Metric Spaces: Tutorial problems: Solutions: Part I

Reference: Introductory Real Analysis by Kolmogorov and Fomin.

Reference: Analysis by Walter Rudin.

Reference: Introduction to Analysis by Rosenlicht.

Reference: Topology by Hocking and Young.

The problems proposed were as follows:

  1. Given a metric space (X,\rho), prove that (1a) |\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u) where x,y,z,u \in X; (1b) |\rho(x,z) -\rho(y,z)| \leq \rho(x,y) where x,y,z \in X.
  2. Verify that (\Sigma_{k=1}^{n}a_{k}b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{n}\Sigma_{j=1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}. Deduce the Cauchy-Schwarz inequality from the above identity. (refer second last blog from here). For convenience, we reproduce it here again: consider the set of all ordered n-tuples such that x=(x_{1}, x_{2}, \ldots, x_{n}) and y=(y_{1}, y_{2}, \ldots, y_{n}) and z=(z_{1}, z_{2}, \ldots, z_{n}) and let a_{k}=x_{k}-y_{k}, b_{k}=y_{k}-z_{k} where k=1,2, \ldots, n: then Cauchy Schwarz inequality tells us that : (\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}\Sigma_{k=1}^{n}b_{k}^{2}
  3. Verify that (\int_{a}^{b}x(t)y(t)dt)^{2} = \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y^{2}(t)dt -\frac{1}{2}\int_{a}^{b}\int_{a}^{b}(x(s)y(t)-x(t)y(s))^{2}dsdt.

Deduce Schwarz’s inequality from this identity. (refer second last blog from here). To aid the memory, we present Schwarz’s inequality below: (\int_{a}^{b}x(t)y(t))^{2} \leq \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y^{2}(t)dt

4. Identify the fallacy in example 10, if p \leq 1? (refer second last blog from here). Hint: Show that Minkowski’s inequality fails for p \leq 1. To help recall, we are reproducing Minkowski’s inequality below: Consider again the set of all ordered n-tuples, and let x=(x_{1}, x_{2}, \ldots, x_{n}), and y=(y_{1}, y_{2}, \ldots, y_{n}) and z=(z_{1}, z_{2}, \ldots, z_{n}) and again further let a_{k}=x_{k}-y_{k} and b_{k}=y_{k}-z_{k}, where k=1,2, \ldots, n then Minkowski’s inequality tells us that: (\Sigma_{k=1}^{n}|a_{k}+b_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{n}|a_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{n}|b_{k}|^{p})^{\frac{1}{p}}

5. Prove that the following metric : Consider two points x=(x_{1}, x_{2}, \ldots, x_{n}) and y=(y_{1}, y_{2}, \ldots, y_{n}) with the metric function: \rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}|

is the limiting case of the following metric: consider the set of all ordered n-tuples x = (x_{1}, x_{2}, \ldots, x_{n}) of real numbers with the metric function now defined by: \rho_{p}(x,y) = (\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}

in the sense that

\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}| = \lim_{p \rightarrow \infty}(\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}

6. Starting from the following inequality: ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}, deduce Holder’s integral inequality as given by:

\int_{a}^{b} x(t)y(t)dt \leq (\int_{a}^{b}|x(t)|^{p}dt)^{\frac{1}{p}}(\int_{a}^{b}|y(t)|^{q})^{\frac{1}{q}} where \frac{1}{p} + \frac{1}{q} =1 valid for any function x(t) and y(t) such that the integrals on the right exist.

7.  Use Holder’s integral inequality to prove Minkowski’s integral inequality:

(\int_{a}^{b}|x(t)+y(t)|^{p}dt)^{\frac{1}{p}} \leq (\int_{a}^{b}|x(t)|^{p}dt)^{\frac{1}{p}} + (\int_{a}^{b}|y(t)|^{p}dt)^{\frac{1}{p}}

where p \geq 1.

8. Exhibit an isometry between the spaces C_{[0,1]} and C_{[1,2]}.

9. Verify that the following is a metric space: all bounded infinite sequences x=(x_{1}, x_{2}, x_{3}, \ldots) of elements of \Re, with \rho(x,y) = lub \{ |x_{1}-y_{1}|,|x_{2}-y_{2}|,|x_{3}-y_{3}|, \ldots\}

10. Verify that (\Re \times \Re, \rho) is a metric space where \rho((x,y),(x^{'}, y^{'})) = |y|+|y^{'}|+|x-x^{'}| when x \neq x^{'} and is |y-y^{'}| when x=x^{'}. Illustrate by diagrams in the plane E^{2} or R^{2} what the open balls of this metric space are.

11. Show that a one-to-one transformation f: S \rightarrow T of a space S onto a space T is a homeomorphism if and only if both f and f^{'} are continuous.

The solutions are as follows:

Problem 1:

  1. Given a metric space (X,\rho), prove that (1a) |\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u) where x,y,z,u \in X; (1b) |\rho(x,z) -\rho(y,z)| \leq \rho(x,y) where x,y,z \in X.

Solution 1:

Part I: TPT: |\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)

Proof: We know that: \rho(x,z) \leq \rho(x,u)+\rho(z,u), which also means that, \rho(x,z) \leq \rho(x,y) + \rho(y,u) + \rho(z,u), that is, \rho(x,z)-\rho(y,u) \leq \rho(x,y)+\rho(z,u). On the other hand, we also know that \rho(y,u) \leq \rho(y,z) + \rho(z,u), which also means that, \rho(y,u) \leq \rho(x,y) + \rho(x,z) + \rho(z,u), that is, -\rho(x,z) + \rho(y,u) \leq \rho(x,y) + \rho(z,u). Combining the two smallish results, we get |\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u). QED.

Part II: TPT: |\rho(x,z)-\rho(y,z)| \leq \rho(x,y). Proof : We know that \rho(x,z) \leq \rho(x,y)+\rho(y,z), that is, \rho(x,z) - \rho(y,z) \leq \rho(x,y). On the other hand, we know that \rho(y,z) \leq \rho(y,z), that is, \rho(y,z) \leq \rho(x,y) +\rho(x,z), that is, -\rho(x,z)+\rho(y,z) \leq \rho(x,y). Combining the two smallish results, we get |\rho(x,z) -\rho(y,z)| \leq \rho(x,y) where x,y,z \in X. QED.

PS: Now that the detailed proof is presented, try to interpret the above two results “geometrically.”

Problem 2 was:

Verify that (\Sigma_{k=1}^{n}a_{k}b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{n}\Sigma_{j=1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}. Deduce the Cauchy-Schwarz inequality from the above identity. (refer second last blog from here). For convenience, we reproduce it here again: consider the set of all ordered n-tuples such that x=(x_{1}, x_{2}, \ldots, x_{n}) and y=(y_{1}, y_{2}, \ldots, y_{n}) and z=(z_{1}, z_{2}, \ldots, z_{n}) and let a_{k}=x_{k}-y_{k}, b_{k}=y_{k}-z_{k} where k=1,2, \ldots, n: then Cauchy Schwarz inequality tells us that : (\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}\Sigma_{k=1}^{n}b_{k}^{2}.

Proof of problem 2: We present a proof based on the principle of mathematical induction. Let us check the proposition for n=1: Then, LHS=(a_{1}b_{1})^{2}=a_{1}^{2}b_{1}^{2} = RHS = a_{1}^{2}b_{1}^{2}-\frac{1}{2} \times 0. So now let the proposition be true for m=n, m \in N, m >1. Then, the following holds true:

(\Sigma_{k=1}^{m}a_{k}b_{k})^{2} = \Sigma_{k=1}^{m}a_{k}^{2}\Sigma_{k=1}^{m}b_{k}^{2} -\frac{1}{2}\Sigma_{i=1}^{m}\Sigma_{j=1}^{m}(a_{i}b_{j}-a_{j}b_{i})^{2}.

Now, we want to prove the proposition for n=m+1, that is, we need to prove that:

(\Sigma_{k=1}^{m+1}a_{k}b_{k})^{2}=\Sigma_{k=1}^{m+1}a_{k}^{2}\Sigma_{k=1}^{m+1}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{m+1}\Sigma_{j=1}^{m+1}(a_{i}b_{j}-a_{j}b_{i})^{2}.

Now, let us see what we need really; we need to add some(same) terms to both sides of the proposition for n=m and manipulate to derive the proposition for n=m+1:

Further notice that

(\Sigma_{k=1}^{m+1}a_{k}b_{k})^{2}=(a_{1}b_{1}+a_{2}b_{2}+ \ldots+a_{m}b_{m}+a_{m+1}b_{m+1})^{2}=\Sigma_{k=1}^{m}a_{k}^{2}b_{k}^{2}+a_{m+1}^{2}b_{m+1}^{2}+2(a_{1}b_{1})(a_{2}b_{2}+a_{3}b_{3}+\ldots+a_{m}b_{m}+a_{m+1}b_{m+1})+2(a_{2}b_{2})(a_{3}b_{3}+a_{4}b_{4}+\ldots+a_{m}b_{m}+a_{m+1}b_{m+1}) + 2a_{3}b_{3}(a_{4}b_{4}+a_{5}b_{5}+ \ldots + a_{m+1}b_{m=1})+ \ldots + 2a_{m}b_{m}a_{m+1}b_{m+1}

Comparing the algebraic statements for proposition for n=m and n=m+1, we can clearly see which terms are to be added to both sides of the proposition for n=m,

(please fill in the few missing last details…a good deal of algebraic manipulation…some concentrated effort …:-))

From this the Cauchy Schwarz inequality follows naturally.

PS: I have no idea how to solve question 3. If any reader helps, I would be obliged.

PS: I have yet to try the other questions.

Cheers,

Nalin Pithwa

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