Exercise 20.9

Question

ewcommand
orm[1]{\left\lVert #1 ightVert}

ewcommand\parens[1]{\left( #1 ight)}

ewcommand\seqfin[2]{\parens{#1_1, \ldots, #1_{#2}}}

ewcommand\seqinf[1]{\parens{#1_1, #1_2, \ldots}}

Show that the euclidean metric $d$ on ${\mathbb{R}}^n$ is a metric, as follows:
  If $\mathbf{x} , \mathbf{y}  \in {\mathbb{R}}^n$ and $c \in {\mathbb{R}}$, define
  \begin{align*}
    \mathbf{x}  + \mathbf{y}  &= (x_1 + y_1, \ldots, x_n + y_n) \,, \
    c\mathbf{x}  &= \seqfin{cx}{n} \,, \
    \mathbf{x}  \cdot \mathbf{y}  &= x_1 y_1 + \cdots + x_n y_n \,.
  \end{align*}
  \begin{enumerate}[label=(\alph*)]
  \item Show that $\mathbf{x}  \cdot (\mathbf{y}  + \mathbf{z} ) = (\mathbf{x}  \cdot \mathbf{y} ) + (\mathbf{x}  \cdot \mathbf{z} )$.
  \item Show that $\left|\mathbf{x}  \cdot \mathbf{y} ight| \leq 
orm{\mathbf{x} } 
orm{\mathbf{y} }$.
    [Hint: If $\mathbf{x} ,\mathbf{y}  
eq 0$, let $a = 1/
orm{\mathbf{x} }$ and $b = 1/
orm{\mathbf{y} }$, and use the fact that $
orm{a\mathbf{x}  \pm b\mathbf{y} } \geq 0$.]
  \item Show that $
orm{\mathbf{x}  + \mathbf{y} } \leq 
orm{\mathbf{x} } + 
orm{\mathbf{y} }$.
    [Hint: Compute $(\mathbf{x}  + \mathbf{y} ) \cdot (\mathbf{x}  + \mathbf{y} )$ and apply (b).]
  \item Verify that $d$ is a metric.
  \end{enumerate}

Dan Whitman · Accepted Answer

\def\a{\alpha}
\def\b{\beta}
\def\d{\delta}
\def\e{\epsilon}
\def{ho}

ewcommand{\vect}[1]{\mathbf{#1}}
\def\va{\vect{a}}
\def\vb{\vect{b}}

ewcommand\parens[1]{\left( #1 ight)}

ewcommand\seqfin[2]{\parens{#1_1, \ldots, #1_{#2}}}

ewcommand\seqinf[1]{\parens{#1_1, #1_2, \ldots}}

ewcommand
orm[1]{\left\lVert #1 ightVert}

First, we show some basic properties of these operations that will be useful:
  \begin{lemma} \label{lem:metric:vprops}
    For any $\mathbf{x} , \mathbf{y}  \in {\mathbb{R}}^n$ and $a \in {\mathbb{R}}$, we assert the following:
    \begin{enumerate}[label=(\arabic*)]
    \item The dot product is commutative, that is $\mathbf{x}  \cdot \mathbf{y}  = \mathbf{y}  \cdot \mathbf{x} $.
    \item $\mathbf{0} \cdot \mathbf{x}  = 0$.
    \item $
orm{\mathbf{x} } = 0$ if and only if $\mathbf{x}  = \mathbf{0}$.
    \item $
orm{\mathbf{x} } \geq 0$
    \item $\mathbf{x}  \cdot \mathbf{x}  = 
orm{\mathbf{x} }^2$
    \item $
orm{a \mathbf{x} } = \left|aight| 
orm{\mathbf{x} }$
    \end{enumerate}
  \end{lemma}
  \begin{proof}
    For assertion (1) clearly
    \begin{gather*}
      \mathbf{x}  \cdot \mathbf{y} = \sum_{i=1}^n x_i y_i = \sum_{i=1}^n y_i x_i = \mathbf{y}  \cdot \mathbf{x} 
    \end{gather*}
    by the definition of the dot product.
    Regarding (2), we clearly have
    \begin{gather*}
      \mathbf{0} \cdot \mathbf{x}  = \sum_{i=1}^n 0 \cdot x_i = \sum_{i=1}^n 0 = 0 \,.
    \end{gather*}

For (3) first suppose that $\mathbf{x}  
eq \mathbf{0}$ so that that there is an $i \in \left\{1, \ldots, night\}$ where $x_k 
eq 0$ so that clearly $x_k^2 > 0$.
    Then we have
    \begin{gather*}
      
orm{\mathbf{x} } = \sum_{i=1}^n x_i^2 = \sum_{i=k} x_i^2 + \sum_{i 
eq k} x_i^2 = x_k^2 + \sum_{i 
eq k} x_i^2 \geq x_k^2 + 0 = x_k^2 > 0
    \end{gather*}
    since each term in the sum $\sum_{i 
eq k} x_i^2$ is non-negative so the overall sum is as well.
    Thus of course $
orm{\mathbf{x} } 
eq 0$.
    This shows the forward implication by contrapositive.
    For the reverse direction, suppose that $\mathbf{x}  = \mathbf{0}$ so that
    \begin{gather*}
      
orm{\mathbf{x} } = 
orm{\mathbf{0}} = \left[\sum_{i=1}^n 0^2ight]^{1/2} = \left[\sum_{i=1}^n 0ight]^{1/2} = \left[0ight]^{1/2} = 0 \,.
    \end{gather*}

Assertion (4) is fairly obvious from the definition.
    Clearly each $x_i^2 \geq 0$ since it is a square, so that $\sum_{i=1}^n x_i^2 \geq 0$ as well.
    It then follows from Corollary~\href{./exercise_20-1}{1} that
    \begin{gather*}
      
orm{\mathbf{x} } = \left[\sum_{i=1}^n x_i^2ight]^{1/2} \geq 0^{1/2} = 0
    \end{gather*}
    as desired.
    Assertion (5) is also easy to show:
    \begin{gather*}
      \mathbf{x}  \cdot \mathbf{x}  = \sum_{i=1}^n x_i x_i = \sum_{i=1}^n x_i^2 = \left(\left[\sum_{i=1}^n x_i^2ight]^{1/2}ight)^2 = 
orm{\mathbf{x} }^2
    \end{gather*}
    by definition.

For part~(6) we have by definition that
    \begin{gather*}
      
orm{a \mathbf{x} } = \left[\sum_{i=1}^n (a x_i)^2ight]^{1/2} = \left[\sum_{i=1}^n a^2 x_i^2ight]^{1/2}
      = \left[a^2 \sum_{i=1}^n x_i^2ight]^{1/2} = \left[a^2ight]^{1/2} \left[\sum_{i=1}^n x_i^2ight]^{1/2} = \left|aight|
orm{\mathbf{x} }
    \end{gather*}
    as desired.
  \end{proof}

extbf{Main Problem.}

(a)
  \begin{proof}
    By definition, we have that
    \begin{gather*}
      \mathbf{y}  + \mathbf{z}  = (y_1 + z_1, \ldots, y_n + y_n)
    \end{gather*}
    so that clearly
    \begin{gather*}
      \mathbf{x}  \cdot (\mathbf{y}  + \mathbf{z} ) = \sum_{i=1}^n x_i (y_i + z_i) = \sum_{i=1}^n (x_i y_i + x_i z_i)
      = \sum_{i=1}^n x_i y_i + \sum_{i=1}^n x_i z_i = (\mathbf{x}  \cdot \mathbf{y} ) + (\mathbf{x}  \cdot \mathbf{z} )
    \end{gather*}
    as desired.
  \end{proof}

(b)
  \begin{proof}
    First, if $\mathbf{x}  = \mathbf{0}$ then clearly by Lemma~ef{lem:metric:vprops} parts (2) and (3)
    \begin{gather*}
      \left|\mathbf{x}  \cdot \mathbf{y} ight| = \left|\mathbf{0} \cdot \mathbf{y} ight| = \left|0ight| = 0 \leq 0 = 0 
orm{\mathbf{y} } = 
orm{\mathbf{0}} 
orm{\mathbf{y} } = 
orm{\mathbf{x} } 
orm{\mathbf{y} } \,.
    \end{gather*}
    Similarly if $\mathbf{y}  = \mathbf{0}$ then by all parts of Lemma~ef{lem:metric:vprops}
    \begin{gather*}
      \left|\mathbf{x}  \cdot \mathbf{y} ight| = \left|\mathbf{x}  \cdot \mathbf{0}ight| = \left|\mathbf{0} \cdot \mathbf{x} ight| = \left|0ight| = 0 \leq 0 = 
orm{\mathbf{x} } 0 = 
orm{\mathbf{x} } 
orm{\mathbf{0}} = 
orm{\mathbf{x} } 
orm{\mathbf{y} } \,.
    \end{gather*}
    So in what follows assume that $\mathbf{x} , \mathbf{y}  
eq \mathbf{0}$.
    Then by Lemma~ef{lem:metric:vprops} part~(3) we have that $
orm{\mathbf{x} }$ and $
orm{\mathbf{y} }$ are both nonzero so that $a = 1/
orm{\mathbf{x} }$ and $b = 1/
orm{\mathbf{y} }$ are defined.
    Then, by Lemma~ef{lem:metric:vprops} part~(4), we of course have
    \begin{gather*}
      
orm{a\mathbf{x}  \pm b\mathbf{y} } \geq 0 \,.
    \end{gather*}
    Since clearly
    \begin{gather*}
      a\mathbf{x}  \pm b\mathbf{y}  = (ax_1 \pm by_1, \ldots, ax_n \pm by_n)
    \end{gather*}
    by the definition of the operations, we have
    \begin{gather*}
      \left[\sum_{i=1}^n (ax_i \pm by_i)^2ight]^{1/2} = 
orm{a\mathbf{x}  \pm b\mathbf{y} } \geq 0 \,.
    \end{gather*}
    It then follows from Lemma~\href{./exercise_20-1}{1} that
    \begin{gather*}
      \sum_{i=1}^n (ax_i \pm by_i)^2 \geq 0^2 = 0 \
      \sum_{i=1}^n (a^2 x_i^2 \pm 2ab x_i y_i + b^2 y_i^2) \geq 0 \
      a^2 \sum_{i=1}^n x_i^2 \pm 2ab \sum_{i=1}^n x_i y_i + b^2 \sum_{i=1}^n y_i^2 \geq 0 \
      a^2 
orm{\mathbf{x} }^2 \pm 2ab (\mathbf{x}  \cdot \mathbf{y} ) + b^2 
orm{\mathbf{y} }^2 \geq 0 \
      \pm 2ab (\mathbf{x}  \cdot \mathbf{y} ) \geq -a^2 
orm{\mathbf{x} }^2 - b^2 
orm{\mathbf{y} }^2 \
      \pm 2\frac{\mathbf{x}  \cdot \mathbf{y} }{
orm{\mathbf{x} } 
orm{\mathbf{y} }} \geq -\frac{
orm{\mathbf{x} }^2}{
orm{\mathbf{x} }^2} - \frac{
orm{\mathbf{y} }^2}{
orm{\mathbf{y} }^2} \
      \pm 2\frac{\mathbf{x}  \cdot \mathbf{y} }{
orm{\mathbf{x} } 
orm{\mathbf{y} }} \geq -1 - 1 \
      \pm 2\frac{\mathbf{x}  \cdot \mathbf{y} }{
orm{\mathbf{x} } 
orm{\mathbf{y} }} \geq -2 \
      \mp \mathbf{x}  \cdot \mathbf{y}  \leq 
orm{\mathbf{x} } 
orm{\mathbf{y} } \,.
    \end{gather*}
    Hence we have that both $\mathbf{x}  \cdot \mathbf{y}  \leq 
orm{\mathbf{x} } 
orm{\mathbf{y} }$ and $- \mathbf{x}  \cdot \mathbf{y}  \leq 
orm{\mathbf{x} } 
orm{\mathbf{y} }$ so that $\mathbf{x}  \cdot \mathbf{y}  \geq -
orm{\mathbf{x} } 
orm{\mathbf{y} }$.
    Hence
    \begin{gather*}
      -
orm{\mathbf{x} } 
orm{\mathbf{y} } \leq \mathbf{x}  \cdot \mathbf{y}  \leq 
orm{\mathbf{x} } 
orm{\mathbf{y} }
    \end{gather*}
    so we can conclude that $\left|\mathbf{x}  \cdot \mathbf{y} ight| \leq 
orm{\mathbf{x} } 
orm{\mathbf{y} }$ as desired.
  \end{proof}

(c)
  \begin{proof}
    We have
    \begin{align*}
      
orm{\mathbf{x}  + \mathbf{y} }^2 &= (\mathbf{x}  + \mathbf{y} ) \cdot (\mathbf{x}  + \mathbf{y} ) & 	ext{(by Lemma~ef{lem:metric:vprops} part~(5))} \
      &= (\mathbf{x}  + \mathbf{y} ) \cdot \mathbf{x}  + (\mathbf{x}  + \mathbf{y} ) \cdot \mathbf{y}  & 	ext{(by part~(a))} \
      &= \mathbf{x}  \cdot (\mathbf{x}  + \mathbf{y} ) + \mathbf{y}  \cdot (\mathbf{x}  + \mathbf{y} ) & 	ext{(by Lemma~ef{lem:metric:vprops} part~(1))} \
      &= \mathbf{x}  \cdot \mathbf{x}  + \mathbf{x}  \cdot \mathbf{y}  + \mathbf{y}  \cdot \mathbf{x}  + \mathbf{y}  \cdot \mathbf{y}  & 	ext{(by part~(a))} \
      &= 
orm{\mathbf{x} }^2 + \mathbf{x}  \cdot \mathbf{y}  + \mathbf{y}  \cdot \mathbf{x}  + 
orm{\mathbf{y} }^2 & 	ext{(by Lemma~ef{lem:metric:vprops} part~(5))} \
      &= 
orm{\mathbf{x} }^2 + \mathbf{x}  \cdot \mathbf{y}  + \mathbf{x}  \cdot \mathbf{y}  + 
orm{\mathbf{y} }^2 & 	ext{(by Lemma~ef{lem:metric:vprops} part~(1))} \
      &\leq 
orm{\mathbf{x} }^2 + 
orm{\mathbf{x} }
orm{\mathbf{y} } + 
orm{\mathbf{x} }
orm{\mathbf{y} } + 
orm{\mathbf{y} }^2 & 	ext{(by part~(b))} \
      &= 
orm{\mathbf{x} }^2 + 2
orm{\mathbf{x} }
orm{\mathbf{y} } + 
orm{\mathbf{y} }^2 \
      &= (
orm{\mathbf{x} } + 
orm{\mathbf{y} })^2 \,.
    \end{align*}
    The desired result that $
orm{\mathbf{x}  + \mathbf{y} } \leq 
orm{\mathbf{x} } + 
orm{\mathbf{y} }$ then follows from Corollary~\href{./exercise_20-1}{1}.
  \end{proof}

(d)
  \begin{proof}
    First recall that the euclidean metric on ${\mathbb{R}}^n$ is defined as
    \begin{gather*}
      d(\mathbf{x} , \mathbf{y} ) = 
orm{\mathbf{x}  - \mathbf{y} } \,.
    \end{gather*}
    Then part~(1) of the definition of a metric follows directly from Lemma~ef{lem:metric:vprops} since of course
    \begin{gather*}
      d(\mathbf{x} , \mathbf{y} ) = 
orm{\mathbf{x}  - \mathbf{y} } \geq 0 \,.
    \end{gather*}
    We also have that $\mathbf{x}  = \mathbf{y} $ if and only if $\mathbf{x}  - \mathbf{y}  = 0$, which is true if and only if
    \begin{gather*}
      d(\mathbf{x} , \mathbf{y} ) = 
orm{\mathbf{x}  - \mathbf{y} } = 0
    \end{gather*}
    by Lemma~ef{lem:metric:vprops} part~(3).
    Similarly part~(2) of the definition follows from Lemma~ef{lem:metric:vprops} part~(6) as follows:
    \begin{gather*}
      d(\mathbf{x} , \mathbf{y} ) = 
orm{\mathbf{x}  - \mathbf{y} } = 
orm{(-1)(\mathbf{y}  - \mathbf{x} )} = \left|-1ight|
orm{\mathbf{y}  - \mathbf{x} } = 
orm{\mathbf{y}  - \mathbf{x} } = d(\mathbf{y} , \mathbf{x} ) \,.
    \end{gather*}
    Lastly, for part~(3) of the definition, we have that
    \begin{align*}
      d(\mathbf{x} , \mathbf{z} ) &= 
orm{\mathbf{x}  - \mathbf{y} } = 
orm{\mathbf{x}  - \mathbf{z}  + \mathbf{y}  - \mathbf{y} } = 
orm{(\mathbf{x}  - \mathbf{y} ) + (\mathbf{y}  - \mathbf{z} )} \
      &\leq 
orm{\mathbf{x}  - \mathbf{y} } + 
orm{\mathbf{x}  - \mathbf{z} } = d(\mathbf{x} , \mathbf{y} ) + d(\mathbf{y} , \mathbf{z} ) \,,
    \end{align*}
    where we have used what was shown in part~(c).
    This shows that $d$ has all the properties required to be a metric.
  \end{proof}

Exercise 20.9

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Exercise 20.9

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