For 1≤p<∞, let us consider the following set of sequences in the linear space K, denoted by lp:
lp = {(x1, x2, … ): xi ∈ K, $\sum_{i=1}^\infty$ |xi|p<∞}.
For x = (x1, x2, …) and y = (y1, y2, …) ∈ lp, define
dp(x, y) = ($\sum_{i=1}^\infty$ |xi – yi|p)1/p.
Next, we prove that dp is a metric on the space lp.
Table of Contents
lp is a Metric Space
Theorem: (lp, dp) is a metric space.
Proof:
Step 1: To show dp(x, y) is finite.
In the Minkowski’s inequality, let ai = xi and bi = -yi, and then taking n→∞ we deduce that dp(x, y) < ∞ for all x, y ∈ lp.
Step 2: The first two conditions of the definition of a metric space are easily satisfied.
Step 3: Check triangle inequality.
Let x, y, z ∈ lp.
For i=1, 2, … we set ai = xi-zi and bi = zi-yi in the Minkowski inequality. Letting n→∞ we get that dp(x, y) ≤ dp(x, z) + dp(z, y), so the triangle inequality is satisfied.
As dp satisfies the three conditions of being a metric, we conclude that lp is a metric space with respect to the metric dp.
l∞ Space
Now, we consider the set of all bounded sequences in K, denoted by l∞.
l∞ = {(x1, x2, … ): xi ∈ K, supi=1,2,… |xi|<∞}.
For x = (x1, x2, …) and y = (y1, y2, …) ∈ l∞, define
d∞(x, y) = supi=1,2,… |xi – yi|.
It can be easily verified that d∞ is a metric on the space l∞, that is l∞ is a metric space with respect to the metric d∞.
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Metric space: definition, examples, properties
Metric space of bounded functions
Let T be a set and let B(T) denote the set of all K-valued bounded functions defined on T. The set B(T) is given below.
B(T) = {x: T → K, supt∈T |x(t)|<∞}
Note that l∞ is a special case of B(T). For x, y ∈ B(T), define
d∞(x, y) = supt∈T |x(t) – y(t)|.
The above function d∞ is a metric on B(T), that is (B(T), d∞) is a metric space. This metric is knnown as sup metric.
FAQs
Answer: Yes, lp is a metric space.
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