Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 2

Let $p={\sum }_{j=1}^m⟨u,v_j⟩v_j$. Orthonormality gives $⟨u-p,v_k⟩=0$ for every $k$, so $u-p$ is orthogonal to $p$. The Pythagorean theorem for inner product spaces givesSince a squared norm is nonnegative,This is Bessel's inequality, with equality exactly when $u$ belongs to the span of the orthonormal set.

Use the Euclidean inner product on ${\mathbb{R}}^n$, with $a=(a_1,\dots ,a_n)$ and $b=(b_1,\dots ,b_n)$. If $b=0$, both sides of the desired inequality are zero. Otherwise take the single unit vector $v=b/\Vert b\Vert$ in part (a). ThensoThis is the real Cauchy-Schwarz inequality. Equality holds when the two vectors are linearly dependent.

In part (i), set every $b_i=1$. Its squared norm is $n$, givingThe equality condition is $a_1=\cdots =a_n$, because $a$ must be a scalar multiple of $(1,\dots ,1)$.

In ${\mathbb{R}}^{2n}$, take the two unit vectorsTheir supports are disjoint, so they are orthonormal. Apply part (a) to $a=(a_1,\dots ,a_{2n})$:Multiplying by $n$ gives

Equality in part (a) requires $a$ to lie in the span of the two chosen unit vectors. Equivalently, there are real constants $r,s$ such thatOne can also see the necessity directly by writing the nonnegative deficit aswhere ${\bar{a}}_e$ and ${\bar{a}}_o$ are the means of the even and odd entries. The deficit vanishes exactly when each group is constant; the two group constants need not be equal.