
# Chapter 2 - Metric Spaces

## 2.5 Basic Concepts

 Main: Index Previous: 1.4 Systems of Sets Next: 2.6 Convergence. Open and Closed Sets

### Results

Definition 1

A metric space is a pair $(X, \rho)$ consisting of a set $X$ and a distance $\rho$, a single-valued, nonnegative, real function $\rho(x,y)$ defined for all $x,y\in X$ which has the following three properties: \begin{align} &\text{1)}\;\; \rho(x, y) = 0 \;\Longleftrightarrow\; x=y \\&\\ &\text{2) Symmetry:}\;\; \rho(x, y) = \rho(y, x) \\&\\ &\text{3) Triangle inequality:}\;\; \rho(x, z) \leq\rho(x, y) + \rho(y, z) \end{align}

Definition 2

A one-to-one mapping $f$ from one metric space $(X, \rho)$ to another metric space $(Y, \rho')$ is said to be an isometric mapping or an isometry if $$\rho(x_1, x_2) = \rho'(f(x_1), f(x_2))$$ for all $x_1,x_2\in X$. Correspondingly, the spaces are said to be isometric to each other.

### Verification 1

Working in $\R^n$ where the points are $n$-tuples $x = (x_1,\ldots,x_n)$ and $y = (y_1,\ldots,y_n)$. The distance defined by the formula: $$\rho_0(x, y) = \max_{1\leq k\leq n}|x_k - y_k|,$$ is a metric.

Proof.
Verifying that the metric satisfies the three defining properties.

(1). If $y = x$ then, $$\rho_0(x, x) = \max_{1\leq k\leq n}|x_k - x_k| = \max_{1\leq k\leq n}|0| = 0.$$ If $\rho_0(x, y) = 0$ then, $$0\leq \max_{1\leq k\leq n}|x_k - y_k| = 0 \;\Rightarrow\; |x_k - y_k| = 0,\,\forall k \;\Rightarrow\; x_k - y_k = 0,\,\forall k \;\Rightarrow\; x_k = y_k,\,\forall k \;\Rightarrow\; x = y.$$

(2) Verifying symmetry, $$\rho_0(x, y) = \max_{1\leq k\leq n}|x_k - y_k| = \max_{1\leq k\leq n}|y_k - x_k| = \rho_0(y, x)$$

(3) Verifying the triangle inequality with some point $z = (z_1,\ldots, z_n)$. $$\rho_0(x, y) = \max_{1\leq k\leq n}|x_k - y_k| = \max_{1\leq k\leq n}|x_k - z_k + z_k - y_k|$$ $$\leq \max_{1\leq k\leq n}|x_k - z_k| + \max_{1\leq k\leq n}|z_k - y_k| = \rho_0(x, z) + \rho_0(z, y).$$

### Verification 2

The set (or function space) $C_{[a,b]}$ of all continuous functions defined on the closed interval $[a,b]$, with distance: $$\rho(f, g) = \max_{a\leq t\leq b}|f(t) - g(t)|,$$ is a metric.

Proof.
Verifying that the metric satisfies the three defining properties.

(1). If $g = f$ then, $$\rho(f, f) = \max_{a\leq t\leq b}|f(t) - f(t)| = \max_{a\leq t\leq b}|0| = 0$$ If $\rho(f, g) = 0$ then, $$0\leq \max_{a\leq t\leq b}|f(t) - g(t)| = 0 \;\Rightarrow\; |f(t) - g(t)| = 0,\,\forall t\in[a,b] \;\Rightarrow\; f(t) - g(t) = 0,\,\forall t\in[a,b]$$ $$\;\Rightarrow\; f(t) = g(t) ,\,\forall t\in[a,b] \;\Rightarrow\; f(t) = g(t).$$

(2) Verifying symmetry, $$\rho(f, g) = \max_{a\leq t\leq b}|f(t) - g(t)| = \max_{a\leq t\leq b}|g(t) - f(t)| = \rho(g, f)$$

(3) Verifying the triangle inequality with some function $h(t)$. $$\rho(f, g) = \max_{a\leq t\leq b}|f(t) - g(t)| = \max_{a\leq t\leq b}|f(t) - h(t) + h(t) - g(t)|$$ $$\leq \max_{a\leq t\leq b}|f(t) - h(t)| + \max_{a\leq t\leq b}|h(t) - g(t)| = \rho(f, h) + \rho(h, g)$$

### Problem 1

Given a metric space $(X, \rho)$, prove that:

(a) $|\rho(x,z) - \rho(y, u)|\leq \rho(x,y) + \rho(z, u)$ for $x,y,z,u\in X$.

Proof.
Verifying with the following steps. Assuming $\rho(x,z) - \rho(y, u) \geq 0$ which means $\rho(x,z) - \rho(y, u) = |\rho(x,z) - \rho(y, u)|$. \begin{align} \rho(x, z) &\leq \rho(x,y) + \rho(y, z) \\&\\ &\leq \rho(x,y) + \rho(y,u) + \rho(u, z) \\&\\ &= \rho(x,y) + \rho(y,u) + \rho(z, u) \end{align} Using the symmetric property on the last term. Subtracting gives us: $$\rho(x, z) - \rho(y,u) \leq \rho(x,y) + \rho(z, u)$$ And since the left hand side is positive; $$|\rho(x, z) - \rho(y,u)| \leq \rho(x,y) + \rho(z, u).$$ Now assume $\rho(x,z) - \rho(y, u) < 0$, so $\rho(y, u) - \rho(x,z) \geq 0$ which means: $$\rho(y, u) - \rho(x, z) = |\rho(y, u) - \rho(x, z)| = |\rho(x,z) - \rho(y, u)|.$$ Now we start by applying the triangle inequality to $\rho(y,u)$. \begin{align} \rho(y, u) &\leq \rho(y,x) + \rho(x, u) \\&\\ &\leq \rho(y,x) + \rho(x, z) + \rho(z, u) \end{align} Subtracting: $$\rho(y, u) - \rho(x, z) \leq \rho(y,x) + \rho(z, u)$$ From our assumption we can conclude that: $$|\rho(y, u) - \rho(x, z)| \leq \rho(y,x) + \rho(z, u).$$ This holds true in both cases which are exhaustive, which proves the inequality.

(b) $|\rho(x,z) - \rho(y, z)| \leq \rho(x,y)$ for $x,y,z\in X$.

Proof.
Verifying with the following steps. Start by applying the triangle inequality to $\rho(x,z)$. \begin{align} \rho(x, z) &\leq \rho(x, y) + \rho(y, z) \;\Longrightarrow \\&\\ \rho(x, z) - \rho(y, z) &\leq \rho(x, y). \end{align} We must consider two cases, first that $\rho(x, z) - \rho(y, z)\geq 0$, in which case: $$\rho(x, z) - \rho(y, z) = |\rho(x, z) - \rho(y, z)|,$$ which means that: $$\rho(x, z) - \rho(y, z) \leq \rho(x, y) \;\Longrightarrow\; |\rho(x, z) - \rho(y, z) | \leq \rho(x, y).$$ The second case is $\rho(x, z) - \rho(y, z) < 0$. Now we apply the triangle inequality to $\rho(y, z)$. \begin{align} \rho(y, z) &\leq \rho(y, x) + \rho(x, z) \;\Longrightarrow \\&\\ \rho(y, z) - \rho(x, z) &\leq \rho(x, y). \end{align} Since we assumed $\rho(x, z) - \rho(y, z) < 0$, then $\rho(y, z) - \rho(x, z)\geq 0$. Now: $$\rho(y, z) - \rho(x, z) \leq \rho(x, y) \;\Longrightarrow\; |\rho(y, z) - \rho(x, z)| \leq \rho(x, y) \;\Longrightarrow\; |\rho(x, z) - \rho(y, z)| \leq \rho(x, y)$$ In summary, we have shown that: $$|\rho(x,z) - \rho(y, z)| \leq \rho(x,y).$$

### Problem 2

Verify that: $$\left(\sum_{k=1}^n a_kb_k\right)^2 = \sum_{k=1}^n a_k^2\sum_{k=1}^n b_k^2 - \frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n (a_ib_j - b_ia_j)^2$$ and deduce the Cauchy-Schwarz inequality.

Proof.
We will verify this in the case when $n=2$. \begin{align} \left(\sum_{k=1}^2 a_kb_k\right)^2 &= (a_1b_1 + a_2b_2)(a_1b_1 + a_2b_2) \\&\\ &= a_1^2b_1^2 + a_2^2b_2^2 + 2a_1b_1a_2b_2 \\&\\ &= a_1^2b_1^2 + a_2^2b_2^2 + 0 + 2a_1b_1a_2b_2 \\&\\ &= a_1^2b_1^2 + a_2^2b_2^2 + (a_1^2b_2^2 + a_2^2b_1^2 - a_1^2b_2^2 - a_2^2b_1^2) + 2a_1b_1a_2b_2 \\&\\ &= a_1^2b_1^2 + a_2^2b_2^2 + a_1^2b_2^2 + a_2^2b_1^2 - a_1^2b_2^2 - a_2^2b_1^2 + 2a_1b_1a_2b_2 \\&\\ &= a_1^2b_1^2 + a_2^2b_2^2 + a_1^2b_2^2 + a_2^2b_1^2 - \frac{1}{2}(2a_1^2b_2^2 + 2a_2^2b_1^2 - 4a_1b_1a_2b_2) \\&\\ &= a_1^2b_1^2 + a_1^2b_2^2 + a_2^2b_2^2 + a_2^2b_1^2 - \frac{1}{2}(2a_1^2b_2^2 + 2a_2^2b_1^2 - 4a_1b_1a_2b_2) \\&\\ &= S_1 - \frac{1}{2}S_2 \end{align} We can treat each of these separately. \begin{align} S_1 &= a_1^2b_1^2 + a_1^2b_2^2 + a_2^2b_2^2 + a_2^2b_1^2 \\&\\ &= a_1^2(b_1^2 + b_2^2) + a_2^2(b_1^2 + b_2^2) \\&\\ &= a_1^2\sum_{k=1}^2b_k^2 + a_2^2\sum_{k=1}^2b_k^2 \\&\\ &= (a_1^2 + a_2^2)\left(\sum_{k=1}^2b_k^2\right) \\&\\ &= \sum_{k=1}^2a_k^2\sum_{k=1}^2b_k^2 \end{align} And for the other part: \begin{align} S_2 &= 2a_1^2b_2^2 + 2a_2^2b_1^2 - 4a_1b_1a_2b_2 \\&\\ &= a_1^2b_2^2 + a_2^2b_1^2 - 2a_1b_1a_2b_2 + a_1^2b_2^2 + a_2^2b_1^2 - 2a_1b_1a_2b_2 \\&\\ &= 0 + a_1^2b_2^2 + a_2^2b_1^2 - 2a_1b_1a_2b_2 + a_1^2b_2^2 + a_2^2b_1^2 - 2a_1b_1a_2b_2 + 0 \\&\\ &= a_1^2b_1^2 + a_1^2b_1^2 - 2a_1^2b_1^2 + a_1^2b_2^2 + a_2^2b_1^2 - 2a_1b_1a_2b_2\\ &\;\;\;\; + a_1^2b_2^2 + a_2^2b_1^2 - 2a_1b_1a_2b_2 + a_2^2b_2^2 + a_2^2b_2^2 - 2a_2^2b_2^2 \\&\\ &= \sum_{i=1}^2a_i^2b_1^2 + a_1^2b_i^2 - 2a_ib_ia_1b_1 + a_i^2b_2^2 + a_2^2b_i^2 - 2a_ib_ia_2b_2 \\&\\ &= \sum_{i=1}^2\sum_{j=1}^2a_i^2b_j^2 + a_j^2b_i^2 - 2a_ib_ia_jb_j \\&\\ &= \sum_{i=1}^2\sum_{j=1}^2(a_ib_j - a_jb_i)^2 \end{align} Combining these steps together, we have shown that: $$\left(\sum_{k=1}^n a_kb_k\right)^2 = \sum_{k=1}^n a_k^2\sum_{k=1}^n b_k^2 - \frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n (a_ib_j - b_ia_j)^2$$ for $n=2$. The same idea can be expanded to general $n\in\N$. Now, on this form we can see that the squared term $(a_ib_j - b_ia_j)^2\geq 0$, so the second term will always be negative. From this we can derive the Cauchy-Schwarz inequality: $$\left(\sum_{k=1}^n a_kb_k\right)^2 = \sum_{k=1}^n a_k^2\sum_{k=1}^n b_k^2 - \frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n (a_ib_j - b_ia_j)^2 \leq \sum_{k=1}^n a_k^2\sum_{k=1}^n b_k^2.$$ Expressed directly: $$\left(\sum_{k=1}^n a_kb_k\right)^2 \leq \sum_{k=1}^n a_k^2\sum_{k=1}^n b_k^2.$$

### Problem 3

Verify that $$\left(\int_a^b x(t)y(t)dt\right)^2 = \int_a^b x^2(t)dt\int_a^b y^2(t)dt - \frac{1}{2}\int_a^b\int_a^b\big[x(s)y(t) - y(s)x(t)\big]^2dsdt$$ and deduce Schwarz's inequality: $$\left(\int_a^b x(t)y(t)dt\right)^2 \leq \int_a^b x^2(t)dt\int_a^b y^2(t)dt$$

Instead of solving this problem as laid out in the book, we will use a different, direct method. We will use the following Theorem not mentioned in the book.

Theorem - Sign of quadratic functions between roots.

Let $a\in\R$ be a strictly positive number, and let $\alpha,\beta$ where $\alpha < \beta$ be roots of the quadratic function: $$Q(x) = ax^2 + bx + c$$ whose discriminant $b^2 - 4ac > 0$. Then: $$\left\{ \begin{matrix} Q(x) < 0 & \alpha < x < \beta \\ Q(x) > 0 & x < \alpha\;\text{or}\; x > \beta \end{matrix} \right.$$ Comment: Stated without proof. The important part is that a positive discriminant implies that the polynomial will have negative values.

Proof.
The functions $f(t)$ and $g(t)$ are defined on the closed interval $[a,b]$. For any real numbers, $0 \leq (xf(t) + g(t))^2$. Since this is always positive: \begin{align} 0&\leq \int_a^b(xf(t) + g(t))^2dt \\&\\ &= \int_a^b(x^2f^2(t) + 2xf(t)g(t) + g^2(t)dt \\&\\ &= x^2\int_a^b(f(t))^2dt + 2x\int_a^b f(t)g(t)dt + \int_a^b (g(t))^2dt \\&\\ &= Ax^2 + Bx + C. \end{align} We define the polynomial $P(x) = Ax^2 + Bx + C$, where $$A = \int_a^b(f(t))^2dt$$ $$B = 2\int_a^b f(t)g(t)dt$$ $$C = \int_a^b (g(t))^2dt.$$ and from the definition $P(x)\geq 0$ for all $x\in\R$. Assume now for contradiction that the discriminant $B^2 - 4AC > 0$. Then there will be two roots such that $P(x) > 0$ and $P(x) < 0$ for certain values. But since $P(x) \geq 0$, this is a contradiction. Hence $B^2 - 4AC\leq 0$ which implies $B^2\leq 4AC$ and that gives us: $$\left(2\int_a^b f(t)g(t)dt\right)^2 \leq 4\left(\int_a^b(f(t))^2dt\right)\left(\int_a^b (g(t))^2dt\right)$$ $$\left(\int_a^b f(t)g(t)dt\right)^2 \leq \int_a^b(f(t))^2dt\int_a^b (g(t))^2dt$$ which is the Schwarz inequality.

### Problem 4

Working with $n$-tuples, $x = (x_1,\ldots, x_n)$ and $y = (y_1,\ldots, y_n)$ and the distance: $$\rho_p(x, y) = \left(\sum_{k=1}^n|x_k - y_k|^p\right)^{1/p}.$$ Show that Minkowski's inequality fails for $p < 1$ (and because of that so does the triangle inequality). Minkowski's inequality: $$\left(\sum_{k=1}^n|a_k + b_k|^pdt\right)^{1/p} \leq \left(\sum_{k=1}^n|a_k|^p\right)^{1/p} + \left(\sum_{k=1}^n|b_k|^p\right)^{1/p}$$

Proof.
Will show a counterexample using $n=2$ and $p=1/2$. Let $x = (1, 2)$, $y = (3, 5)$ and $z = (6, 6)$. We define: $$a = x - y = (-2, -3), \quad b = y - z = (-3, -1)$$

First showing that it works for $p=1$. We can calculate the left hand side: $$\left(\sum_{k=1}^n|a_k + b_k|^pdt\right)^{1/p} = (|-2 - 3| + |-3 - 1|) = |-5| + |-4| = 5 + 4 = 9$$ Now, each of the sums on the right hand side: $$\left(\sum_{k=1}^n|a_k|^p\right)^{1/p} = |-2| + |-3| = 2 + 3 = 5$$ $$\left(\sum_{k=1}^n|b_k|^p\right)^{1/p} = |-3| + |-1| = 3 + 1 = 4$$ In this case the right hand side sums to 9 which gives us equality.

Now, we set $p=1/2$, so $1/p = 2$. We can calculate the left hand side: $$\left(\sum_{k=1}^n|a_k + b_k|^pdt\right)^{1/p} = \left(\sqrt{|-2 - 3|} + \sqrt{|-3 - 1|}\right)^2 = \left(\sqrt{|-5|} + \sqrt{|-4|}\right)^2 = (\sqrt{5} + 2)^2 \approx 17.944$$ Now, each of the sums on the right hand side: $$\left(\sum_{k=1}^n|a_k|^p\right)^{1/p} = \left(\sqrt{|-2|} + \sqrt{|-3|}\right)^2 = (\sqrt{2} + \sqrt{3})^2 \approx 9.898$$ $$\left(\sum_{k=1}^n|b_k|^p\right)^{1/p} = \left(\sqrt{|-3|} + \sqrt{|-1|}\right)^2 = (\sqrt{3} + 1)^2 \approx 7.464$$ Now the right hand sums up to 17.363 < 17.944, which shows that Minkowski's inequality fails for $p = 1/2$. (In fact, for $p < 1$ the reverse inequality holds and can be proved).

### Problem 5

Prove that the following metric: $$\rho_0(x, y) = \max_{1\leq k\leq n}|x_k - y_k|$$ is the limiting case of this metric: $$\rho_p(x, y) = \left(\sum_{k=1}^n|x_k - y_k|^p\right)^{1/p}$$ in the sense that: $$\rho_0(x, y) = \max_{1\leq k\leq n}|x_k - y_k| = \lim_{p\rightarrow\infty}\left(\sum_{k=1}^n|x_k - y_k|^p\right)^{1/p}$$

Proof.
Let $1\leq M\leq n$ be the index such that $|x_M-y_M|$ is the largest value. Calculating the limit: \begin{align} \lim_{p\rightarrow\infty}\left(\sum_{k=1}^n|x_k - y_k|^p\right)^{1/p} &= \lim_{p\rightarrow\infty}\left(|x_1-y_1|^p+|x_2-y_2|^p+\ldots+|x_n-y_n|^p\right)^{1/p} \\&\\ &= \lim_{p\rightarrow\infty}\left(|x_M-y_M|^p\bigg[\frac{|x_1-y_1|^p}{|x_M-y_M|^p}+\frac{|x_2-y_2|^p}{|x_M-y_M|^p} +\ldots+\frac{|x_n-y_n|^p}{|x_M-y_M|^p}\bigg]\right)^{1/p} \\&\\ &= \lim_{p\rightarrow\infty}\left(|x_M-y_M|^p\bigg[\left(\frac{|x_1-y_1|}{|x_M-y_M|}\right)^p +\left(\frac{|x_2-y_2|}{|x_M-y_M|}\right)^p +\ldots + \left(\frac{|x_n-y_n|}{|x_M-y_M|}\right)^p\bigg]\right)^{1/p} \\&\\ &= \lim_{p\rightarrow\infty}\left(|x_M-y_M|^p\left[ \alpha_1^p +\alpha_2^p +\ldots + 1 + \ldots + \alpha_n^p\right]\right)^{1/p} \\&\\ &= \lim_{p\rightarrow\infty}\left(|x_M-y_M|^p\right)^{1/p} \cdot \lim_{p\rightarrow\infty}\left( \alpha_1^p +\alpha_2^p +\ldots + 1 + \ldots + \alpha_n^p\right)^{1/p} \\&\\ &= |x_M-y_M|\cdot 1 \\&\\ &= |x_M-y_M| \end{align} which can also be expressed as $\max_{1\leq k\leq n}|x_k - y_k|$. (All the terms $\alpha^p_i < 1$ and so tend to 0 as $p\rightarrow\infty$).

### Problem 6

Starting from Young's inequality: $$ab\leq \frac{a^p}{p} + \frac{b^q}{q},$$ which applies to arbitrary positive $a$ and $b$ whenever $p>1$ and $q>1$ and $$\frac{1}{p} + \frac{1}{q} = 1,$$ deduce Hölder's integral inequality: $$\int_a^b x(t)y(t)dt \leq \left(\int_a^b|x(t)|^pdt\right)^{1/p}\left(\int_a^b|y(t)|^qdt\right)^{1/q}$$ which is valid for any functions $x(t)$ and $y(t)$ such that the integrals on the right exist.

Proof.
Define: $$a = \frac{|x(t)|}{\left(\int_a^b |x(t)|^p dt\right)^{1/p}}, \quad b = \frac{|y(t)|}{\left(\int_a^b |y(t)|^q dt\right)^{1/q}}$$ which are both positive values. Applying Young's inequality: $$ab = \frac{|x(t)y(t)|}{\left(\int_a^b |x(t)|^p dt\right)^{1/p}\left(\int_a^b |y(t)|^q dt\right)^{1/q}} \leq \frac{|x(t)|^p}{p\int_a^b |x(t)|^p dt} + \frac{|y(t)|^q}{q\int_a^b |y(t)|^q dt}$$ Using a property not mentioned in the book, but for positive, measurable functions, integrals are monotone, i.e. they preserve inequalities. We take the integral on both sides, and note that the denominators are constants with regard to the integrals, so we get: $$\frac{\int_a^b|x(t)y(t)|dt}{\left(\int_a^b |x(t)|^p dt\right)^{1/p}\left(\int_a^b |y(t)|^q dt\right)^{1/q}} \leq \frac{\int_a^b|x(t)|^pdt}{p\int_a^b |x(t)|^p dt} + \frac{\int_a^b|y(t)|^qdt}{q\int_a^b |y(t)|^q dt} = \frac{1}{p} + \frac{1}{q} = 1$$ From this inequality, we get: $$\int_a^b|x(t)y(t)|dt \leq \left(\int_a^b |x(t)|^p dt\right)^{1/p}\left(\int_a^b |y(t)|^q dt\right)^{1/q}$$ Since $x(t)y(t)\leq|x(t)y(t)|$ and by the monotoncity of integrals, we get Hölder's inequality as stated in the book: $$\int_a^b x(t)y(t) dt \leq \left(\int_a^b |x(t)|^p dt\right)^{1/p}\left(\int_a^b |y(t)|^q dt\right)^{1/q}$$ which holds when: $$\frac{1}{p} + \frac{1}{q} = 1.$$

### Problem 7

Use Hölder's integral inequality to prove Minkowski's integral inequality. $$\left(\int_a^b|x(t) + y(t)|^pdt\right)^{1/p} \leq \left(\int_a^b|x(t)|^p\right)^{1/p} + \left(\int_a^b|y(t)|^p\right)^{1/p},\quad (p\geq 1).$$

Proof.
For Hölder's inequality, note that if we fix $1/p$, then: $$\frac{1}{q} = 1 - \frac{1}{p} \;\;\Longrightarrow\;\; q = \frac{p}{p-1}.$$ Taking the $p$-th power of the integral. \begin{align} I^p = \int_a^b|x(t) + y(t)|^pdt &= \int_a^b|x(t) + y(t)||x(t) + y(t)|^{p-1}dt \\&\\ &\leq \int_a^b\Big(|x(t)| + |y(t)|\Big)|x(t) + y(t)|^{p-1}dt \\&\\ &= \int_a^b |x(t)||x(t) + y(t)|^{p-1}dt + \int_a^b |y(t)||x(t) + y(t)|^{p-1}dt \\&\\ &\leq \left(\int_a^b|x(t)|^pdt\right)^{1/p} \left(\int_a^b|x(t) + y(t)|^p\right)^{1-1/p} \tag{Hölder's ineq.} \\ &\;\;\;+ \left(\int_a^b|y(t)|^pdt\right)^{1/p} \left(\int_a^b|x(t) + y(t)|^p\right)^{1-1/p} \end{align} When applying Hölder's inequality, we get $$\left(|x(t) + y(t)|^{p-1}\right)^q = \left(|x(t) + y(t)|^{p-1}\right)^{\frac{p}{p-1}} = |x(t) + y(t)|^p$$ We have shown: $$\int_a^b|x(t) + y(t)|^pdt \leq \left(\int_a^b|x(t) + y(t)|^p\right)^{1-1/p} \Bigg( \left(\int_a^b|x(t)|^pdt\right)^{1/p} + \left(\int_a^b|y(t)|^pdt\right)^{1/p} \Bigg)$$ Rewriting the integral as a fraction. $$\int_a^b|x(t) + y(t)|^pdt \leq \frac{\int_a^b|x(t) + y(t)|^pdt}{\left(\int_a^b|x(t) + y(t)|^p\right)^{1/p}} \Bigg( \left(\int_a^b|x(t)|^pdt\right)^{1/p} + \left(\int_a^b|y(t)|^pdt\right)^{1/p} \Bigg)$$ From this, Minkowski's inequality follows immediately. $$\left(\int_a^b|x(t) + y(t)|^pdt\right)^{1/p} \leq \left(\int_a^b|x(t)|^p\right)^{1/p} + \left(\int_a^b|y(t)|^p\right)^{1/p}$$

### Problem 8

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

Recall that the metric for function spaces is: $$\rho_A(h,g) = \max_{t\in[0,1]}|h(t) - g(t)|$$ $$\rho_B(h,g) = \max_{t\in[1,2]}|h(t) - g(t)|$$ We need to find some function $f$ such that $$\rho_A(h,g) = \rho_B(f(h),f(g))$$ for all functions $h,g\in C_{[1,2]}$. If we define: $$f(k(t)) = k(t-1),$$ for any continuous function $k(t)$, we map all function values from $C_{[1,2]}$ back to $C_{[0,1]}$ and get an isometry.