Solution.

The law

Expanding both terms with \(||v||^2=(v,v)\) and using linearity, \[||x+y||^2=||x||^2+2\operatorname {Re}(x,y)+||y||^2,\] \[||x-y||^2=||x||^2-2\operatorname {Re}(x,y)+||y||^2 .\] Adding, the cross terms cancel and the rest doubles.

The supremum norm fails it

Take \(f(t)=1\) and \(g(t)=t\) on \([0,1]\). Then \(||f||_\infty =||g||_\infty =1\), and \[f+g=1+t,\quad ||f+g||_\infty =2;\qquad f-g=1-t,\quad ||f-g||_\infty =1 .\] So the left side is \(4+1=5\), while the right side is \(2(1+1)=4\). The law fails, so \(||\cdot ||_\infty \) is not induced by any inner product.

The parallelogram law is exactly the dividing line: a norm satisfies it if and only if it comes from an inner product, the inner product then being recovered by the polarisation identity.

Solution. Fix \(n\) and let \(s_n=\sum _{k=1}^{n}(x,e_k)e_k\) be the partial projection. The key computation is that \(x-s_n\) is orthogonal to each \(e_j\) with \(j\leq n\): \[(x-s_n,e_j)=(x,e_j)-\sum _{k=1}^{n}(x,e_k)(e_k,e_j)=(x,e_j)-(x,e_j)=0 ,\] using orthonormality to kill every term but \(k=j\). Hence \(x-s_n\perp s_n\), and by Pythagoras \[||x||^2=||x-s_n||^2+||s_n||^2\geq ||s_n||^2 .\] Also \(||s_n||^2=\sum _{k=1}^{n}|(x,e_k)|^2\), again by orthonormality. So \[\sum _{k=1}^{n}|(x,e_k)|^2\leq ||x||^2\] for every \(n\). The left side is a non-decreasing sequence bounded above, so it converges, and its limit satisfies the same bound.

The inequality says the coefficients cannot carry more than the total length of \(x\). Equality for all \(x\) — Parseval’s relation — is exactly the statement that the orthonormal set is complete.

Solution.

Existence

The set \(x-M=\{x-u:u\in M\}\) is closed and convex, and \(H\) is complete, so by the minimum-norm theorem proved above it contains a unique element \(n\) of smallest norm. Write \(n=x-m\) with \(m\in M\).

\(n\) is orthogonal to \(M\)

Let \(u\in M\) be a unit vector and \(\lambda \) a scalar. Since \(m+\lambda u\in M\), minimality gives \[||n||^2\leq ||x-(m+\lambda u)||^2=||n-\lambda u||^2 =||n||^2-2\operatorname {Re}\big (\overline {\lambda }(n,u)\big )+|\lambda |^2 .\] Choosing \(\lambda =(n,u)\) reduces this to \(0\leq -|(n,u)|^2\), forcing \((n,u)=0\). As \(u\) was arbitrary, \(n\perp M\).

Uniqueness

If \(x=m'+n'\) is another such splitting then \(m-m'=n'-n\) lies in both \(M\) and \(M^{\perp }\), hence is orthogonal to itself and therefore zero.

\(m\) is nearest

For any \(u\in M\), \(x-u=(x-m)+(m-u)\) with \(x-m\perp m-u\), so by Pythagoras \[||x-u||^2=||x-m||^2+||m-u||^2\geq ||x-m||^2 ,\] with equality only when \(u=m\).

This is the projection theorem, and the last line is why it underlies least squares: the best approximation from a subspace is the orthogonal projection onto it.

Solution.

First vector

\(e_1=1\), with \(||e_1||^2=\int _{-1}^{1}1\,dt=2\).

Second

\((t,1)=\int _{-1}^{1}t\,dt=0\) by oddness, so no correction is needed and \[e_2=t,\qquad ||e_2||^2=\int _{-1}^{1}t^2dt=\frac {2}{3}.\]

Third

\((t^2,1)=\int _{-1}^{1}t^2dt=\frac 23\) and \((t^2,t)=\int _{-1}^{1}t^3dt=0\), again by oddness. So \[e_3=t^2-\frac {2/3}{2}\cdot 1-0\cdot t=t^2-\frac 13 .\]

Normalising, \[\frac {1}{\sqrt 2},\qquad \sqrt {\frac 32}\,t,\qquad \sqrt {\frac {45}{8}}\left (t^2-\frac 13\right ),\] the last from \(||e_3||^2=\int _{-1}^{1}\left (t^2-\frac 13\right )^2dt=\frac {8}{45}\).

These are the Legendre polynomials up to scaling, and the oddness shortcuts are worth noticing: on a symmetric interval, an odd function is automatically orthogonal to an even one.

Solution.

Statement

For every bounded linear functional \(u\in H^*\) there is a unique \(y\in H\) with \[u(x)=(x,y)\qquad \text {for all }x\in H ,\] and moreover \(||u||=||y||\).

Uniqueness

Suppose \((x,y_1)=(x,y_2)\) for all \(x\). Then \((x,y_1-y_2)=0\) for all \(x\), and choosing \(x=y_1-y_2\) gives \(||y_1-y_2||^2=0\), so \(y_1=y_2\).

Why \(||u||=||y||\)

By Cauchy–Schwarz \(|u(x)|=|(x,y)|\leq ||x||\,||y||\), so \(||u||\leq ||y||\). Taking \(x=y\) gives \(u(y)=||y||^2\), so \(||u||\geq ||y||^2/||y||=||y||\) when \(y\neq 0\). Hence equality.

The theorem says a Hilbert space is its own dual in a concrete way, which is what makes the sesquilinear form corollary above — and with it much of operator theory — possible.