Functional Analysis Exercises 1 : Distance Metrics

Avishek Sen Gupta on 21 September 2021

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This post lists solutions to many of the exercises in the Distance Metrics section 1.1 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is definitely a work in progress, and proofs may be refined or added over time.

1.1.2. Does $d(x,y)={(x-y)}^2$ define a metric on the set of all real numbers?

Proof:

For the distance metric $d(x,y)={(x-y)}^2$, we need to prove or disprove the Triangle Inequality:

$$d(x,z) \leq d(x,y) + d(y,z)$$

We start with the term ${(x-z)}^2$, as follows:

$${(x-z)}^2 = {((x-y)+(y-z))}^2 \\ \Rightarrow {(x-z)}^2 = {(x-y)}^2 + {(y-z)}^2 + 2 (x-y) (y-z) \\ \Rightarrow d(x,z) = d(x,y) + d(y,z) + \underbrace{2 (x-y) (y-z)}_\text{positive or negative}$$

When the term $2 (x-y) (y-z)$ is positive:

$$d(x,z) > d(x,y) + d(y,z)$$

When the term $2 (x-y) (y-z)$ is negative or zero:

$$d(x,z) \leq d(x,y) + d(y,z)$$

Thus, the Triangle Inequality can only be satisfied for specific values of $x$, $y$, and $z$; hence $d(x,y)={(x-y)}^2$ is not a valid distance metric.

$$\blacksquare$$

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1.1.3. Show that $d(x,y)=\sqrt{|x-y|}$ defines a metric on the set of all real numbers.

Proof:

For the distance metric $d(x,y)=\sqrt{\vert x-y \vert}$, we need to prove the Triangle Inequality:

$$d(x,z) \leq d(x,y) + d(y,z)$$

We start with the basic Triangle Inequality for $\mathbb{R}$:

$$|x-z| \leq |x-y| + |y-z|$$

Adding and subtracting $2 \sqrt{\vert x-y \vert \vert y-z \vert}$ on the RHS we get:

$$|x-z| \leq |x-y| + |y-z| + 2 \sqrt{\vert x-y \vert \vert y-x \vert} - 2 \sqrt{\vert x-y \vert \vert y-z \vert} \\ \Rightarrow |x-z| \leq {(\sqrt{\vert x-y \vert} + \sqrt{\vert y-z \vert})}^2 - \underbrace{2 \sqrt{\vert x-y \vert \vert y-z \vert}}_\text{positive}$$

Setting $C=2 \sqrt{\vert x-y \vert \vert y-z \vert}>0$, we get:

$$|x-z| \leq {\left(\sqrt{\vert x-y \vert} + \sqrt{\vert y-z \vert}\right)}^2 - C \\ \Rightarrow {\left(\sqrt{\vert x-z \vert}\right)}^2 \leq {\left(\sqrt{\vert x-y \vert} + \sqrt{\vert y-z \vert}\right)}^2 - C \\ \Rightarrow {\left(\sqrt{\vert x-z \vert}\right)}^2 + C \leq {\left(\sqrt{\vert x-y \vert} + \sqrt{\vert y-z \vert}\right)}^2 \\ \Rightarrow {\left(\sqrt{\vert x-z \vert}\right)}^2 \leq {\left(\sqrt{\vert x-y \vert} + \sqrt{\vert y-z \vert}\right)}^2 \\ \Rightarrow \sqrt{\vert x-z \vert} \leq \sqrt{\vert x-y \vert} + \sqrt{\vert y-z \vert} \\ \Rightarrow d(x,z) \leq d(x,y) + d(y,z)$$

Hence, this proves the Triangle Inequality, and consequently, $d(x,y)=\sqrt{\vert x-y \vert}$ is a valid distance metric.

$$\blacksquare$$

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1.1.5. Let $d$ be a metric on $X$. Determine all constants $k$ such that (i) $kd$, (ii) $d+k$ is a metric on $X$.

(i) Proof:

Let $\bar{d}(x,y)=kd(x,y)$ be a candidate metric on $X$. For it to be a valid distance metric, it must satisfy the four axioms of a metric, i.e.:

• $\bar{d}(x,y)>0$ if $x \neq y$: For this to hold for $x \neq y$, we must have $k>0, k \in \mathbb{R}$.
• $\bar{d}(x,y)=0$ if and only if $x=y$: For this to hold, we can have $k \in \mathbb{R}$.
• $\bar{d}(x,y)=\bar{d}(y,x)$: For this to hold, we can have $k \in \mathbb{R}$.

• $\bar{d}(x,z) \leq \bar{d}(x,y) + \bar{d}(y,z)$: For this, if we multiply the original, valid Triangle Inequality by $k$ on both sides, we have the following:

  $$kd(x,z) \leq k[d(x,y) + d(y,z)] \\ \Rightarrow kd(x,z) \leq kd(x,y) + kd(y,z) \\ \Rightarrow \bar{d}(x,z) \leq \bar{d}(x,y) + \bar{d}(y,z)$$

  proving that the Triangle Inequality holds for $\bar{d}$ for any $k \in \mathbb{R}$.

Putting all of these together, we get the condition that $k>0, k \in \mathbb{R}$.

(ii) Proof:

Let $\bar{d}(x,y)=d(x,y) + k$ be a candidate metric on $X$. For it to be a valid distance metric, it must satisfy the four axioms of a metric, i.e.:

• $\bar{d}(x,y)>0$ if $x \neq y$: For this to hold for $x \neq y$, we must have $k>0, k \in \mathbb{R}$.
• $\bar{d}(x,y)=0$ if and only if $x=y$: For this to hold, we must have $d(x,y)+k=0$. Since $d(x,y)=0$ already, $k=0, k \in \mathbb{R}$.
• $\bar{d}(x,y)=\bar{d}(y,x)$: For this to hold, we can have $k \in \mathbb{R}$.

• $\bar{d}(x,z) \leq \bar{d}(x,y) + \bar{d}(y,z)$: For this, we can need to find the condition for which the following holds:

  $$d(x,z) + k \leq [d(x,y) + k] + [d(y,z) + k] \\ \Rightarrow d(x,z) + k \leq d(x,y) + d(y,z) + 2k \\ \Rightarrow d(x,z) \leq d(x,y) + d(y,z) + k$$

  Since $d(x,z) \leq d(x,y) + d(y,z)$ already, we must have $k \geq 0$ for the above inequality to always hold; this shows that the Triangle Inequality holds for $\bar{d}$ for $k \geq 0 \in \mathbb{R}$.

Putting all of these together, we get the condition that $k=0, k \in \mathbb{R}$.

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1.1.6. Show that $d$ in 1.1-6 satisfies the triangle inequality.

Answer: $l^\infty$ is the set of all bounded sequences of complex numbers. The metric under consideration is $d(x,y)=sup \vert \zeta_i - eta_i\vert$.

We need to prove that $d(x,y) \leq d(x,z) + d(z,y)$. We know that: $\vert \eta_i - \zeta_i \vert \leq \vert \eta_i - \theta_i \vert + \vert \theta_i - \zeta_i \vert$. Taking $sup$ on both sides we get:

$$\begin{equation} sup \vert \eta_i - \zeta_i \vert \leq sup [\vert \eta_i - \theta_i \vert + \vert \theta_i - \zeta_i \vert] \label{eq:1-1-6-1} \end{equation}$$

For two sequences $(a_i)$ and $(b_i)$, we have: $a_i \leq sup(a_i) \\ b_i \leq sup(b_i)$

Adding the two inequalities, we get:

$$a_i + b_i \leq sup(a_i) + sup(b_i) \\ \Rightarrow sup(a_i + b_i) \leq sup(a_i) + sup(b_i)$$

The above is because any $a_i+b_i$ is less than or equal to some constant, so the supremum is also less than or equal to that constant.

Setting $a_i = \vert \eta_i - \theta_i \vert$ and $b_i = \vert \theta_i - \zeta_i \vert$, we get:

$$\begin{equation} sup[\vert \eta_i - \theta_i \vert + \vert \theta_i - \zeta_i \vert] \leq sup \vert \eta_i - \theta_i \vert + sup \vert \theta_i - \zeta_i \vert \label{eq:1-1-6-2} \end{equation}$$

Putting $\eqref{eq:1-1-6-1}$ and $\eqref{eq:1-1-6-2}$ together we get:

$$sup \vert \eta_i - \zeta_i \vert \leq sup [\vert \eta_i - \theta_i \vert + \vert \theta_i - \zeta_i \vert] \leq sup \vert \eta_i - \theta_i \vert + sup \vert \theta_i - \zeta_i \vert \\ \Rightarrow sup \vert \eta_i - \zeta_i \vert \leq sup \vert \eta_i - \theta_i \vert + sup \vert \theta_i - \zeta_i \vert \\$$

This proves the Triangle Inequality for the given distance metric.

$$\blacksquare$$

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1.1.8. Show that another metric $\bar{d}$ on the set $X$ in 1.1-7 is defined by $\bar{d}(x,y)=\displaystyle\int\limits_a^b |x(t) - y(t)| dt$.

Proof:

The distance metric given is: $d(x,y)=\displaystyle\int\limits_a^b \vert x(t) - y(t) \vert$ We know that:

$$\vert x(t) - y(t) \vert \leq \vert x(t) - z(t) \vert + \vert z(t) - y(t) \vert$$

Integrating both sides with respect to $t$ from $a$ to $b$, we get:

$$\displaystyle\int\limits_a^b \vert x(t) - y(t) \vert \leq \displaystyle\int\limits_a^b \vert x(t) - z(t) \vert + \displaystyle\int\limits_a^b \vert z(t) - y(t) \vert \\ \Rightarrow d(x,y) \leq d(x,z) + d(z,y)$$ $$\blacksquare$$

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1.1.9. Show that $d$ in 1.1-8 is a metric.

For reference, the axioms (M1) to (M4) are as follows:

• (M1) $0 \leq d(x,y)<\infty, d(x,y)\in \mathbb{R}$
• (M2) $d(x,y)=0$ if and only if $x=y$
• (M3) $d(x,y)=d(y,x)$
• (M4) $d(x,z) \leq d(x,y) + d(y,z)$

The discrete metric is: $d(x,x)=0 \\ d(x,y)=1, \text{ for } x \neq y$

This satisfies (M1), since $d(x,y) \in \{0,1\}$.

(M2) also follows from $d(x,x) = 0$.

(M3) also follows from $d(x,y) = d(y,x) = 1$ if $x \neq y$ and $d(x,x) = 0$.

Let’s prove the Triangle Inequality. We have:

$$d(x,z) \geq 0 \\ d(z,y) \geq 0$$

Adding the above inequalities, we get:

$$d(x,z) + d(z,y) \geq 0$$

Case 1

If $x=y$, then $d(x,y) = 0$, and we have:

$$d(x,z) + d(z,y) \geq d(x,y) \text{ for } x=y$$

Case 2

If $x \neq y$, then $d(x,y) = 1$. Then, we have 3 sub-cases:

(2.1) $z=x, z \neq y$. Then $d(x,z) + d(z,y) = 0 + 1 = 1 \geq d(x,y)$

(2.2) $z=y, z \neq x$. Then $d(x,z) + d(z,y) = 1 + 0 = 1 \geq d(x,y)$

(2.3) $z \neq y, z \neq x$. Then $d(x,z) + d(z,y) = 1 + 1 = 2 \geq d(x,y)$

In all the above cases, we have $d(x,y) \leq d(x,z) + d(z,y)$, thus proving (M4) (the Triangle Inequality).

$$\blacksquare$$

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1.1.10. (Hamming Distance) Let $X$ be the set of all ordered triples of zeros and ones. Show that $X$ consists of eight elements and a metric $d$ on $X$ is defined by $d(x,y)=$ number of places where $x$ and $y$ have different entries. (This space and similar spaces of $n$-tuples play a role in switching and automata theory and coding. $d(x,y)$ is called the Hamming distance between $x$ and $y$; cf. the paper by R. W. Hamming (1950) listed in Appendix 3.)

Proof:

The set of all ordered triples of zeros and ones, forms a sequence of numbers $X=[0,7], x_i \in X, x_i \in \mathbb{Z}$ in their binary form. Thus, it is evident that the number of triples is $2^3=8$.

Let $a,b,c \in {0,1}$, then each number in this set, can be represented as $x_i=a+2b+4c$. Then the suggested distance metric is:

$$d(x,y)=|a_x-a_y|+|b_x-b_y|+|c_x-c_y|$$

Let $x_1$, $x_2$, $x_3$ be defined as follows:

$$x=a_x+2b_x+4c_x \\ y=a_y+2b_y+4c_y \\ z=a_z+2b_z+4c_z$$ $$\begin{align*} d(x,z) &= |a_x-a_z|+|b_x-b_z|+|c_x-c_z| \\ &=|a_x-a_y+a_y-a_z|+|b_x-b_y+b_y-b_z|+|c_x-c_y+c_y-c_z| \end{align*}$$

Now we have the following inequalities:

$$|a_x-a_y+a_y-a_z| \leq |a_x-a_y|+|a_y-a_z| \\ |b_x-b_y+b_y-b_z| \leq |b_x-b_y|+|b_y-b_z| \\ |c_x-c_y+c_y-c_z| \leq |c_x-c_y|+|c_y-c_z|$$

Summing up these inequalities, and noting that the LHS resolves to $d(x,z)$, we get:

$$d(x,z) \leq |a_x-a_y|+|a_y-a_z| + |b_x-b_y|+|b_y-b_z| + |c_x-c_y|+|c_y-c_z| \\ \Rightarrow d(x,z) \leq (|a_x-a_y|+|b_x-b_y|+|c_x-c_y|) + (|a_y-a_z|+|b_y-b_z|+|c_y-c_z|) \\ \Rightarrow d(x,z) \leq d(x,y) + d(y,z)$$

Thus, this proves the Triangle Inequality for the Hamming Distance as a metric.

$$\blacksquare$$

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1.1.12. (Triangle inequality) The triangle inequality has several useful consequences. For instance, using the generalised triangle inequality, show that $|d(x,y) - d(z,w)| \leq d(x,z) + d(y,w)$.

We write the two following triangle inequalities. One involves $x$, $y$, $z$. The other one involves $w$, $y$, $z$.

$$d(x,y) \leq d(x,z) + d(z,y)$$ $$\begin{equation} d(z,y) \leq d(z,w) + d(w,y) \label{eq:1-1-12-1} \end{equation}$$

Adding $d(x,z)$ to $\eqref{eq:1-1-12-1}$, we get:

$$d(x,z) + d(z,y) \leq d(x,z) + d(z,w) + d(w,y)$$

Thus, we have:

$$d(x,y) \leq d(x,z) + d(z,y) \leq d(x,z) + d(z,w) + d(w,y) \\ \Rightarrow d(x,y) \leq d(x,z) + d(z,w) + d(w,y) \\ \Rightarrow d(x,y) \leq d(x,z) + d(z,w) + d(y,w) \text{(by the Symmetry property of a Distance Metric)}$$ $$\begin{equation} d(x,y) - d(z,w) \leq d(x,z) + d(w,y) \label{eq:1-1-12-abs-1} \end{equation}$$

We write the two following triangle inequalities. One involves $x$, $z$, $w$. The other one involves $x$, $y$, $w$.

$$d(z,w) \leq d(z,x) + d(x,w)$$ $$\begin{equation} d(x,w) \leq d(x,y) + d(y,w) \label{eq:1-1-12-2} \end{equation}$$

Adding $d(z,x)$ to $\eqref{eq:1-1-12-2}$, we get:

$$d(z,x) + d(x,w) \leq d(z,x) + d(x,y) + d(y,w)$$

Thus, we have:

$$d(z,w) \leq d(z,x) + d(x,w) \leq d(z,x) + d(x,y) + d(y,w) \\ \Rightarrow d(z,w) \leq d(z,x) + d(x,y) + d(y,w) \\ \Rightarrow d(z,w) \leq d(x,z) + d(x,y) + d(y,w) \text{(by the Symmetry property of a Distance Metric)}$$ $$\begin{equation} d(z,w) - d(x,y) \leq d(z,x) + d(y,w) \label{eq:1-1-12-abs-2} \end{equation}$$

Summarising $\eqref{eq:1-1-12-abs-1}$ and $\eqref{eq:1-1-12-abs-2}$, we get:

$$d(x,y) - d(z,w) \leq d(x,z) + d(y,w) \\ d(z,w) - d(x,y) \leq d(x,z) + d(y,w) \\ \Rightarrow \mathbf{ |d(x,y) - d(z,w)| \leq d(x,z) + d(y,w) }$$ $$\blacksquare$$

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1.1.13. Using the triangle inequality, show that $|d(x,z) - d(y,z)| \leq d(x,y)$.

Proof:

We have to show that:

$$|d(x,z) - d(y,z)| \leq d(x,y)$$

We write the following Triangle Inequality:

$$\begin{equation} d(x,z) \leq d(x,y) + d(y,z) \\ \Rightarrow d(x,z) - d(y,z) \leq d(x,y) \label{eq:1-1-13-abs-1} \end{equation}$$

The other Triangle Inequality we write is:

$$\begin{equation} d(y,z) \leq d(y,x) + d(x,z) \\ \Rightarrow d(y,z) - d(x,z) \leq d(y,x) \\ \Rightarrow d(y,z) - d(x,z) \leq d(x,y) \text{(by the Symmetry property of a Distance Metric)} \label{eq:1-1-13-abs-2} \end{equation}$$

Summarising the results of $\eqref{eq:1-1-13-abs-1}$ and $\eqref{eq:1-1-13-abs-2}$, we get:

$$d(x,z) - d(y,z) \leq d(x,y) \\ d(y,z) - d(x,z) \leq d(x,y) \\ \Rightarrow \mathbf{|d(x,z) - d(y,z)| \leq d(x,y)}$$ $$\blacksquare$$

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1.1.14. (Axioms of a metric) (M1) to (M4) could be replaced by other axioms (without changing the definition). For instance, show that (M3) and (M4) could be obtained from (M2) and $d(x,y) \leq d(z,x) + d(z,y)$.

For reference, the axioms (M1) to (M4) are as follows:

• (M1) $0 \leq d(x,y)<\infty, d(x,y)\in \mathbb{R}$
• (M2) $d(x,y)=0$ if and only if $x=y$
• (M3) $d(x,y)=d(y,x)$
• (M4) $d(x,z) \leq d(x,y) + d(y,z)$

Proof for (M3):

The allowed assumptions are:

• (A1) $d(x,y)=0$ if and only if $x=y$
• (A2) $d(x,y) \leq d(z,x) + d(z,y)$

Set $z=y$ in (A2), so that we get:

$$\require{cancel} \begin{equation} d(x,y) \leq d(y,x) + \cancel{d(y,y)} \text{ (by (A1))} \\ \Rightarrow d(x,y) \leq d(y,x) \label{eq:1-1-14-abs-1} \end{equation}$$

From (A2), we get $d(y,x)$ as:

$$\begin{equation} d(y,x) \leq d(z,y) + d(z,x) \label{eq:1-1-14-y-x} \end{equation}$$

Set $z=x$ in $\eqref{eq:1-1-14-y-x}$ again, so that we get:

$$\begin{equation} d(y,x) \leq d(x,y) + \cancel{d(x,x)} \text{ (by (A1))} \\ \Rightarrow d(y,x) \leq d(x,y) \\ \Rightarrow d(x,y) \geq d(y,x) \label{eq:1-1-14-abs-2} \end{equation}$$

Summarising the results of $\eqref{eq:1-1-14-abs-1}$ and $\eqref{eq:1-1-14-abs-2}$, we get:

$$d(x,y) \leq d(y,x) \\ d(x,y) \geq d(y,x)$$

This implies that:

$$\begin{equation} d(x,y)=d(y,x) \label{eq:1-1-14-symmetry} \end{equation}$$ $$\blacksquare$$

Proof for (M4):

(M4) should immediately follow from $\eqref{eq:1-1-14-symmetry}$, since:

$$d(x,y) \leq d(z,x) + d(z,y) \Rightarrow d(x,y) \leq d(x,z) + d(z,y)$$ $$\blacksquare$$

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1.1.15. Show that nonnegativity of a metric follows from (M2)to (M4).

Proof:

We have to prove that: $d(x,y) \geq 0$ follows from (M2) and (M4).

• (M2) $d(x,y)=0$ if and only if $x=y$
• (M3) $d(x,y)=d(y,x)$
• (M4) $d(x,z) \leq d(x,y) + d(y,z)$

From the Triangle Inequality (M4), we have:

$$d(x,y) \leq d(x,z) + d(z,y)$$

Set $x=y$, then we get:

$$d(y,y) \leq d(y,z) + d(z,y) \\ \Rightarrow \underbrace{\cancel{d(y,y)}}_\text{by (M2)} \leq d(y,z) + \underbrace{d(y,z)}_\text{by (M3)} \\ \Rightarrow 2d(y,z) \geq 0 \\ \Rightarrow d(y,z) \geq 0$$

This proves (M1).

$$\blacksquare$$

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