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\pset{6}{Dan Ports}{2003/04/16}{drkp@mit.edu}

\begin{ppart}{Problem 1}
Let $f$ be a continuous mapping from the closed unit interval $I = [0,1]$ into $I$. We will show that there exists some $x \in I$ where $f(x) = x$.
\paragraph{}
First, note that if $f(0) = 0$ or $f(1) = 1$, then we are trivially done. Otherwise, the range of $f$ requires that $f(0) > 0$ and $f(1) < 1$. Consider $g(x) = x - f(x)$. Then $g(0) = 0 - f(0) < 0$ and $g(1) = 1 - f(1) > 0$, applying the previous inequalities. We know that $g$ is continuous because it is a linear combination of $f(x)$ and $x$, which are both continuous. Since $g(0) < 0 < g(1)$, the intermediate value theorem indicates that there exists some $x \in (0,1)$ such that $g(x) = 0$. At this point $x$, $g(x) = x - f(x) = 0$. Therefore, $f(x) = x$ at this point $x$.
\end{ppart}

\begin{ppart}{Problem 2}
Let $f$ be a function defined on $(a,b)$ such that $f'(x) > 0$. We show by contradiction that $f$ is strictly increasing. Suppose it is not. Then there exists some $x,y \in (a,b)$ such that $x < y$ but $f(x) \ge f(y)$. The mean value theorem tells us that there is some point $p \in (x,y) \subseteq (a,b)$ such that $f'(p) = \frac{f(y)-f(x)}{y-x}$. But this fraction is either zero or negative, so $f'(p) \leq 0$, contradicting the assumption that $\forall x \in (a,b) ~ f'(x) > 0$.
\paragraph{}
Now let $g(x)=f^{-1}(x)$; that is, $g$ is defined by $g(f(x))=x$. We will first show that $g$ is continuous. Consider any point $y$ in the domain of $g$. Then there is some $x \in (a,b)$ such that $y = f(x)$ (and thus $g(y) = g(f(x)) = x$).Since $(a,b)$ is an open interval that contains $x$, we can choose points $p,q$ such that $a < p < x < q < b$. Then $[p,q]$ is a compact set (closed and bounded in $\mathbb{R}$) that contains $x$. Applying Theorem 4.17, $g = f^{-1}$ is continuous on the image of $[p,q]$, which contains $y$. Thus $g$ is continuous at $y$. Since $y$ was an arbitrary point in $g$'s domain, $g$ is everywhere continuous.
\paragraph{}
We will next show that $g$ is differentiable and that $g'(f(x)) = \frac{1}{f'(x)}$. Take $\{y_n\}$ to be a sequence in the image of $(a,b)$ under $f$ that converges to a point $y$ also in that image; we also require that no element $y_n$ actually equals $y$. Then let $x = g(y)$ (and hence $y = f(x)$). By the continuity of $g$, the sequence $\{x_n\} = \{g(y_n)\}$ converges to $g(y) = x$. Now we consider the sequence of difference quotients:
\[
 \frac{g(y_n)-g(y)}{y_n-y} = \frac{x_n-x}{f(x_n)-f(x)} = \frac{1}{\frac{f(x_n)-f(x)}{x_n-x}}
\]
Since ${x_n} \to x$, this sequence converges to $\frac{1}{f'(x)}$ by definition of the derivative of $f$. We have thus shown (by the sequential definition of a limit) that $\lim_{t \to x}\frac{f(t)-f(x)}{t-x} = \frac{1}{f'(x)}$, so by definition $g'(x) = \frac{1}{f'(x)}$.
\end{ppart}

\begin{ppart}{Problem 3}
Consider the function $f(x) = C_0x+\frac{C_1}{2}x^2+\dots+\frac{C_{n-1}}{n}x^n + \frac{C_n}{n+1}x^{n+1}$. Note that every term contains at least one power of $x$, so $f(0) = 0$. Also note that $f(1) = C_0 + \frac{C_1}{2} + \dots + \frac{C_{n-1}}{n} + \frac{C_n}{n+1}$ which we are given equals zero. Because $f$ is a polynomial function, it is continuous and differentiable. Differentiating $f$, we obtain: $f'(x) = C_0 + C_1x + \dots + C_{n-1}x^{n-1} + C_nx^n$. Observe that the equation we are given is $f'(x) = 0$. By the mean value theorem, there exists some point $x \in (0,1)$ where $f'(x) = 0$. Therefore, the equation has some real root between 0 and 1.
\end{ppart}

\begin{ppart}{Problem 4}
Let $f$ be a twice-differentiable real function on $(a, \infty)$, and $M_0, M_1, M_2$ be the suprema of $\abs{f(x)}$, $\abs{f'(x)}$, $\abs{f''(x)}$ respectively. We will show that $M_1^2 \leq 4 M_0 M_2$.
\paragraph{}
Consider the second Taylor polynomial $P_2(t) = f(t) + \frac{f'(t)}{2}x$. Let $h = \sqrt{\frac{M_0}{M_2}}$. For some $x \in (a, \infty)$, we apply Taylor's theorem on the interval $(x,x+2h) \subset (a, \infty)$, and find that for some $\xi \in (x,x+2h)$,
\begin{eqnarray*}
f(x+2h) &=& P(x+2h) + \frac{f''(\xi)}{2}(x+2h-x)^2\ \\
        &=& f(x) + f'(x)2h + \frac{f''(\xi)}{2}(2x)^2 \\
f'(x)2h &=& f(x+2h) - f(x) - 2f''(\xi)h^2 \\
f'(x)   &=& \frac{1}{2h}[f(x+2h)-f(x)]-hf''(\xi)
\end{eqnarray*}
Since $M_0 = \sup \abs{f(x)}$, $\abs{f(x+2h)-f(x)} \leq 2M_0$; also $f''(\xi) \leq \sup \abs{f''(x)} = M_2$. Hence,
\begin{eqnarray*}
\abs{f'(x)} &\leq& \frac{M_0}{h} + hM_2 \\
&=& M_0 \sqrt{\frac{M_2}{M_0}} + M_2 \sqrt{\frac{M_0}{M_2}} = \sqrt{M_0M_2} + \sqrt{M_0M_2} = 2 \sqrt{M_0M_2} \\
\abs{f'(x)}^2 &\leq& 4 M_0 M_2\\
M_1^2 &\leq& 4 M_0 M_2
\end{eqnarray*}
\paragraph{}
We now show an example where $M_1^2=4M_0M_2$. With $a = -1$, let
\[f(x) = 
 \begin{cases} 2x^2-1 & (-1 < x < 0) \\
 \frac{x^2-1}{x^2+1} & (0 \leq x < \infty) \end{cases} \]
Differentiating each piece separately,
\[f'(x) = \begin{cases} 4x & (-1 < x < 0) \\ \frac{4x}{(x^2+1)^2} & (0 \leq x < \infty) \end{cases} ~~~~~~
f''(x) = \begin{cases} 4 & (-1 < x < 0) \\ \frac{-4(3x^2-1)}{(x^2+1)^3} & (0 \leq x < \infty) \end{cases} \]
Note that $2x^2 - 1$ and $\frac{x^2-1}{x^2+1}$ are both always less than 1. $\lim_{x\to-1}2x^2-1 = \lim_{x\to\infty}\frac{x^2-1}{x^2+1} = 1$, so $M_0 = 1$. For $M_1$, $\sup_{(-1,0)} \abs{4x}$ = 4 since $\lim_{x\to-1} 4x = -4$, and it is easy to show that $\frac{4x}{(x^2+1)^2}$ is everywhere less than 4, so $M_1 = \sup\abs{f'(x)} = 4$. For $M_2$, clearly $\sup_{(-1,0)}\abs{4} = 4$, and $\abs{\frac{-4(3x^2-1)}{(x^2+1)^3}}$ is also everywhere less than $4$, so $M_2 = \sup\abs{f''(x)} = 4$. So $M_1^2 = 4^2 = 16 = 4~4~1 = 4M_0M_2$. The inequality can in fact be equal.
\paragraph{}
This argument does not necessarily hold if the interval $(a, \infty)$ is replaced with $(0,1)$. To obtain the inequality, we required that the function (and $M_0, M_1, M_2$) be defined on the interval $[x,x+2h]$. Since $h = \sqrt{\frac{M_0}{M_2}}$, if, say, $M_0 > M_2$, then $h > 1$ and there can be no interval $[x,x+2h]$ contained in $(0,1)$.

\end{ppart}

\begin{ppart}{Problem 5}
Let $f$ be a function differentiable on $[a,b]$ such that for some $A \in \mathbb{R}$, $\abs{f'(x)} \leq A \abs{f(x)}$ on $[a,b]$. We will show that $\forall x \in [a,b] ~ f(x) = 0$.
\paragraph{}
Consider first the case when $b-a < \frac{1}{A}$. Let $M_0 = \sup\abs{f(x)}$ and $M_1 = \sup\abs{f'(x)}$. Note that, since $\abs{f'(x)} \leq A\abs{f(x)}$, $M_1 \leq AM_0$. We will show that $\forall x \in [a,b] ~ \abs{f(x)} \leq M_1(b-a)$. Suppose this is not true. Then there exists some point $p \in [a,b]$ such that $\abs{f(p)} > M_1(b-a)$. Then:
\begin{eqnarray*}
\abs{f(p)} &>& M_1 (b-a) \\
\frac{\abs{f(p)}}{b-a} &>& M_1 \\
\frac{\abs{f(p)}-\abs{f(a)}}{b-a} &>& M_1 \mbox{  since $f(a) = 0$} \\
\frac{\abs{f(p)-f(a)}}{\abs{b-a}} &>& M_1 \\
\frac{\abs{f(p)-f(a)}}{\abs{p-a}} &\ge& \frac{\abs{f(p)-f(a)}}{\abs{b-a}} > M_1 \mbox{  since $p \le b$} \\
\end{eqnarray*}
By the mean value theorem, we find some point $q \in [a,p] \subseteq [a,b]$ with $f'(q) = \frac{f(p)-f(a)}{p-a}$. Then $\abs{f'(q)} = \frac{\abs{f(p)-f(a)}}{\abs{p-a}} > M_1$, contradicting our definition of $M_1$ as $\sup\abs{f'(x)}$. Therefore, no such point $p$ can exist and $\forall x \in [a,b] ~ \abs{f(x)} \leq M_1(b-a)$.
\paragraph{}
Applying the inequality between $M_0$ and $M_1$, we obtain $\forall x \in [a,b] ~ \abs{f(x)} \leq M_1(b-a) \leq A M_0(b-a)$. Because $\abs{f(x)}$ is a continous function defined on the compact set $[a,b]$, it achieves its supremum $M_0$ at some point. So $M_0 \leq A(b-a)M_0$. Since we have restricted ourselves to cases where $b-a < \frac{1}{A}$, $A(b-a) < 1$. $M_0$ cannot be greater than zero, because if it were, then $M_0 > A(b-a)M_0$, contradicting the inequality above. $M_0$ cannot be negative by definition, so $M_0 = 0$. Therefore, $\abs{f(x)} = 0$ on every point in the interval $[a,b]$: if $\abs{f(p)}$ were non-zero at some point $p \in [a,b]$, it would contradict our definition of $M_0$. So $f(x) = 0$ for every $x \in [a,b]$.
\paragraph{}
We now generalize this result to the general case in which the interval $[a,b]$ has arbitrary length. Let $K = \frac{1}{2A}$. We will use induction (on $n$), dividing $[a,b]$ into successive intervals of length $K$ and showing that $f(x) = 0$ on those intervals. Let the induction hypothesis be that $f(x) = 0$ for every point on the interval $[a+nK,\min\{a+(n+1)K~,~b\}]$ (we must use the minimum in that expression to account for the final subinterval, where $a+(n+1)K$ may be greater than $b$). The base case $n=0$ follows from our previous result, since $f(a) = 0$ is given and $K$ was chosen so $K < \frac{1}{A}$. We assume the inductive hypothesis is true for $n-1$ and show it is then also true for $n$. By the inductive hypothesis, we know $f(a+nK) = 0$. We then consider the interval $[a+nK,\min\{a+(n+1)K,b\}]$. The interval has length $\min\{a+(n+1)K,b\}-(a+nK) \leq a+(n+1)K-(a+nK) = K < \frac{1}{A}$, so we can apply our previous result and show that $f(x) = 0$ on the interval. This proves the induction hypothesis for $n$. By induction, therefore, the induction hypothesis is true for all $n$. Thus $f(x) = 0$ on the entire interval $[a,b]$.
\end{ppart}

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