Mathematical Immaturity

1.13 Exercises

• Let $V$ consist of all infinite sequences $\{x_n\}$ of real numbers for which the series $\sum x_n^2$ converges. If $x = \{x_n\}$ and $y = \{y_n\}$ are two elements of $V,$ define $$(x, y) = \sum_{n=1}^\infty x_n y_n.$$
  $\text{(a)}\quad$ Prove that this series converges absolutely.
  $\quad$[Hint: Use the Cauchy-Schwarz inequality to estimate the sum $\sum_{n=1}^M |x_n y_n|.$]
  $\text{(b)}\quad$ Prove that $V$ is a linear space with $(x, y)$ as an inner product.
  $\text{(c)}\quad$ Compute $(x, y)$ if $x_n = 1/n$ and $y_n = 1/(n + 1)$ for $n \geq 1.$
  $\text{(d)}\quad$ Compute $(x, y)$ if $x_n = 2^n$ and $y_n = 1/n!$ for $n \geq 1.$

• $\text{(a)} \quad$ Applying the Cauchy-Schwarz inequality to the sum $\sum_{n=1}^M |x_ny_n|$ we have \begin{align*} \\ \left(\sum_{n=1}^M |x_ny_n|\right)^2 &= \left(\sum_{n=1}^M |x_n||y_n|\right)^2 \\ &\leq \left(\sum_{n=1}^M |x_n|^2\right)\left(\sum_{n=1}^M |y_n|^2\right) \\ &= \left(\sum_{n=1}^M x_n^2\right)\left(\sum_{n=1}^M y_n^2\right) \end{align*} But because $\sum x_n^2$ and $\sum y_n^2$ both converge and because their partial sums are monotonically increasing for all $M \geq 1,$ we know that $\sum_{n=1}^M x^2$ and $\sum_{n=1}^M y^2$ are bounded for all $M \geq 1.$ If we denote $\sum x_n^2 = S_x$ and $\sum y_n^2 = S_y,$ the above result implies that: \begin{align*} \\ \sum_{n=1}^M |x_ny_n| &\leq \sqrt{S_xS_y} \end{align*} for all $M \geq 1.$ Hence, $(x, y) = \sum_{n=1}^{\infty} x_ny_n$ is absolutely convergent. $\quad \blacksquare$
  
  $\text{(b)} \quad$ To prove that $V$ is a linear space, we must verify that $V$ satisfies the ten axioms of a linear space:
  
  Closure axioms
  Axiom 1.$\quad$ Closure Under Addition.
  For every pair of elements $x$ and $y$ in $V,$ there corresponds a unique element in $V$ called the sum of $x$ and $y,$ denoted by $x + y.$
  
  Proof.$\quad$ Let $x$ and $y$ be two elements of $V.$ If we denote by $x + y$ the infinite sequence $\{x_n + y_n\},$ then the series $\sum (x_n + y_n)^2$ must converge for $x + y$ to be an element of $V,$ thus satisfying closure under addition. We can rewrite $\sum (x_n + y_n)^2$ as follows: \begin{align*} \sum_{n = 1}^{\infty}(x_n + y_n)^2 &= \sum_{n = 1}^{\infty} \left(x_n^2 + 2x_ny_n + y_n^2\right) \\ &= \sum_{n = 1}^{\infty} x_n^2 + 2\sum_{n = 1}^{\infty}x_ny_n + \sum_{n = 1}^{\infty}y_n^2. \end{align*} By definition, the series $\sum x_n^2$ and $\sum y_n^2$ are convergent, and in part (a) we showed that the series $\sum_{n=1}^{\infty} x_ny_n$ is absolutely convergent. As such, $\sum(x_n+y_n)^2$ must also be convergent. Hence, $x + y$ is an element of $V. \quad \blacksquare$
  
  Axiom 2.$\quad$ Closure Under Multiplication.
  For every $x$ in $V$ and every real number $a$ there corresponds an element in $V$ called the product of $a$ and $x,$ denoted by $ax.$
  
  Proof.$\quad$ Let $x$ be an element of $V$ and $a$ be a real scalar. Then, denote by $ax$ the infinite sequence $\{ax_n\}.$ To satisfy closure under multiplication by real scalars, $ax$ must be an element of $V.$ In other words, the series $\sum a^2x_n^2$ must converge. But since $\sum a^2x_n^2 = a^2\sum x_n^2,$ if we set $M = \sum_{n=1}^{\infty} x^2$ then we can see that the series $\sum a^2x_n^2$ converges to $a^2M.$ Thus, $ax$ is an element of $V. \quad \blacksquare$
  
  Axioms for addition
  Axiom 3.$\quad$ Commutative Law.
  For all $x$ and $y$ in $V,$ we have $x + y = y + x.$
  Proof.$\quad$ Let $x + y =\{x_n + y_n\}$ be an element of $V.$ For each $n = 1, 2, \dots,$ $x_n$ and $y_n$ are real numbers. As such, $x_n + y_n = y_n + x_n$ for each $n,$ which in turn implies that $\{x_n + y_n\} = \{y_n + x_n\}$ and thus $x + y = y + x. \quad \blacksquare$
  
  Axiom 4.$\quad$ Associative Law.
  For all $x,$ $y,$ and $z$ in $V,$ we have $(x + y) + z = x + (y + z).$
  
  Proof.$\quad$ Let $x,$ $y,$ and $z$ be elements of $V,$ and denote $(x + y) = \{x_n + y_n\}$ and $(y + z) = \{y_n + z_n\}.$ For $n = 1, 2, \dots,$ $x_n,$ $y_n,$ and $z_n$ are real numbers, so we have \begin{align*} (x + y) + z &= \{(x_n + y_n) + z_n\} \\ &= \{x_n + (y_n + z_n)\} \\ &= x + (y + z) \quad \blacksquare \end{align*}
  
  Axiom 5.$\quad$ Existence of Zero Element.
  There is an element in $V,$ denoted by $O,$ such that \begin{align*} x + O = x \quad \text{for all $x$ in $V.$} \end{align*}
  Proof.$\quad$ We denote $O = \{0\}.$ That is, $O$ is the infinite sequence of zeros. Since $\sum 0^2 = 0,$ $O$ is a member of $V.$ Then, for any $x = \{x_n\}$ in $V,$ we have \begin{align*} x + O &= \{x_n + 0\} \\ &= \{x_n\} \\ &= x \quad \blacksquare \end{align*} Axiom 6.$\quad$ Existence of Negatives.
  For every $x$ in $V,$ the element $(-1)x$ has the property \begin{align*} x + (-1)x = O \end{align*}
  Proof.$\quad$ Let $x = \{x_n\}$ be an element of $V.$ Then, denote by $(-1)x$ the infinite sequence $\{-x_n\}.$ We can see that $\sum(-x_n)^2 = \sum x_n^2,$ and since $\sum x_n^2$ converges, so does $\sum(-x_n)^2.$ Thus, $(-1)x$ is an element of $V.$ Then, we have \begin{align*} x + (-1)x &= \{x_n - x_n\} \\ &= \{0\} \\ &= O \quad \blacksquare \end{align*}
  
  Axioms for multiplication by numbers
  Axiom 7.$\quad$ Associative Law.
  For every $x$ in $V$ and all real numbers $a$ and $b,$ we have \begin{align*} a(bx) = (ab)x. \end{align*}
  Proof.$\quad$ Let $x = \{x_n\}$ be an element of $V$ and let $a$ and $b$ be real scalars. From the proof of Axiom 2, we know that $bx = \{bx_n\}$ is also an element of $V.$ Then, applying multiplication by scalars as defined previously, we have \begin{align*} a(bx) &= \{a(bx_n)\} \\ &= \{(ab)x_n\} \\ &= (ab)x \quad \blacksquare \end{align*}
  
  Axiom 8.$\quad$ Distributive Law for Addition in $V.$
  For all $x$ and $y$ in $V$ and all real $a,$ we have \begin{align*} a(x + y) = ax + ay. \end{align*}
  Proof.$\quad$ Let $x + y = \{x_n + y_n\}$ be an element of $V$ and let $a$ be a real scalar. Then, applying multiplication by real scalars as defined previously, we have \begin{align*} a(x + y) &= \{a(x_n + y_n)\} \\ &= \{ax_n + ay_n\} \\ &= ax + ay \quad \blacksquare \end{align*}
  
  Axiom 9.$\quad$ Distributive Law for Addition of Numbers.
  For all $x$ in $V$ and all real $a$ and $b,$ we have \begin{align*} (a + b)x = ax + bx. \end{align*}
  Proof.$\quad$ Let $x = \{x_n\}$ be an element of $V,$ and let $a$ and $b$ be real scalars. Then, with addition and multiplication by scalars as defined previously, we have \begin{align*} (a + b)x &= \{(a + b)x_n\} \\ &= \{ax_n + bx_n\} \\ &= ax + bx \quad \blacksquare \end{align*}
  
  Axiom 10.$\quad$ Existence of Identity.
  For every $x$ in $V,$ we have $1x = x.$
  Proof.$\quad$ Let $x = \{x_n\}$ be an element of $V.$ Then, \begin{align*} 1x &= \{1(x_n)\} \\ &= \{x_n\} \\ &= x \quad \blacksquare \end{align*}
  Thus, we have proven that the set $V$ of all infinite sequences $\{x_n\}$ of real numbers for which the series $\sum x_n^2$ converges is a real linear space.

  To prove that $(x, y) = \sum_{n = 1}^{\infty}x_ny_n$ is an inner product for the linear space $V,$ we will show that $(x, y)$ satisfies the four axioms of an inner product.

  
  
  

  $\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry).
  Proof. \begin{align*} (x, y) &= \sum_{n=1}^{\infty}x_n y_n \\ &= \sum_{n=1}^{\infty}y_n x_n \\ &= (y, x) \quad \blacksquare \end{align*}
  $\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  Proof. \begin{align*} (x, y + z) &= \sum_{n=1}^{\infty}x_n(y_n+z_n) \\ &= \sum_{n=1}^{\infty}x_n y_n+ x_n z_n \\ &= \sum_{n=1}^{\infty}x_n y_n + \sum_{n=1}^{\infty}x_n z_n \\ &= (x, y) + (x, z) \quad \blacksquare \end{align*}
  $\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity).
  Proof. \begin{align*} c(x, y) &= c\sum_{n=1}^{\infty}x_n y_n \\ &= \sum_{n=1}^{\infty} (cx_n)y_n \\ &= (cx, y) \quad \blacksquare \end{align*}
  $\text{(4)}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  Proof.$\quad$ \begin{align*} (x, x) &= \sum_{n=1}^{\infty}x_n^2 \\ &\geq 0 \end{align*} But if $x \neq O,$ there exists some $n$ in $\{1, 2, \dots\}$ such that $x_n \neq 0$ and $x_n^2 \gt 0.$ Hence, $(x, x) \gt 0$ if $x \neq O. \quad \blacksquare$

  This completes the proof.

  $\text{(c)} \quad$ If $x_n = 1/n$ and $y_n = 1/(n + 1)$ for $n \geq 1,$ we have \begin{align*} (x, y) &= \sum_{n=1}^{\infty}x_ny_n \\ &= \sum_{n=1}^{\infty}\frac{1}{n(n + 1)} \\ &= \sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{n + 1 - n}{n + 1}\right) \\ &= \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n + 1} \\ &= \sum_{n=1}^{\infty}\frac{1}{n} - \sum_{n=2}^{\infty}\frac{1}{n} \end{align*} As we can see, this is a telescoping series whose terms cancel for $n \geq 2$ leaving us with $(x, y) = 1. \quad \blacksquare$

  $\text{(d)} \quad$ $x = \{2^n\}$ is not an element of $V$ because $\sum 2^{2n}$ does not converge. Nevertheless, we refer to Exercise 22 of Volume 1, Section 10.9 which tells us that for all $x,$ \begin{align*} \sum_{n = 0}^{\infty} \frac{x^n}{n!} &= e^x \end{align*} Setting $x = 2$ we get: \begin{align*} \sum_{n = 0}^{\infty} \frac{2^n}{n!} &= e^2 \end{align*} But since the sum $(x, y)$ starts at index $n = 1,$ we have \begin{align*} (x, y) &= \sum_{n = 1}^{\infty} \frac{2^n}{n!} \\ &= \left(\sum_{n = 0}^{\infty} \frac{2^n}{n!}\right) - \frac{2^0}{0!} \\ &= e^2 - 1. \quad \blacksquare \end{align*}