Inner Product Spaces

(a) An inner product. Symmetry: swapping $x$ and $y$ turns $x_1{y}_2+x_2{y}_1$ into $y_1{x}_2+y_2{x}_1$, the same expression, and the two square terms are visibly symmetric. Additivity and homogeneity in the first slot hold because every term is a product of exactly one coordinate of $x$ with one coordinate of $y$, and each such product is linear in $x$. For positive definiteness complete the square:

$$⟨x,x⟩=2x_1^2+2x_1{x}_2+3x_2^2=2{\left(x_1+\frac{1}{2}{x}_2\right)}^2+\frac{5}{2}{x}_2^2.$$

A sum of two nonnegative terms, so $⟨x,x⟩\ge 0$; and it is $0$ only if $x_2=0$ and $x_1+\frac{1}{2}{x}_2=0$, i.e. only if $x=0$. All four axioms hold.

(b) Not an inner product. Symmetry and bilinearity hold for the same reasons as in (a), so the only candidate for failure is positive definiteness, and it does fail: with $x=(1,-1)$,

$$⟨x,x⟩=1-2-2+1=-2<0.$$

A nonzero vector with negative "squared length" violates axiom 4. (Note $⟨(1,1),(1,1)⟩=6>0$, so the form is neither positive nor negative — it is indefinite, and no rescaling rescues it.)

(c) An inner product on ${\mathcal{P}}_1$. Evaluation at $0$ and differentiation are both linear, so $p\mapsto p(0)$ and $p\mapsto p^{\prime }(0)$ are linear maps, and $⟨p,q⟩$ is a sum of products of one linear functional of $p$ with one of $q$: bilinear and symmetric. For positivity write $p(t)=a+bt$; then $p(0)=a$ and $p^{\prime }(0)=b$, so

$$⟨p,p⟩=a^2+b^2\ge 0,$$

with equality only when $a=b=0$, i.e. $p=0$. Axiom 4 holds.

On ${\mathcal{P}}_2$ the same two numbers no longer determine the polynomial: $p(t)=t^2$ has $p(0)=0$ and $p^{\prime }(0)=0$, so $⟨p,p⟩=0$ although $p\ne 0$. Positive definiteness collapses and we have only a semi-inner product. The smallest repair is to add one more independent functional, for instance $⟨p,q⟩=p(0)q(0)+p^{\prime }(0)q^{\prime }(0)+p^{\prime \prime }(0)q^{\prime \prime }(0)$: for $p=a+bt+c{t}^2$ this gives $a^2+b^2+4c^2$, which vanishes only for $p=0$. The moral matches the sampling example: a formula is an inner product exactly when the quantities it reads off are enough to pin the vector down.

(a) Three integrals.

$$\Vert f{\Vert }^2={\int }_0^1{t}^{4}dt=\frac{1}{5},\Vert g{\Vert }^2={\int }_0^1{t}^{2}dt=\frac{1}{3},⟨f,g⟩={\int }_0^1{t}^{3}dt=\frac{1}{4}.$$

So $\Vert f\Vert =1/\sqrt{5}$ and $\Vert g\Vert =1/\sqrt{3}$. Cauchy–Schwarz demands $|⟨f,g⟩|\le \Vert f\Vert \Vert g\Vert$, i.e. $\frac{1}{4}\le \frac{1}{\sqrt{15}}$. Squaring, $\frac{1}{16}\le \frac{1}{15}$, which is true — and only just, which is reasonable because $t$ and $t^2$ are both increasing on $[0,1]$ and "agree" strongly.

(b) Distance two ways. Directly, $(f-g)(t)=t^2-t$, so

$$d(f,g{)}^2={\int }_0^1{\left(t^2-t\right)}^{2}dt={\int }_0^1\left(t^4-2t^3+t^2\right)dt=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}=\frac{6-15+10}{30}=\frac{1}{30}.$$

By the expansion, $\Vert f-g{\Vert }^2=\frac{1}{5}-2\cdot \frac{1}{4}+\frac{1}{3}=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}=\frac{1}{30}$ as well. Hence $d(f,g)=1/\sqrt{30}$. Sanity check: $|t^2-t|\le \frac{1}{4}$ on $[0,1]$, so the distance must be below $\frac{1}{4}$, and $1/\sqrt{30}\approx 0.18$ is.

(c) Normalising. From (b), $\Vert h\Vert =1/\sqrt{30}$, so

$$\frac{h}{\Vert h\Vert }=\sqrt{30}\left(t^2-t\right).$$

For $-4h$: the norm is homogeneous, $\Vert cv\Vert =|c|\Vert v\Vert$, so $\Vert -4h\Vert =4\Vert h\Vert$ and

$$\frac{-4h}{\Vert -4h\Vert }=\frac{-4h}{4\Vert h\Vert }=-\frac{h}{\Vert h\Vert }=-\sqrt{30}\left(t^2-t\right).$$

The absolute value in the homogeneity rule is what makes the sign survive: normalising a vector keeps its direction and forgets only its length, so scaling by a negative number flips the unit vector.

(a) Work in ${\mathbb{R}}^3$ with the dot product and take $u=(a,b,c)$, $v=(1,1,1)$. Then $⟨u,v⟩=a+b+c=1$, $\Vert u{\Vert }^2=a^2+b^2+c^2$ and $\Vert v\Vert =\sqrt{3}$. Cauchy–Schwarz gives

$$1=|⟨u,v⟩|\le \Vert u\Vert \Vert v\Vert =\sqrt{3}\sqrt{a^2+b^2+c^2},$$

and squaring, $a^2+b^2+c^2\ge \frac{1}{3}$. Equality in Cauchy–Schwarz holds exactly when $u$ and $v$ are linearly dependent, i.e. $(a,b,c)=\lambda (1,1,1)$; the constraint $a+b+c=1$ then forces $\lambda =\frac{1}{3}$. So equality holds precisely for $a=b=c=\frac{1}{3}$, and indeed $3\cdot \frac{1}{9}=\frac{1}{3}$.

(b) Upper bound: by the triangle inequality applied to $u$ and $-v$,

$$\Vert u-v\Vert \le \Vert u\Vert +\Vert -v\Vert =5+3=8.$$

Lower bound: by the reverse triangle inequality, $\Vert u-v\Vert \ge |\Vert u\Vert -\Vert v\Vert |=2$. Both are attained. Take $v=\frac{3}{5}u$: then $\Vert v\Vert =\frac{3}{5}\cdot 5=3$ and $\Vert u-v\Vert =\Vert \frac{2}{5}u\Vert =2$. Take $v=-\frac{3}{5}u$: again $\Vert v\Vert =3$, and $\Vert u-v\Vert =\Vert \frac{8}{5}u\Vert =8$. So the range of possible values is exactly $[2,8]$ (every value between is reached by rotating $v$ continuously from one position to the other, but the question only asks for the extremes).

(c) Work in $C[0,1]$ with $⟨g,h⟩={\int }_0^{1}gh$. Because $f>0$ and continuous, both $g=\sqrt{f}$ and $h=1/\sqrt{f}$ are continuous, so they are legitimate vectors. Their inner product is

$$⟨g,h⟩={\int }_0^1\sqrt{f(t)}\cdot \frac{1}{\sqrt{f(t)}}dt={\int }_0^{1}1dt=1,$$

while $\Vert g{\Vert }^2={\int }_0^{1}f$ and $\Vert h{\Vert }^2={\int }_0^{1}1/f$. Cauchy–Schwarz, $⟨g,h{⟩}^2\le \Vert g{\Vert }^2\Vert h{\Vert }^2$, reads

$$1\le \left({\int }_0^{1}f\right)\left({\int }_0^1\frac{1}{f}\right),$$

which is the claim. Equality holds iff $g$ and $h$ are linearly dependent, i.e. $\sqrt{f}=\lambda /\sqrt{f}$ for a constant $\lambda$, i.e. $f\equiv \lambda$ is constant. Sanity check with $f(t)=1+t$: the left factor is $\frac{3}{2}$ and the right is $\ln 2\approx 0.69$, product $\approx 1.04\ge 1$, and strictly greater because $f$ is not constant.

(a) The trace inner product is the entrywise dot product, so $⟨A,I⟩=1\cdot 1+1\cdot 0+1\cdot 0+1\cdot 1=2$. The norms are $\Vert A\Vert =\sqrt{1+1+1+1}=2$ and $\Vert I\Vert =\sqrt{1+0+0+1}=\sqrt{2}$. Therefore

$$\cos \theta =\frac{⟨A,I⟩}{\Vert A\Vert \Vert I\Vert }=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}},\theta =\frac{\pi }{4}.$$

(Geometrically: as vectors in ${\mathbb{R}}^4$, $A=(1,1,1,1)$ and $I=(1,0,0,1)$, and $I$ is "half of" $A$ in two of the four coordinates.)

(b) Orthogonality requires

$$⟨1+ct,t⟩={\int }_0^1\left(t+c{t}^2\right)dt=\frac{1}{2}+\frac{c}{3}=0,\text{so }c=-\frac{3}{2}.$$

With $u=1-\frac{3}{2}t$ and $v=t$:

$$\Vert u{\Vert }^2={\int }_0^1\left(1-3t+\frac{9}{4}{t}^2\right)dt=1-\frac{3}{2}+\frac{3}{4}=\frac{1}{4},\Vert v{\Vert }^2={\int }_0^1{t}^{2}dt=\frac{1}{3},$$

and $u+v=1-\frac{1}{2}t$, so

$$\Vert u+v{\Vert }^2={\int }_0^1\left(1-t+\frac{1}{4}{t}^2\right)dt=1-\frac{1}{2}+\frac{1}{12}=\frac{7}{12}.$$

Since $\frac{1}{4}+\frac{1}{3}=\frac{3}{12}+\frac{4}{12}=\frac{7}{12}$, the Pythagorean theorem $\Vert u+v{\Vert }^2=\Vert u{\Vert }^2+\Vert v{\Vert }^2$ is confirmed — as it must be, since the cross term $2⟨u,v⟩$ was arranged to vanish.

(c) Every norm induced by an inner product satisfies the parallelogram law $\Vert u+v{\Vert }^2+\Vert u-v{\Vert }^2=2\Vert u{\Vert }^2+2\Vert v{\Vert }^2$, so a norm that violates it cannot be induced by one. Test $u=(1,0)$, $v=(0,1)$. Then $\Vert u{\Vert }_4=\Vert v{\Vert }_4=1$, while $u+v=(1,1)$ and $u-v=(1,-1)$ both have $\Vert \cdot {\Vert }_4=(1+1{)}^{1/4}=2^{1/4}$, whose square is $\sqrt{2}$. So

$$\Vert u+v{\Vert }_4^2+\Vert u-v{\Vert }_4^2=2\sqrt{2}\approx 2.83,2\Vert u{\Vert }_4^2+2\Vert v{\Vert }_4^2=4.$$

The two sides differ, so the parallelogram law fails and $\Vert \cdot {\Vert }_4$ is not induced by any inner product. Only the $2$-norm among the $p$-norms passes this test, which is why Euclidean geometry is special.

(a) The unit "circle" is an ellipse. By definition $\Vert x{\Vert }_w^2=⟨x,x{⟩}_w=x_1^2+4x_2^2$, so the unit circle is the curve

$$x_1^2+4x_2^2=1,$$

an ellipse centred at the origin crossing the horizontal axis at $(\pm 1,0)$ and the vertical axis at $(0,\pm \frac{1}{2})$. It lies inside the Euclidean unit circle except at $(\pm 1,0)$, where the two curves touch: a vertical displacement is charged four times as heavily as a horizontal one, so vertical unit vectors are shorter on the page. Draw the ellipse solid and the Euclidean circle dashed.

(b) A pair that is orthogonal but does not look it.

$$⟨u,v{⟩}_w=2\cdot 2+4\cdot (1)(-1)=4-4=0,$$

so $u{\perp }_{w}v$. Their Euclidean dot product is $2\cdot 2+1\cdot (-1)=3\ne 0$; with $|u|=|v|=\sqrt{5}$ the Euclidean angle satisfies $\cos \theta =\frac{3}{5}$, about $53^{\circ }$. Draw both arrows from the origin — $u$ rising to the right, $v$ falling to the right, symmetric about the horizontal axis — and label the angle between them $\arccos \frac{3}{5}$. Their $w$-lengths are $\Vert u{\Vert }_w=\Vert v{\Vert }_w=\sqrt{4+4}=2\sqrt{2}$.

(c) Two different perpendicular lines. The vector $x$ is $w$-orthogonal to $(1,1)$ iff $x_1+4x_2=0$, i.e. iff $x_2=-\frac{1}{4}{x}_1$: a line through the origin of slope $-\frac{1}{4}$. Euclidean orthogonality to $(1,1)$ is $x_1+x_2=0$, the line of slope $-1$. Draw the arrow $(1,1)$ and both lines, labelled by their equations. Notice the $w$-perpendicular is much flatter: to cancel the heavily-weighted second coordinate you need only a quarter as much of it.

(d) The stretch that explains everything. Take $\Phi (x_1,x_2)=(x_1,2x_2)$. Then

$$\Phi (x)\cdot \Phi (y)=x_1{y}_1+(2x_2)(2y_2)=x_1{y}_1+4x_2{y}_2=⟨x,y{⟩}_w.$$

So $\Phi$ carries the weighted inner product to the dot product: every $w$-length, $w$-angle and $w$-orthogonality is the Euclidean one after stretching the plane vertically by a factor $2$. Equivalently, the sketch is the ordinary Euclidean picture squashed vertically by $\frac{1}{2}$. Check each feature: $\Phi$ sends the ellipse $x_1^2+4x_2^2=1$ onto the unit circle; it sends $u=(2,1)$ and $v=(2,-1)$ to $(2,2)$ and $(2,-2)$, which are visibly perpendicular; and it sends $(1,1)$ to $(1,2)$, whose Euclidean perpendicular direction is $(2,-1)$, which pulls back under ${\Phi }^{-1}$ to $(2,-\frac{1}{2})$ — slope $-\frac{1}{4}$, exactly the line drawn in (c). Every weighted inner product with positive weights is a Euclidean inner product viewed through such a stretch; that is why none of the theorems of the chapter can tell the two apart.

(a) Gram–Schmidt. Start with $w_1=1$, and $\Vert w_1{\Vert }^2={\int }_0^{1}1dt=1$.

For $w_2$: $⟨t,w_1⟩={\int }_0^{1}tdt=\frac{1}{2}$, so

$$w_2=t-\frac{1/2}{1}\cdot 1=t-\frac{1}{2},\Vert w_2{\Vert }^2={\int }_0^1{\left(t-\frac{1}{2}\right)}^{2}dt={\left[\frac{(t-\frac{1}{2}{)}^3}{3}\right]}_0^1=\frac{1}{24}+\frac{1}{24}=\frac{1}{12}.$$

For $w_3$ we need two inner products: $⟨t^2,w_1⟩={\int }_0^1{t}^{2}dt=\frac{1}{3}$ and

$$⟨t^2,w_2⟩={\int }_0^1\left(t^3-\frac{1}{2}{t}^2\right)dt=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}.$$

Hence

$$w_3=t^2-\frac{1/3}{1}\cdot 1-\frac{1/12}{1/12}\left(t-\frac{1}{2}\right)=t^2-\frac{1}{3}-t+\frac{1}{2}=t^2-t+\frac{1}{6}.$$

Check against both earlier vectors: $⟨w_3,1⟩=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0$ and $⟨w_3,t⟩=\frac{1}{4}-\frac{1}{3}+\frac{1}{12}=\frac{3-4+1}{12}=0$; orthogonality to $t$ and to $1$ gives orthogonality to $w_2=t-\frac{1}{2}$. Good.

For the norm, the cheapest route uses orthogonality: since $w_3=t^2-(\text{something in }{\mathcal{P}}_1)$ and $w_3\perp {\mathcal{P}}_1$,

$$\Vert w_3{\Vert }^2=⟨w_3,w_3⟩=⟨w_3,t^2⟩={\int }_0^1\left(t^4-t^3+\frac{1}{6}{t}^2\right)dt=\frac{1}{5}-\frac{1}{4}+\frac{1}{18}=\frac{36-45+10}{180}=\frac{1}{180}.$$

(Expanding $(t^2-t+\frac{1}{6}{)}^2$ and integrating term by term gives $\frac{1}{5}-\frac{1}{2}+\frac{4}{9}-\frac{1}{6}+\frac{1}{36}=\frac{1}{180}$ as well.)

(b) Normalising. $e_1=w_1/1=1$. Next $\Vert w_2\Vert =1/\sqrt{12}=1/(2\sqrt{3})$, so $e_2=2\sqrt{3}\left(t-\frac{1}{2}\right)=\sqrt{3}(2t-1)$. Finally $\Vert w_3\Vert =1/\sqrt{180}=1/(6\sqrt{5})$, so

$$e_3=6\sqrt{5}\left(t^2-t+\frac{1}{6}\right)=\sqrt{5}\left(6t^2-6t+1\right).$$

Check: ${\int }_0^{1}5(6t^2-6t+1{)}^{2}dt=5\cdot 36\cdot \frac{1}{180}=1$. Three orthonormal vectors in the $3$-dimensional space ${\mathcal{P}}_2$ form an orthonormal basis.

(c) Projection and distance for free. Gram–Schmidt subtracts from $t^2$ precisely its projection onto $\mathrm{span}\{w_1,w_2\}$, and that span equals $\mathrm{span}\{1,t\}={\mathcal{P}}_1$ by the span property of the algorithm. So the subtracted part is the projection:

$${\mathrm{proj}}_{{\mathcal{P}}_1}\left(t^2\right)=t^2-w_3=t-\frac{1}{6},$$

and the leftover $w_3$ is the residual, which is orthogonal to ${\mathcal{P}}_1$. By the best-approximation theorem the distance from $t^2$ to ${\mathcal{P}}_1$ is the length of that residual:

$$d\left(t^2,{\mathcal{P}}_1\right)=\Vert w_3\Vert =\frac{1}{6\sqrt{5}}=\frac{\sqrt{5}}{30}\approx \mathrm{0.075.}$$

Sanity check: $t-\frac{1}{6}$ is the line that best fits the parabola $t^2$ on $[0,1]$ in mean square. It passes through $(\frac{1}{6},0)$ and $(1,\frac{5}{6})$, undershooting at both ends and overshooting in the middle, which is what a balanced fit to a convex curve should do.

(a) The step for $t^3$. Gram–Schmidt says

$$w_3=t^3-\frac{⟨t^3,w_0⟩}{\Vert w_0{\Vert }^2}{w}_0-\frac{⟨t^3,w_1⟩}{\Vert w_1{\Vert }^2}{w}_1-\frac{⟨t^3,w_2⟩}{\Vert w_2{\Vert }^2}{w}_2.$$

On the symmetric interval the integral of an odd function is zero. The products $t^3\cdot 1=t^3$ and $t^3\left(t^2-\frac{1}{3}\right)=t^5-\frac{1}{3}{t}^3$ are odd, so $⟨t^3,w_0⟩=⟨t^3,w_2⟩=0$ without computation. Only the middle term survives:

$$⟨t^3,t⟩={\int }_{-1}^1{t}^{4}dt=\frac{2}{5},w_3=t^3-\frac{2/5}{2/3}t=t^3-\frac{3}{5}t.$$

Check: $⟨w_3,t⟩=\frac{2}{5}-\frac{3}{5}\cdot \frac{2}{3}=\frac{2}{5}-\frac{2}{5}=0$. Since $w_3(1)=1-\frac{3}{5}=\frac{2}{5}$, rescale by $\frac{5}{2}$:

$$P_3(t)=\frac{5}{2}\left(t^3-\frac{3}{5}t\right)=\frac{5t^3-3t}{2}.$$

(b) Roots. Factor: $P_3(t)=\frac{1}{2}t\left(5t^2-3\right)$, so the roots are

$$t=0,t=\pm \sqrt{\frac{3}{5}}=\pm \frac{\sqrt{15}}{5}\approx \pm \mathrm{0.775.}$$

Three distinct real roots, all strictly inside $(-1,1)$ — exactly what the theorem promises for a degree-$3$ polynomial orthogonal to all polynomials of lower degree.

Why the hypothesis holds: by construction $w_3$ is orthogonal to each of $w_0,w_1,w_2$, and by the span property of Gram–Schmidt, $\mathrm{span}\{w_0,w_1,w_2\}=\mathrm{span}\{1,t,t^2\}={\mathcal{P}}_2$. A vector orthogonal to a spanning set of a subspace is orthogonal to the whole subspace (only the span matters), so $w_3$, and hence its multiple $P_3$, is orthogonal to every polynomial of degree at most $2$. That is the full hypothesis of the theorem.

(c) Testing the two-point rule. For $f(t)=t^2$ the exact integral is ${\int }_{-1}^1{t}^{2}dt=\frac{2}{3}$, and the rule gives ${\left(-\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$. Exact, as promised for degree $2\le 3$. (Degree $3$ is exact too, trivially: any odd $f$ integrates to $0$ and the rule returns $f(-a)+f(a)=0$.)

For $f(t)=t^4$ the exact integral is $\frac{2}{5}$, but the rule gives $\frac{1}{9}+\frac{1}{9}=\frac{2}{9}$. Since $\frac{2}{9}\ne \frac{2}{5}$, exactness first fails at degree $4$, one beyond the guarantee. The pattern — $n$ nodes at the roots of $P_n$ give exactness up to degree $2n-1$ — is what makes the roots of orthogonal polynomials the sample points of choice in numerical integration, and part (b) is the reason those roots are real, distinct and inside the interval, as sample points must be.

(a) Two inclusions.

$\subseteq$. Let $v\in (U+W{)}^{\perp }$, so $⟨v,x⟩=0$ for every $x\in U+W$. Every $u\in U$ is in $U+W$ (write $u=u+0$, using $0\in W$), so $⟨v,u⟩=0$ for all $u\in U$: $v\in U^{\perp }$. Symmetrically $v\in W^{\perp }$. Hence $v\in U^{\perp }\cap W^{\perp }$.

$\supseteq$. Let $v\in U^{\perp }\cap W^{\perp }$ and let $x\in U+W$, say $x=u+w$. By additivity in the second slot,

$$⟨v,u+w⟩=⟨v,u⟩+⟨v,w⟩=0+0=0,$$

so $v\in (U+W{)}^{\perp }$.

This part uses only that $U$ and $W$ are subspaces (so that $0\in W$ and $U\subseteq U+W$) and the bilinearity of the inner product. Finite-dimensionality is not used anywhere, and the argument is valid in any inner product space. $■$

(b) Apply (a) to the subspaces $U^{\perp }$ and $W^{\perp }$ (each is a subspace, since every orthogonal complement is):

$${\left(U^{\perp }+W^{\perp }\right)}^{\perp }={\left(U^{\perp }\right)}^{\perp }\cap {\left(W^{\perp }\right)}^{\perp }=U\cap W,$$

where the last equality is the double-complement corollary $(X^{\perp }{)}^{\perp }=X$, valid because $U$ and $W$ are finite-dimensional. Now take orthogonal complements of both sides:

$${\left({\left(U^{\perp }+W^{\perp }\right)}^{\perp }\right)}^{\perp }=(U\cap W{)}^{\perp }.$$

The left side is $U^{\perp }+W^{\perp }$ by the double-complement corollary once more, applied to the finite-dimensional subspace $U^{\perp }+W^{\perp }$. Hence $(U\cap W{)}^{\perp }=U^{\perp }+W^{\perp }$. $■$

Finite-dimensionality enters exactly twice, both times through the double complement, which in turn rests on the orthogonal decomposition theorem $V=X\oplus X^{\perp }$. Part (a) needed neither.

(c) A vector $x$ lies in $(U+W{)}^{\perp }$ iff it is orthogonal to a spanning set of $U+W$, i.e. to both $(1,1,1)$ and $(1,0,-1)$:

$$x_1+x_2+x_3=0,x_1-x_3=0.$$

From the second, $x_3=x_1$; substituting, $x_2=-2x_1$. So $(U+W{)}^{\perp }=\mathrm{span}\{(1,-2,1)\}$.

Separately, $U^{\perp }$ is the plane $x_1+x_2+x_3=0$ and $W^{\perp }$ is the plane $x_1-x_3=0$; their intersection is the solution set of the same two equations, again $\mathrm{span}\{(1,-2,1)\}$. The two computations agree, as (a) says they must. (Sanity check: $(1,-2,1)\cdot (1,1,1)=0$ and $(1,-2,1)\cdot (1,0,-1)=0$.)

Dimensions: $(1,1,1)$ and $(1,0,-1)$ are independent, so $\dim (U+W)=2$, and the decomposition theorem predicts $\dim (U+W{)}^{\perp }=3-2=1$. Correct. Note also $\dim U^{\perp }+\dim W^{\perp }=2+2=4>3$, which is fine: two planes through the origin in ${\mathbb{R}}^3$ always share a line, and that line is the answer.

(a) The basis. Pairwise inner products: $⟨1,t⟩={\int }_{-1}^{1}tdt=0$ and $⟨t,t^2-\frac{1}{3}⟩={\int }_{-1}^1\left(t^3-\frac{1}{3}t\right)dt=0$, both because the integrand is odd; and

$$⟨1,t^2-\frac{1}{3}⟩={\int }_{-1}^1{t}^{2}dt-\frac{1}{3}{\int }_{-1}^{1}1dt=\frac{2}{3}-\frac{2}{3}=0.$$

Three pairwise orthogonal nonzero vectors are independent, and ${\mathcal{P}}_2$ has dimension $3$, so they form a basis. Squared norms:

$$\Vert 1{\Vert }^2=2,\Vert t{\Vert }^2=\frac{2}{3},{\Vert t^2-\frac{1}{3}\Vert }^2={\int }_{-1}^1\left(t^4-\frac{2}{3}{t}^2+\frac{1}{9}\right)dt=\frac{2}{5}-\frac{4}{9}+\frac{2}{9}=\frac{8}{45}.$$

The formula ${\mathrm{proj}}_{W}v={\sum }_i\frac{⟨v,w_i⟩}{\Vert w_i{\Vert }^2}{w}_i$ is derived from the orthogonal decomposition and is only valid when the $w_i$ are pairwise orthogonal. The basis $\{1,t,t^2\}$ is not: $⟨1,t^2⟩=\frac{2}{3}\ne 0$. Applying the formula to it would produce some element of ${\mathcal{P}}_2$, but not the nearest one, and the residual would fail to be orthogonal to ${\mathcal{P}}_2$.

(b) The projection. Three inner products with $f(t)=|t|$, which is even:

$$⟨f,1⟩=2{\int }_0^{1}tdt=1,⟨f,t⟩=0(\text{odd integrand}),⟨f,t^2-\frac{1}{3}⟩=2{\int }_0^1\left(t^3-\frac{1}{3}t\right)dt=2\left(\frac{1}{4}-\frac{1}{6}\right)=\frac{1}{6}.$$

Therefore

$$\hat{f}=\frac{1}{2}\cdot 1+0\cdot t+\frac{1/6}{8/45}\left(t^2-\frac{1}{3}\right)=\frac{1}{2}+\frac{15}{16}\left(t^2-\frac{1}{3}\right)=\frac{15}{16}{t}^2+\frac{3}{16},$$

since $\frac{1}{6}\cdot \frac{45}{8}=\frac{45}{48}=\frac{15}{16}$ and $\frac{1}{2}-\frac{15}{48}=\frac{24-15}{48}=\frac{3}{16}$. Sanity check: $\hat{f}(0)=\frac{3}{16}$ sits a little above the corner at $0$, and $\hat{f}(\pm 1)=\frac{18}{16}=\frac{9}{8}$ a little above the endpoints — a parabola cannot make a corner, so it overshoots at the ends to sink toward it in the middle.

(c) The error. Since $\hat{f}\in {\mathcal{P}}_2$ and $f-\hat{f}\perp {\mathcal{P}}_2$, the Pythagorean theorem gives $\Vert f{\Vert }^2=\Vert \hat{f}{\Vert }^2+\Vert f-\hat{f}{\Vert }^2$. Here $\Vert f{\Vert }^2={\int }_{-1}^1{t}^{2}dt=\frac{2}{3}$, and by the coordinate formula for an orthogonal basis

$$\Vert \hat{f}{\Vert }^2=\sum _i\frac{⟨f,w_i{⟩}^2}{\Vert w_i{\Vert }^2}=\frac{1^2}{2}+0+\frac{(1/6{)}^2}{8/45}=\frac{1}{2}+\frac{1}{36}\cdot \frac{45}{8}=\frac{1}{2}+\frac{5}{32}=\frac{21}{32}.$$

Hence

$$\Vert f-\hat{f}{\Vert }^2=\frac{2}{3}-\frac{21}{32}=\frac{64-63}{96}=\frac{1}{96},\Vert f-\hat{f}\Vert =\frac{1}{\sqrt{96}}=\frac{1}{4\sqrt{6}}=\frac{\sqrt{6}}{24}\approx \mathrm{0.10.}$$

Residual check: $⟨f-\hat{f},1⟩=⟨f,1⟩-\frac{15}{16}⟨t^2,1⟩-\frac{3}{16}⟨1,1⟩=1-\frac{15}{16}\cdot \frac{2}{3}-\frac{3}{16}\cdot 2=1-\frac{5}{8}-\frac{3}{8}=0$. Good.

The best constant is the projection onto $\mathrm{span}\{1\}$, namely $\frac{⟨f,1⟩}{\Vert 1{\Vert }^2}=\frac{1}{2}$ (the mean of $|t|$), with squared error $\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\approx 0.167$. Admitting $t$ contributes nothing, because $|t|$ is even; admitting $t^2$ cuts the squared error from $\frac{1}{6}$ to $\frac{1}{96}$, a factor of $16$. That drop is the coefficient $\frac{5}{32}$ in the Bessel sum above — the energy of $|t|$ that lives in the $P_2$ direction.

(a) Step 1 is the error. Bessel's inequality ${\sum }_i⟨v,e_i{⟩}^2\le \Vert v{\Vert }^2$ requires the set $\{e_i\}$ to be orthonormal, not merely orthogonal. Here $\Vert \sin t{\Vert }^2=\Vert \sin 2t{\Vert }^2=\pi$, so neither vector is a unit vector, and the coefficients $⟨f,\sin mt⟩$ are not the coordinates the theorem is about. Steps 2 and 3 are arithmetically fine; they compute the wrong quantity.

For an orthogonal set $\{w_i\}$ of nonzero vectors, normalise first: $e_i=w_i/\Vert w_i\Vert$ gives $⟨v,e_i⟩=⟨v,w_i⟩/\Vert w_i\Vert$, and Bessel's inequality becomes

$$\sum _i\frac{⟨v,w_i{⟩}^2}{\Vert w_i{\Vert }^2}\le \Vert v{\Vert }^2.$$

The missing denominators are exactly what inflated the student's sum by a factor of $\pi$.

(b) Checking the integrals. Both integrands $t\sin mt$ are even, so integrate over $[0,\pi ]$ and double. With $u=t$, $dv=\sin mtdt$:

$${\int }_0^{\pi }t\sin mtdt={\left[-\frac{t\cos mt}{m}\right]}_0^{\pi }+\frac{1}{m}{\int }_0^{\pi }\cos mtdt=-\frac{\pi \cos m\pi }{m}+0=\frac{\pi (-1{)}^{m+1}}{m}.$$