October 2, 2020 – Physicspupil

In $C[0,1]$ , determine the values of $d_{\infty}(x,y)$ and $d_{1}(x,y)$ , when

(a) $x(t)=t^3+t+1$ and $y(t)=t^3+t^2+\frac{1}{2}t+1$ ;

(b) $x(t)=\sin t$ and $y(t)=t$ ;
(c) $x(t)=\sin t$ and $y(t)=t-\displaystyle{\frac{t^3}{6}}$ ;
(d) $x(t)=\mathrm{exp}(t)$ and $y(t)=\displaystyle{\sum_{m=0}^n}\frac{t^m}{m!}$ .

Recall.

Definition. (uniform metric) Let $C[a,b]$ be the set of all real-valued continuous functions defined on $[a,b]$ . For any $x,y\in C[a,b]$ , define the uniform metric $d_{\infty}$ :

$d_{\infty}(x,y)=\displaystyle{\max_{t\in [a,b]}}|x(t)-y(t)|$ .

N.b. If we let $B[a,b]$ be the set of all real-valued functions defined and bounded on $[a,b]$ , the uniform metric is then defined

$d_{\infty}(x,y)=\displaystyle{\sup_{t\in [a,b]}}|x(t)-y(t)|$ .

_{(cited from Examples 14 and 15, pg. 13, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Definition. For any $x, y\in C[a,b]$ , define

$d_1(x,y)=\displaystyle{\int_a^b}|x(t)-y(t)|\,\mathrm{d}t$

N.b. $d_1(x,y)$ represents the absolute area between the functions $x$ and $y$ as a measure of the distance between these two functions.

_{(cited from Example 16, pg. 14, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Solution.

(a)

$\begin{aligned} & \quad\, d_{\infty} (x,y) \\ &= \max_{t\in [0,1]} |x(t)-y(t)| \\ \end{aligned}$

Roughwork.

Approach.

To know the maximum value of $|x(t)-y(t)|$ , apply differentiation to

$f(t)=-t^2+\displaystyle{\frac{1}{2}}t$

and attain

$\begin{aligned} f(t) & = -t^2 + \frac{1}{2}t \\ f'(t) & = -2t+\frac{1}{2} \\ f'(t) & = 0 \Leftrightarrow t=\frac{1}{4} \\ \end{aligned}$

If the quadratic function $f(t)$ is plotted in a graph, a parabola admits of no inflexion points, needless to check on $f''(x)=0$ . So,

$\begin{array}{c|c|c|c|c|c} & t=0 & 0<t<\frac{1}{4} & t=\frac{1}{4} & \frac{1}{4}<t<1 & t=1 \\ &&&&&\\ \hline &&&&&\\ f(t) & 0 & \dots & \displaystyle{\frac{1}{16}} & \dots & -\displaystyle{\frac{1}{2}} \\ &&&&&\\ \hline &&&&&\\ f'(t) & \displaystyle{\frac{1}{2}}\enspace (>0) & \dots & 0\enspace (=0) & \dots & -\displaystyle{\frac{3}{2}}\enspace (<0) \\ &&&&&\\ \hline &&&&&\\ \textrm{plot} & \diagup & \dots & --- & \dots & \diagdown \\ &&&&&\\ \end{array}$

The continuous function $f(t)$ in the closed interval $[0,1]$ attains its maximum value $f(\frac{1}{4})=\frac{1}{16}$ when $t=\frac{1}{4}$ .

$\begin{aligned} & d_{\infty} (x,y) \\ = & \max_{t\in [0,1]} |x(t)-y(t)| \\ = & \max_{t\in [0,1]} |f(t)| \\ = & \frac{1}{16} \qquad\qquad\qquad \checkmark\\ \end{aligned}$

and should you think of what follows as quite right

$\begin{aligned} d_1(x,y) & = \int_0^1 |x(t)-y(t)| \,\mathrm{d}t \\ & = \int_0^1 f(t)\,\mathrm{d}t \\ & = \int_0^1 \bigg| -t^2 + \frac{1}{2}t \bigg| \, \mathrm{d}t \\ & = \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_0^1 \\ & = -\frac{1}{12}\qquad\qquad\qquad \times \\ \end{aligned}$

you might have rather mistaken calculus.

Correction.

Get back to the basics,

$\begin{aligned} f(t) & = 0 \\ \bigg| -t^2+\frac{1}{2}t \bigg| & = 0 \\ t^2-\frac{1}{2}t &= 0 \\ (t-\frac{1}{2})t & = 0 \\ t & = 0\quad \textrm{\scriptsize{OR}}\quad \frac{1}{2} \\ \end{aligned}$

From the previous graph of $C[0,1]$ , $f(t)$ is found to be positive when $t\in (0,0.5)$ , zero when $t\in \{ 0\} \cup\{ 0.5\}$ , and negative when $t\in (0.5,1]$ .

Doing it step-by-step,

$\begin{aligned} d_1(x,y) & = \int_0^1 \bigg| -t^2+\frac{1}{2}t \bigg| \,\mathrm{d}t \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg| -t^2+\frac{1}{2}t \bigg| + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg| -t^2+\frac{1}{2}t \bigg| \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg( - t^2 + \frac{1}{2}t \bigg) + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg( t^2-\frac{1}{2}t\bigg) \\ \end{aligned}$

Evaluating term-by-term, the first term being

$\begin{aligned} & \int_{0}^{0.5} \bigg( -t^2 + \frac{1}{2}t \bigg) \,\mathrm{d}t \\ = & \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_{0}^{0.5} \\ = & \bigg[ -\frac{(0.5)^3}{3} + \frac{(0.5)^2}{4} \bigg] - \bigg[ -\frac{(0)^3}{3} + \frac{(0)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \frac{1}{48} \\ \end{aligned}$

and the second term being

$\begin{aligned} & \int_{0.5}^{1}\bigg( t^2-\frac{1}{2}t\bigg) \,\mathrm{d}t \\ = & \bigg[ \frac{t^3}{3} - \frac{t^2}{4} \bigg]\bigg|_{0.5}^{1} \\ = & \bigg[ \frac{(1)^3}{3} - \frac{(1)^2}{4} \bigg] - \bigg[ \frac{(0.5)^3}{3} - \frac{(0.5)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \bigg( \frac{1}{12} \bigg) - \bigg( -\frac{1}{48} \bigg) \\ = & \frac{5}{48} \\ \end{aligned}$

In sum,

$d_{1}(x,y)=\displaystyle{\frac{1}{48}+\frac{5}{48}=\frac{6}{48}=\frac{1}{8}}=0.125$ .

Part (b), (c), and (d) are not chosen.

Let $(X,d)$ be a metric space and let $k$ be a fixed positive real number. For $x,\, y\in X$ , define

$d^{*}=kd(x,y)$ .

Prove that $d^{*}$ is a metric on $X$ .

Recall.

Definition. (metric) Let $X$ be a non-empty set. A metric on $X$ is a real-valued function $d:\enspace X\times X\rightarrow \mathbb{R}$ satisfying the following conditions i– iv:

i. $d(x,y)\ge 0$ ;
ii. $d(x,y)=0\Leftrightarrow x=y$ ;
iii. (Symmetry) $d(x,y)=d(y,x)$ ;
iv. (Triangle Inequality) $d(x,y)\le d(x,z)+d(z,y)$ for any $x,\, y,\, z\in X$ .

N.b. Given $x,\,y\in X$ , $d(x,y)$ is sometimes called the distance between $x$ and $y$ with respect to $d$ .

Proof.

i.

WTS (wish to show)

$d^{*}(x,y)\ge 0$

By definition $d^{*}(x,y)=kd(x,y)$ and in that the metric $d$ is let clear ( $\therefore d(x,y)\ge 0$ ) and $k$ a fixed positive real number ( $\therefore k>0$ ),

one can see

$\begin{aligned} d^{*}(x,y) & = kd(x,y) \\ \textrm{\dots because\enspace} & k>0 \enspace \textrm{and}\enspace d(x,y)\ge 0\textrm{\enspace \dots}\\ d^{*}(x,y) & \geqslant 0 \\ \end{aligned}$

$\therefore$ Condition i. is made.

ii.

$\begin{aligned} d^{*}(x,y) & = 0 \\ \Leftrightarrow kd(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace k>0 \enspace & \textrm{so}\enspace k\neq 0\enspace \dots \\ \Leftrightarrow d(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace d \enspace \textrm{was} &\enspace\textrm{foretold to be a metric}\enspace \dots \\ \Leftrightarrow x & = y \\ \end{aligned}$

$\therefore$ Condition ii. is made.

iii.

NTS (need to show)

$d^{*}(x,y)=d^{*}(y,x)$

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \enspace & \textrm{by the symmetric property of }d\enspace \dots \\ & =kd(y,x) \\ & = d^{*}(y,x) \\ & = \textrm{RHS} \\ \end{aligned}$

$\therefore$ Condition iii. is made.

iv.

RTP (required to prove)

$d^{*}(x,y)\leqslant d^{*}(x,z)+d^{*}(z,y)$

One starts with the left hand side,

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \textrm{as does}\enspace & d(x,y)\le d(x,z)+d(z,y)\enspace\textrm{the metric}\enspace d\enspace \textrm{do} \dots \\ & \le k\Big( d(x,z) + d(z,y) \Big) \\ & = kd(x,z) + kd(z,y) \\ & = d^{*}(x,z) + d^{*}(z,y) \\ & = \textrm{RHS} \\ \end{aligned}$

Condition iv. is made.

In conclusion, $(X,d^{*})$ is a metric space metered by a well-defined metric $d^{*}$ . This metric space shall simply be called $X$ hence.

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Day: October 2, 2020

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