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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