To show that d is a metric, we need to show that it satisfies the following properties:
1. Non-negativity: d(f,g) >= 0 for all f,g in X.
2. Identity of indiscernibles: d(f,g) = 0 if and only if f = g for all f,g in X.
3. Symmetry: d(f,g) = d(g,f) for all f,g in X.
4. Triangle inequality: d(f,g) + d(g,h) >= d(f,h) for all f,g,h in X.
Let's go through each property to show that d satisfies them:
1. Non-negativity:
Since the supremum of absolute values is always non-negative, it follows that d(f,g) = sup|x in X|f(x) - g(x)| >= 0.
2. Identity of indiscernibles:
If d(f,g) = 0, then sup|x in X|f(x) - g(x)| = 0 which means that for every ε > 0, there exists an x in X such that |f(x) - g(x)| < ε. This implies that f(x) = g(x) for all x in X, so f = g.
Conversely, if f = g, then for every x in X, |f(x) - g(x)| = 0. Therefore, sup|x in X|f(x) - g(x)| = 0, so d(f,g) = 0.
3. Symmetry:
For any f,g in X, d(f,g) = sup|x in X|f(x) - g(x)| = sup|x in X|g(x) - f(x)| = d(g,f).
4. Triangle inequality:
Let f,g,h be in X. We want to show that d(f,g) + d(g,h) >= d(f,h).
By the definition of suprema, for every ε > 0, there exists an x1 in X such that |f(x1) - g(x1)| < d(f,g) + ε, and there exists an x2 in X such that |g(x2) - h(x2)| < d(g,h) + ε.
Now consider the expression |f(x1) - g(x1) + g(x2) - h(x2)|. We can split this into |f(x1) - g(x1) + g(x2) - h(x2) + f(x2) - f(x2)| and use triangle inequality for the last two terms:
|f(x1) - g(x1) + g(x2) - h(x2) + f(x2) - f(x2)| <= |f(x1) - f(x2)| + |f(x2) - g(x2)| + |g(x2) - h(x2)|
Since |f(x1) - f(x2)| <= sup|x in X|f(x) - g(x)| = d(f,g), |f(x2) - g(x2)| <= sup|x in X|f(x) - g(x)| = d(f,g), and |g(x2) - h(x2)| <= sup|x in X|g(x) - h(x)| = d(g,h), we have:
|f(x1) - g(x1) + g(x2) - h(x2) + f(x2) - f(x2)| <= d(f,g) + d(f,g) + d(g,h) = 2d(f,g) + d(g,h).
Therefore, d(f,g) + d(g,h) >= 2d(f,g) + d(g,h) = d(f,g) + d(g,h), as desired.
Since d satisfies all the properties of a metric, we can conclude that d is indeed a metric.
Let B(x) be the set of all bounded functions defined on X defined by d(f,g)=sup x belong to X |f(x)-g(x)|, for all f, g that belong to X. Show that d is a metric
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