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A Quick Note on Proving the Triangle Inequality on a Derived Distance Metric using Monotonicity

Avishek Sen Gupta on 20 October 2022

This is a quick note on proving the Triangle Inequality criterion of the following claim:

If \(d(x,y)\) is a distance metric, then \(\bar{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}\) is also a valid distance metric.

The four criteria for satisfying a distance metric are:

(M1) to (M3) follow quite readily. Let us look at proving (M4).

Observing the form of \(\bar{d}(x,y)\), let us assume the function \(f(t)=\displaystyle\frac{t}{1+t}\). Differentiating with respect to \(t\), we get:

\[\frac{df(t)}{dt}=\frac{1}{1+t} - \frac{t}{ {(1+t)}^2} \\ = \frac{1}{ {(1+t)}^2}\]

This shows that \(f(t)\) is monotonically increasing. A function \(f(x)\) is monotonically increasing if for \(x_1 \leq x_2\), we have \(f(x_1) \leq f(x_2)\). Since our \(f(t)\) is monotonically increasing, we can write that for \(t_1 \leq t_2\):

\[\begin{equation} f(t_1) \leq f(t_2) \\ \Rightarrow \frac{t_1}{1+t_1} \leq \frac{t_2}{1+t_2} \label{eq:1} \end{equation}\]

Set \(t_1=d(x,y)\) and \(t_2=d(x,z) + d(z,y)\). We can immediately see that \(t_1 \leq t_2\). Thus substituting these values into \(\eqref{eq:1}\), we get:

\[\begin{equation} \frac{d(x,y)}{1+d(x,y)} \leq \frac{d(x,z) + d(z,y)}{1+d(x,z) + d(z,y)} \\ = \frac{d(x,z)}{1+d(x,z) + d(z,y)} + \frac{d(z,y)}{1+d(x,z) + d(z,y)} \label{eq:2} \end{equation}\]

We see that:

\[\displaystyle\frac{d(x,z)}{1+d(x,z) + d(z,y)} \leq \frac{d(x,z)}{1+d(x,z)} \\ \displaystyle\frac{d(z,y)}{1+d(x,z) + d(z,y)} \leq \frac{d(z,y)}{1+d(z,y)}\]

Thus we can rewrite the above inequality in \(\eqref{eq:2}\) as:

\[\frac{d(x,y)}{1+d(x,y)} \leq \frac{d(x,z)}{1+d(x,z)} + \frac{d(z,y)}{1+d(z,y)} \\ \Rightarrow \bar{d(x,y)} \leq \bar{d}(x,z) + \bar{d}(z,y)\]

thus, proving the Triangle Inequality.

This is the central idea in proving that the distance metric in a sequence space of all bounded and unbounded complex numbers (Kreyszig 1.2-1) has a metric defined by:

\[d(x,y)=\displaystyle\sum_{j=1}^\infty \frac{1}{2^j} \frac{\vert \zeta_j - \eta_j\vert}{1 + \vert \zeta_j - \eta_j\vert}\]

where \(x=(\eta_j)\) and \(y=(\zeta_j)\).


tags: Mathematics - Proof - Functional Analysis - Pure Mathematics - Kreyszig