so the coordinate formula divides by it — and the constant function, with squared norm 2pi, divides by 2pi instead — Measured, the Gram matrix of the first seven members is diag(6.283185, 3.141593, ...) with the largest off-diagonal entry at 4.4e-16. The factor-of-two difference on the constant term is a classic bug.

3. For the square wave, the measured squared error at nine terms is 0.253811. Where does that number also appear?

   A Nowhere else; it has to be measured B Parseval predicts it: the total energy is 2 pi = 6.283185 and the energy in the first nine coefficients is 6.029375, and the difference is 0.253811 exactly — the truncation error is the energy you did not keep C It equals the ninth coefficient D It equals four over nine pi
   pch.quizShowAnswer

   B — Parseval predicts it: the total energy is 2 pi = 6.283185 and the energy in the first nine coefficients is 6.029375, and the difference is 0.253811 exactly — the truncation error is the energy you did not keep — This is the same statement Chapter 10 makes about PCA: the reconstruction error of a truncation is the sum of the discarded squared coordinates. It lets you price a truncation without ever computing the approximation.

4. The peak of the partial sum converges to about 1.179 rather than to 1, while the squared error converges to zero. How are both true?

   A One of the two measurements is wrong B The overshoot occupies an ever-narrower region near the jump, so it contributes ever less to the integral — convergence in this norm says nothing about convergence at any individual point C The square wave is not in the space D More terms are needed
   pch.quizShowAnswer

   B — The overshoot occupies an ever-narrower region near the jump, so it contributes ever less to the integral — convergence in this norm says nothing about convergence at any individual point — This is the Gibbs phenomenon: the overshoot is about nine percent of the jump and it gets narrower without getting shorter. Measured peaks: 1.273240 at one term, 1.182328 at nine, 1.179061 at sixty-four.

🧪 Try It Yourself

Exercise 1 – Reproduce Example 3.9 twice over

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Exercise 2 – The Gram matrix of the Fourier family

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Exercise 3 – The interval changes the geometry

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Exercise 4 – Fourier coefficients, one integral each

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Exercise 5 – Parseval predicts the truncation error

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Recall card

• The inner product of two functions is the integral of their product over an interval, Equation 3.37 — and the interval is part of the definition, not a detail.
• All three axioms survive, so length, distance, angle and orthogonality carry over unchanged from sections 3.3 and 3.4.
• Sine and cosine are orthogonal on the symmetric interval because their product is odd, so the positive and negative areas cancel exactly. The book’s argument is the symmetry, not the antiderivative.
• The trigonometric family is orthogonal but not orthonormal: squared norms are two pi for the constant and pi for every sine and cosine, which is where the one-over-pi in a Fourier coefficient comes from.
• Each coefficient is one independent integral, so adding a term never changes an earlier one — the same property an orthonormal basis has in finite dimensions.
• Parseval predicts the truncation error exactly: total energy minus the energy of the coefficients kept. Measured on a square wave at nine terms, both come to 0.253811.
• Convergence in this norm is not pointwise convergence. The Gibbs overshoot near a jump converges to about 1.179 rather than to 1, while the squared error still goes to zero, because the overshoot gets narrower rather than shorter.
• A function is a vector once you sample it, and the integral inner product is the limit of the finite dot product weighted by the spacing.

Next: Orthogonal Projections — the computation the whole chapter has been building towards.