CSIR JUNE 2011 PART C QUESTION 63 SOLUTION (When $d(x,y) = |f(x) - f(y)|$ defines a metric)

Which of the following is/are metric on $\Bbb R$?
1)$d(x,y) = min\{x,y\}$,
2)$d(x,y) = |x-y|$,
3) $d(x,y) = |x^2 - y^2|$,
4)$d(x,y) = |x^3 - y^3|$.
Solution

option (1): NO. $d(-1,0) = -1$ which is negative.
option (2): YES. Given metric is the usual metric.
option (3): NO. $d(-1,1) = 0$. In a metric space $d(x,y) =0$ if and only if $x=y$.
option (4):YES.  We will verify all the axioms.
axiom 1: Clearly $d(x,y) \ge 0$,
axiom 2: Let $x,y \in \Bbb R$, then $d(x,y) = 0$ if and only if $|x^3-y^3|=0$ if and only if x^3 = y^3 if and only if $x=y$. [Since f(x)=x^3 is injective]
axiom 3: $d(x,y) = d(y,x)$ follows.
axiom 4: (Triangle inequality)
Let $x,y,z \in \Bbb R$. $d(x,y) = |x^3-y^3| = |x^3-z^3+z^3-y^3| \le |x^3-z^3| + |z^3-y^3|$ and this is equal to $d(x,z) + d(z,y)$. This proves that $d$ is a metric.
Bonus: Define a function $d$ on $\Bbb R$ by $d(x,y) = |f(x) - f(y)|$where $f: \Bbb R \to \Bbb R$ is a function. What is the necessary and sufficient condition on $f$ to make $d$ a metric. The answer is it is enough $f$ to be injective. 
Proof: Proof of option (4) work for this general case!!!
In option (3), $f(x) =x^2$ which is not injective so not a metric.
In option (4), $f(x) = x^3$ is injective, so it is a metric. 
Exercises:
Which of the following define metric on $\Bbb R$?
1) $d(x,y) = |sin\,x - \sin y|$.  ($f(x) = sin\,x$).
2) $d(x,y) = |e^x - e^y|$. ($f(x) = e^x$).
3) $d(x,y) = |log \,x - log \,y|$ on $\Bbb R^+$. ($f(x) = log \,x$).
4) $d(x,y) = |\sqrt x - \sqrt y|$ on $\Bbb R^+$. ($f(x) = \sqrt x$).
5) $d(x,y) = |tan\,x - tan\,y|$ on $(-\frac{\pi}{2},\frac{\pi}{2})$. ($f(x) = tan\,x$).
6) $d(x,y) = |\frac{1}{x} - \frac{1}{y}|$ on $\Bbb R^+$. ($f(x) =\frac{1}{x}$).
You can ask this question for any function and you know the answer!!!
 Click here for more problems. 
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