Theorems on Continuous Functions

PROOF. Let $S$ be the set of all $x$ such that $a\le x\le b$ and $f(x)\le y_0\cdot$ Then $a$ is in $S$, so that $S$ is not empty. Also, $b$ is an upper bound for $S\cdot$ Therefore $S$ has a least upper bound $x_0\cdot$ In the case of the function graphed below, $x_0$ is as indicated:

Now either , or $f(x_0)=y_0\cdot$ We show that $f(x_0)=y_0$ by showing that the other two are impossible.

First of all, suppose Let Then we can find a such that whenever , we have Now we must have $x_0\ne b$, since $f(b)\ge y_0\cdot$ Therefore $h=\text{Min}(b-x_0,\frac{\delta }{2})$ is positive, and if we set $x=x_0+h$, , and

\begin{aligned} f(x) - f(x_0) &\le |f(x) - f(x_0)| < \epsilon \\ &= y_0 - f(x_0), \quad \text{or} \quad f(x) < y_0. \end{aligned}

Thus $x=x_0+h$ is greater than $x_0$ and is in $S\cdot$ But $x_0$ was an upper bound for $S\cdot$ This is a contradiction, and therefore $f(x_0)\ge y_0\cdot$ Here we have used only the property that $x_0$ is an upper bound for $S\cdot$

Now suppose We shall show that this is impossible by the fact that $x_0$ is the least upper bound for $S\cdot$ Clearly, $x_0\ne a\cdot$ Let $ϵ=f(x_0)-y_0\cdot$ Then , and we can find a such that whenever , then , so

\begin{aligned} f(x_0) - f(x) &\le |f(x_0) - f(x)| < \epsilon = f(x_0) - y_0, \\ -f(x) &< -y_0, \quad \text{or} \quad f(x) > y_0. \end{aligned}

Therefore no values of $x$ between $x_0-\frac{\delta }{2}$ and $x_0$ are in $S$, and since no values of $x\ge x_0$ are in $S$, no values of $x\ge x_0-\frac{\delta }{2}$ are in $S\cdot$ If $h=\text{Min}(x_0-a,\frac{\delta }{2})$, then $x_0-h$ is an upper bound for $S$, and This contradicts the fact that $x_0$ was the **least** upper bound. Therefore is impossible, and we must have $f(x_0)=y_0\cdot$ This completes the proof.

PROOF. Let $S$ be the set of all $x$ in the interval $a\le x\le b$ such that for some x' > x, f(x') > f(x'') whenever a \le x'' \le x \cdot In the diagram below, all points to the left of, and not including $x_0$ are in $S$:

Since $b$ is an upper bound for $S,S$ has a least upper bound unless $S$ is empty. But if $S$ is empty, then $a$ is not in $S$; i.e., for every x' > a, we have f(x') \le f(a); this is simply the case where $a$ is a maximum for $f(x)$, and in this case we are done. Thus there remains the case where $S$ is not empty, and therefore has a least upper bound $x_0\cdot$ We shall prove that $x_0$ is a maximum for $f(x)$ on $a\le x\le b$, as in the diagram.

First we claim that $x_0$ is not in $S\cdot$ If $x_0=b$, this is clear, since there is no x' of the interval such that x' > b \cdot Thus we may assume $x_0\ne b\cdot$ If $x_0$ were in $S$, we could find x' > x_0 such that f(x') > f(x) whenever $a\le x\le x_0\cdot$ In particular, f(x') > f(x_0) \cdot Then we could find a such that whenever |x - x_0| < \delta, f(x) - f(x_0) < f(x') - f(x_0), or f(x) < f(x'), by using \epsilon = f(x') - f(x_0) in the continuity of $f(x)$ at $x_0\cdot$ We may also assume Then consider $x_0+\frac{\delta }{2}\cdot$ For all $x$ such that $a\le x\le x_0+\frac{\delta }{2}$, we have f(x) < f(x'), or $x_0+\frac{\delta }{2}$ is in $S$ and is greater than $x_0\cdot$ This contradicts the fact that $x_0$ is an upper bound for $S\cdot$ Therefore $x_0$ is not in $S\cdot$

Now let $x$ be any point of the interval $a\le x\le b\cdot$ We shall show $f(x)\le f(x_0)\cdot$ Suppose first that If , there is a such that f(x') < f(x) whenever |x' - x_0| < \delta, as before. By the fact that $x_0$ is the least upper bound, there is a number $x_1$ in $S$ such that Consequently there is an such that f(x_2) > f(x') whenever a \le x' \le x_1 \cdot Either there is such an $x_2$ with or there is not. If we do have one with , then , and so f(x) > f(x') whenever a \le x' \le x_0, or $x_0$ is in $S\cdot$ But this is impossible. Therefore we may assume $x_2\ge x_0+\delta \cdot$ Then if $f(x)\ge f(x_2)$, let $x_3=x$; if , let $x_3=x_2\cdot$ In either case, f(x_3) > f(x') whenever a \le x' \le x_0, so $x_0$ is in $S$, a contradiction. Thus we must conclude that we must not have , and therefore that $f(x)\le f(x_0)$ whenever $x\ge x_0\cdot$

Finally, we show that $f(x)\le f(x_0)$ when and that Then there is an interval |x' - x_0| < \delta where f(x') < f(x) \cdot But there is a point $x_1$ of $S$ such that Therefore there is an such that f(x_2) > f(x'') whenever a \le x'' < x_1; in particular, Therefore $x_2\ge x_0+\delta$, or , and we have But we have shown that this is impossible. Therefore $f(x)\le f(x_0),x_0$ is a maximum for $f(x)$, and the theorem is proved.