Mathematical Immaturity

1.13 Exercises

• Let $V$ be the set of all real functions $f$ continuous on $[0, +\infty)$ and such that the integral $\int_0^\infty e^{-t} f^2(t) \, dt$ converges. Define $(f, g) = \int_0^\infty e^{-t} f(t) g(t) \, dt.$
  $\text{(a)}\quad$ Prove that the integral for $(f, g)$ converges absolutely for each pair of functions $f$ and $g$ in $V.$
  $\quad$[Hint: Use the Cauchy-Schwarz inequality to estimate the integral $\int_0^M e^{-t} |f(t) g(t)| \, dt.$]
  $\text{(b)}\quad$ Prove that $V$ is a linear space with $(f, g)$ as an inner product.
  $\text{(c)}\quad$ Compute $(f, g)$ if $f(t) = e^{-t}$ and $g(t) = t^n,$ where $n = 0, 1, 2, \ldots.$

• $\text{(a)} \quad$ Applying the Cauchy-Schwarz inequality to the integral $\int_0^M e^{-t}|f(t)g(t)|\,dt,$ we get \begin{align*} \left(\int_0^M e^{-t}|f(t)g(t)|\,dt\right)^2 &= \left(\int_0^M e^{-t/2}|f(t)|e^{-t/2}|g(t)|\,dt\right)^2 \\ &\leq \left(\int_0^M e^{-t}|f(t)|^2\,dt\right)\left(\int_0^M e^{-t}|g(t)|^2\,dt\right) \\ &\leq \left(\int_0^{\infty} e^{-t}f^2(t)\,dt\right)\left(\int_0^{\infty} e^{-t}g^2(t)\,dt\right) \end{align*} But we know that $\left(\int_0^{\infty} e^{-t}f^2(t)\,dt\right)$ and $\left(\int_0^{\infty} e^{-t}g^2(t)\,dt\right)$ converge to nonnegative sums. If we denote these sums as $S_f$ and $S_g,$ respectively, we can see that \begin{align*} \int_0^M e^{-t}|f(t)g(t)|\,dt \leq \sqrt{S_fS_g} \end{align*} for all $M \geq 0.$ And since $\int_0^M e^{-t}|f(t)g(t)|\,dt$ is bounded for all $M \geq 0,$ we conclude that $(f, g)$ is absolutely convergent for each pair of functions $f$ and $g$ in $V. \quad \blacksquare$

  $\text{(b)} \quad$ 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 $f$ and $g$ be two elements of $V,$ with $(f + g)(t) = f(t) + g(t).$ To show that $V$ satisfies closure under addition, it will suffice to show that $\int_0^{\infty} e^{-t}[f + g]^2(t) \,dt$ converges for arbitrary $f$ and $g$ in $V.$ First, we expand the integral as follows: \begin{align*} \int_0^{\infty} e^{-t}[f + g]^2(t) \,dt &= \int_0^{\infty}e^{-t} \left[f(t) + g(t)\right]^2 \,dt \\ &= \int_0^{\infty}e^{-t}\left[f^2(t) + 2f(t)g(t) + g^2(t)\right] \,dt \\ &= \int_0^{\infty}e^{-t}f^2(t) \,dt + 2\int_0^{\infty}e^{-t}f(t)g(t) \,dt + \int_0^{\infty}e^{-t}g^2(t)\,dt \\ \end{align*} Because $f$ and $g$ are elements of $V,$ we know by definition that the integrals $\int_0^{\infty}e^{-t}f^2(t) \,dt$ and $\int_0^{\infty}e^{-t}g^2(t) \,dt$ are convergent. And from part (a), we showed that the improper integral $\int_0^{\infty}e^{-t}f(t)g(t) \,dt$ is absolutely convergent for all $f$ and $g$ in $V.$ Hence, $\int_0^{\infty} e^{-t}[f + g]^2(t) \,dt$ is convergent and $f + g$ 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 $f$ be an element of $V$ and $a$ be a real scalar. Then, denote by $af$ the function $af(t)$ defined on the same domain as $f.$ To satisfy closure under multiplication by real scalars, $af$ must be an element of $V.$ In other words, $af$ must be continuous and the improper integral $\int_0^{\infty}e^{-t}[af(t)]^2\,dt$ must converge.
  
  We know from Theorem 3.2 (Volume 1, Section 3.4) that for all real $a,$ $af$ is continuous. And for all real $a,$ we have: \begin{align*} \int_0^{\infty}e^{-t}[af(t)]^2\,dt &= a^2\int_0^{\infty}e^{-t}f^2(t)\,dt \end{align*} But if $\int_0^{\infty}e^{-t}f^2(t)\,dt$ converges to some finite value then so does $a^2\int_0^{\infty}e^{-t}f^2(t)\,dt.$ Thus, $\int_0^{\infty}e^{-t}[af(t)]^2\,dt$ is convergent and $af$ 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 $f$ and $g$ be elements of $V.$ Because $f$ and $g$ are real-valued, they satisfy the field axioms for real numbers. Hence $f + g = g + f$ for any $f$ and $g$ in $V. \quad \blacksquare$
  
  Axiom 4.$\quad$ Associative Law.
  For all $x,$ $y,$ and $z$ in $V,$ we have $(x + y) + z = x + (y + z).$
  
  Proof.$\quad$ Let $f,$ $g,$ and $h$ be elements of $V.$ Because all elements of $V$ are real-valued for all $t$ in $[0, +\infty),$ these elements satisfy the field axioms of real numbers, hence $(f + g)(t) + h(t) = f(t) + (g + h)(t)$ for all $t. \quad \blacksquare$
  
  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 $O$ to be the function $O(t) = 0$ for all $t$ in $[0, +\infty).$ We can see that $O$ is a member of $V$ since $O$ is continuous for all $t$ and $\int_0^{\infty}e^{-t}O^2(t)\,dt$ converges to $0.$ Then, for all $f$ in $V,$ we have $(f + O)(t) = f(t) + 0 = f(t)$ for all $t \in [0, +\infty). \quad \blacksquare$
  
  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 $f$ be an element of $V.$ Then, denote by $(-1)f$ the function $-f(t).$ Because $f$ is continuous on $[0, +\infty),$ $(-1)f$ is also continuous on $[0, +\infty)$ (Refer to Volume 1, Theorem 3.2 and let $g(t) = -1$). Then, because $[-f(t)]^2 = f^2(t),$ the integral $\int_0^{\infty}e^{-t}[-f(t)]^2\,dt$ converges, which means $(-1)f$ is an element of $V.$ Then, for all $t \in [0, +\infty)$ we have: \begin{align*} f(t) + (-1)f(t) &= f(t) - f(t) \\ &= O \end{align*} Thus, $(-1)f$ is the negative of $f. \quad \blacksquare$
  
  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 $f$ be an element of $V$ and let $a$ and $b$ be real scalars. Because $f$ is real-valued, we have $a(bf) = (ab)f \quad \blacksquare.$
  
  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 $x$ and $y$ be elements of $V,$ and let $a$ be a real scalar. We know from the proof of Axiom 1 that $(x + y)(t) = x(t) + y(t).$ Then, because $x$ and $y$ are real-valued for $t \in [0, +\infty),$ we have \begin{align*} a(x + y)(t) &= ax(t) + ay(t) \end{align*} for all $t \in [0, +\infty).$ Hence, $a(x + y) = ax + ay. \quad \blacksquare$
  
  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 $x$ be an element of $V,$ and let $a$ and $b$ be real scalars. Then, because $x$ is real-valued for all $t \in [0, +\infty),$ we have $(a + b)x(t) = ax(t) + bx(t)$ and hence $(a + b)x = ax + bx. \quad \blacksquare.$
  
  Axiom 10.$\quad$ Existence of Identity.
  For every $x$ in $V,$ we have $1x = x.$
  
  Proof.$\quad$ Let $x$ be an element of $V.$ Then, for each $t \in [0, +\infty)$ we have $1x(t) = x(t)$ and thus $1x = x. \quad \blacksquare$
  
  Thus we have proven that $V$ is a real linear space. To prove that $(f, g) = \int_0^\infty e^{-t} f(t) g(t) \, dt$ is an inner product for the linear space $V,$ we will show that $(f, g)$ satisfies the four axioms of an inner product.

  $\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry).
  Proof. \begin{align*} (f, g) &= \int_0^\infty e^{-t} f(t) g(t) \, dt \\ &= \int_0^\infty e^{-t} g(t) f(t) \, dt \\ &= (g, f) \quad \blacksquare \end{align*}
  $\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  Proof. \begin{align*} (f, g + h) &= \int_0^\infty e^{-t} f(t) [g+h](t) \, dt \\ &= \int_0^\infty e^{-t} f(t) g(t) + f(t)h(t) \, dt \\ &= \int_0^\infty e^{-t} f(t) g(t)\,dt + \int_0^\infty e^{-t}f(t)h(t) \, dt \\ &= (f, g) + (f, h) \quad \blacksquare \end{align*}
  $\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity).
  Proof. \begin{align*} c(f, g) &= c\int_0^\infty e^{-t} f(t) g(t) \, dt \\ &= \int_0^\infty e^{-t} [cf(t)] g(t) \, dt \\ &= (cf, g) \quad \blacksquare \end{align*}
  $\text{(4)}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  Proof.$\quad$ Let $f \neq O$ be an element of $V.$ By definition, the integral $\int_0^{\infty}e^{-t}f^2(t)\,dt$ is convergent. And since $e^{-t}f^2(t) \geq 0$ for $t \in [0, +\infty),$ we have $\int_0^\infty e^{-t} f^2(t) \, dt \geq 0.$ But, since $e^{-t} \gt 0$ for all real $t,$ and $f(t)$ is a continuous function with $f \neq O$, there exists some $t_0 \in [0, \infty)$ and some $h \gt 0$ such that $e^{-t}f^2(t) \gt 0$ on the interval $[t_0 - h, t_0 + h]$ and thus \begin{align*} \int_{t_0-h}^{t_0+h}e^{-t}f^2(t)\, dt \gt 0 \end{align*} But since $e^{-t}f^2(t) \geq 0$ for all $t \in [0, +\infty),$ this implies that $(f, f) \gt 0$ if $f \neq O. \quad \blacksquare$

  Thus, we have shown that the set $V$ of all real functions $f$, continuous on $[0, \infty),$ such that $\int_0^{\infty}e^{-t}f^2(t)\,dt$ converges, is a real linear space with $(f, g) = \int_0^{\infty}e^{-t}f(t)g(t)\,dt$ as an inner product. This completes the proof.

  $\text{(c)} \quad$ Recall that $(f, g)$ is given by: \begin{align*} (f, g) &= \int_0^{\infty}e^{-t}f(t)g(t)\,dt \end{align*} Thus, with $f(t) = e^{-t},$ if we denote $g_n(t) = t^n$ for $n = 0, 1, 2, \dots,$ we can write: \begin{align*} (f, g_n) &= \int_0^{\infty}e^{-2t}t^n\,dt \end{align*} Suppose $n = 0.$ Then $g_0(t) = 1$ and $(f, g_0)$ is given by: \begin{align*} (f, g_0) &= \int_0^{\infty}e^{-2t}\,dt \\ &= -\frac{1}{2}e^{-t}\,\Biggr|_0^{\infty} \\ &= \frac{1}{2} \end{align*} If $n = 1,$ then $g_1(t) = t$ and \begin{align*} (f, g_1) &= \int_0^{\infty} te^{-2t}\,dt \end{align*} Integrating by parts, with $u = t$ and $dv = e^{-2t}\,dt,$ we get: \begin{align*} (f, g_1) &= \int_0^{\infty} te^{-2t}\,dt \\ &= -\frac{1}{2}te^{-2t}\,\Biggr|_0^{\infty} + \frac{1}{2}\int_0^{\infty}e^{-2t}\,dt \\ &= 0 + \frac{1}{2}(f, g_0) \\ &= \frac{1}{2}\left(\frac{1}{2}\right) \\ &= \frac{1!}{2^2} = \frac{1}{4} \end{align*} If $n = 2,$ then $g_2(t) = t^2$ and we have \begin{align*} (f, g_2) &= \int_0^{\infty} t^2e^{-2t}\,dt \end{align*} Integrating by parts, with $u = t^2$ and $dv = e^{-2t},$ we get: \begin{align*} (f, g_2) &= \int_0^{\infty} t^2e^{-2t}\,dt \\ &= -\frac{1}{2}t^2e^{-2t}\,\Biggr|_0^{\infty} + \frac{1}{2}\int_0^{\infty}2te^{-2t}\,dt \\ &= 0 + \frac{2}{2}(f, g_1) \\ &= \frac{2}{2}\left[\frac{1}{2}(f, g_0)\right] \\ &= \frac{2}{2}\left[\frac{1}{2}\left(\frac{1}{2}\right)\right] \\ &= \frac{2!}{2^3} = \frac{1}{4} \end{align*} To compute $(f, g_n)$ for $n \geq 1,$ we perform an induction step. Let $m$ be an integer $\geq 0$ and let $n = m + 1.$ By hypothesis, we have \begin{align*} (f, g_m) &= \int_0^{\infty}t^m e^{-2t}\,dt \\ &= \frac{m!}{2^{m + 1}} \end{align*} We wish to show that $(f, g_n) = (m + 1)!/2^{m + 2} = n!/2^{n + 1}$. We begin by evaluating the integral for $(f, g_n):$ \begin{align*} (f, g_n) &= \int_0^{\infty}t^n e^{-2t}\,dt \\ &= -\frac{1}{2}t^ne^{-2t}\,\Biggr|_0^{\infty} + \frac{n}{2}\int_{0}^{\infty}t^{n -1}e^{-2t}\,dt \\ &= 0 + \frac{n}{2}\int_0^{\infty}t^me^{-2t}\,dt \end{align*} But the rightmost integral is $(f, g_m),$ which gives us: \begin{align*} (f, g_n) &= \frac{n}{2}\left(\frac{m!}{2^{m + 1}}\right) \\ &= \frac{n!}{2^{n + 1}} \quad \blacksquare \end{align*}