The Euclidean Metric on \R^n

Proving that the triangle inequality is satisfied is often the most difficult part of proving that a function is a metric. We will work through this proof with the help of the Cauchy-Schwarz Inequality.

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Lemma 3.9. Cauchy-Schwarz Inequality.

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Let

n

be a positive integer and

x = (x_{1}, x_{2}, \dots, x_{n}),

y = (y_{1}, y_{2}, \dots, y_{n})

be in

R^{n} .

Then

\begin{matrix} (3.1) & \sum_{i = 1}^{n} x_{i} y_{i} \leq (\sqrt{\sum_{i = 1}^{n} x_{i}^{2}}) (\sqrt{\sum_{i = 1}^{n} y_{i}^{2}}) . \end{matrix}

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Activity 3.4.

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Before we prove the Cauchy-Schwarz Inequality, let us analyze it in two specific situations.

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(a)

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Let

x = (1, 4)

and

y = (3, 2)

R^{2} .

Verify the Cauchy-Schwarz Inequality in this case.

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(b)

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Let

x = (1, 2, - 3)

and

y = (- 4, 0, - 1)

R^{3} .

Verify the Cauchy-Schwarz Inequality in this case.

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Now we prove the Cauchy-Schwarz Inequality.

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Proof.

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Let

n

be a positive integer and

x = (x_{1}, x_{2}, \dots, x_{n}),

y = (y_{1}, y_{2}, \dots, y_{n})

be in

R^{n} .

To verify (3.1) it suffices to show that

{(\sum_{i = 1}^{n} x_{i} y_{i})}^{2} \leq (\sum_{i = 1}^{n} x_{i}^{2}) (\sum_{i = 1}^{n} y_{i}^{2}) .

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This is difficult to do directly, but there is a nice trick one can use. Consider the expression

\sum (x_{i} - λ y_{i})^{2} .

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(All of our sums are understood to be from 1 to

n,

so we will omit the limits on the sums for the remainder of the proof.) Now

\begin{aligned} 0 & \leq \sum (x_{i} - λ y_{i})^{2} \\ = \sum (x_{i}^{2} - 2 λ x_{i} y_{i} + λ^{2} y_{i}^{2}) \\ (3.2) & = (\sum y_{i}^{2}) λ^{2} - 2 (\sum x_{i} y_{i}) λ + (\sum x_{i}^{2}) . \end{aligned}

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To interpret this last expression more clearly, let

a = \sum y_{i}^{2},

b = - 2 \sum x_{i} y_{i}

and

c = \sum x_{i}^{2} .

The inequality defined by (3.2) can then be written in the form

p (λ) = a λ^{2} + b λ + c \geq 0 .

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So we have a quadratic

p (λ)

that is never negative. This implies that the quadratic

p (λ)

can have at most one real zero. The quadratic formula gives the roots of

p (λ)

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .

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b^{2} - 4 a c > 0,

then

p (λ)

has two real roots. Therefore, in order for

p (λ)

to have at most one real zero we must have

0 \geq b^{2} - 4 a c = 4 {(\sum x_{i} y_{i})}^{2} - 4 (\sum y_{i}^{2}) (\sum x_{i}^{2})

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(\sum y_{i}^{2}) (\sum x_{i}^{2}) \geq {(\sum x_{i} y_{i})}^{2} .

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This establishes the Cauchy-Schwarz Inequality.

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One consequence of the Cauchy-Schwarz Inequality that we will need to show that

d_{E}

is a metric is the following.

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Corollary 3.10.

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Let

n

be a positive integer and

x = (x_{1}, x_{2}, \dots, x_{n}),

y = (y_{1}, y_{2}, \dots, y_{n})

be in

R^{n} .

Then

\sqrt{\sum_{i = 1}^{n} (x_{i} + y_{i})^{2}} \leq \sqrt{\sum_{i = 1}^{n} x_{i}^{2}} + \sqrt{\sum_{i = 1}^{n} y_{i}^{2}} .

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Activity 3.5.

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Before we prove the corollary, let us analyze it in two specific situations.

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(a)

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Let

x = (1, 4)

and

y = (3, 2)

R^{2} .

Verify Corollary 3.10 in this case.

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(b)

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Let

x = (1, 2, - 3)

and

y = (- 4, 0, - 1)

R^{3} .

Verify Corollary 3.10 in this case.

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Now we prove Corollary 3.10.

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Proof.

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Let

n

be a positive integer and

x = (x_{1}, x_{2}, \dots, x_{n}),

y = (y_{1}, y_{2}, \dots, y_{n})

be in

R^{n} .

Now

\begin{aligned} \sum {(x_{i} + y_{i})}^{2} & = \sum (x_{i}^{2} + 2 x_{i} y_{i} + y_{i}^{2}) \\ = \sum x_{i}^{2} + 2 \sum x_{i} y_{i} + \sum y_{i}^{2} \\ \leq \sum x_{i}^{2} + 2 (\sqrt{\sum x_{i}^{2}}) (\sqrt{\sum y_{i}^{2}}) + \sum y_{i}^{2} \\ = {(\sqrt{\sum x_{i}^{2}} + \sqrt{\sum y_{i}^{2}})}^{2} . \end{aligned}

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Taking the square roots of both sides yields the desired inequality.

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We can now complete the proof that

d_{E}

is a metric.

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Activity 3.6.

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Let

n

be a positive integer and

x = (x_{1}, x_{2}, \dots, x_{n}),

y = (y_{1}, y_{2}, \dots, y_{n}),

and

z = (z_{1}, z_{2}, \dots, z_{n})

be in

R^{n} .

Use Corollary 3.10 to show that

d_{E} (x, y) \leq d_{E} (x, z) + d_{E} (z, y) .

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This concludes our proof that the Euclidean metric is in fact a metric.

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We have seen several metrics in this section, some of which are given special names. Let

x = (x_{1}, x_{2}, \dots, x_{n})

and

y = (y_{1}, y_{2}, \dots, y_{n})

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The Euclidean metric $d_{E},$ where

$\begin{aligned} d_{E} (x, y) & = \sqrt{(x_{1} - y_{1})^{2} + (x_{2} - y_{2})^{2} + \dots (x_{n} - y_{n})^{2}} \\ = \sqrt{\sum_{i = 1}^{n} (x_{i} - y_{i})^{2}} . \end{aligned}$
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The Taxicab metric $d_{T},$ where

$d_{T} (x, y) = | x_{1} - y_{1} | + | x_{2} - y_{2} | + \dots + | x_{n} - y_{n} | = \sum_{i = 1}^{n} {| x_{i} - y_{i} |} .$
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The max metric $d_{M},$ where

$\begin{aligned} d_{M} (x, y) & = max {| x_{1} - y_{1} |, | x_{2} - y_{2} |, | x_{3} - y_{3} |, \dots, | x_{n} - y_{n} |} \\ = max_{1 \leq i \leq n} {| x_{i} - y_{i} |} . \end{aligned}$

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We have only shown that

d_{T}

and

d_{M}

are metrics on

R^{2},

but similar arguments apply in

R^{n} .

Proofs are left to Exercise 5 and Exercise 6. In addition, the discrete metric

d (x, y) = {\begin{cases} 0 & if x = y \\ 1 & if x \neq y \end{cases}

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makes any set

X

into a metric space. The proof is left to Exercise 1.

Topology: An Inquiry-Based Approach

Search Results:

Section The Euclidean Metric on $R^{n}$

Activity 3.3.

(a)

(b)

(c)

(d)

Lemma 3.9. Cauchy-Schwarz Inequality.

Activity 3.4.

(a)

(b)

Proof.

Corollary 3.10.

Activity 3.5.

(a)

(b)

Proof.

Activity 3.6.

Section The Euclidean Metric on Rn

Activity 3.3.

(a)

(b)

(c)

(d)

Lemma 3.9. Cauchy-Schwarz Inequality.

Activity 3.4.

(a)

(b)

Proof.

Corollary 3.10.

Activity 3.5.

(a)

(b)

Proof.

Activity 3.6.

Section The Euclidean Metric on $R^{n}$