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

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 /∥ b ∥

in part (a). Then

∥ a ∥^{2} \geq \frac{⟨ a , b ⟩ ^{2}}{∥ b ∥ ^{2}},

(i = 1 \sum n a_{i}^{2}) (i = 1 \sum n b_{i}^{2}) \geq (i = 1 \sum n a_{i} b_{i})^{2} .

This is the real Cauchy-Schwarz inequality. Equality holds when the two vectors are linearly dependent.

ii

In part (i), set every

b_{i} = 1

. Its squared norm is

n

n i = 1 \sum n a_{i}^{2} \geq (i = 1 \sum n a_{i})^{2} .

The equality condition is

a_{1} = \dots = a_{n}

, because

a

(1, \dots, 1)

R^{2 n}

, take the two unit vectors

v_{even} = \frac{1}{n} (0, 1, 0, 1, \dots, 0, 1), v_{odd} = \frac{1}{n} (1, 0, 1, 0, \dots, 1, 0) .

Their supports are disjoint, so they are orthonormal. Apply part (a) to

a = (a_{1}, \dots, a_{2 n})

\sum_{i = 1}^{2 n} a_{i}^{2} \geq \frac{1}{n} (\sum_{i = 1}^{n} a_{2 i})^{2} + \frac{1}{n} (\sum_{i = 1}^{n} a_{2 i - 1})^{2} .

Multiplying by

n

gives

n i = 1 \sum 2 n a_{i}^{2} \geq (i = 1 \sum n a_{2 i})^{2} + (i = 1 \sum n a_{2 i - 1})^{2} .

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