- #1

Eclair_de_XII

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- Homework Statement
- Define a continuous function ##f:[a,b]\longrightarrow \mathbb{R}## such that ##f(b)>0## and ##f(a)<0##. Prove that there is a real number ##c\in (a,b)## such that ##f(c)=0##.

- Relevant Equations
- A function ##f## is continuous at ##a\in \textrm{dom}(f)## if for all positive ##\epsilon##, there is a ##\delta>0## such that for ##x\in \textrm{dom}(f)##, if ##0<|x-a|<\delta## then ##|f(x)-f(a)|<\epsilon##

Proof goes like this:

(1) Prove the existence of open intervals centered around the end-points of the domain such that the image of the points in these intervals through ##f## has the same sign as the image of the end-point through ##f##. In other words, prove that there is a ##\delta>0## such that if ##b>x>b-\delta##, ##f(x)f(b)>0##; and similarly for ##a##.

(2) Conclude there is a point ##c\in (a,b)## such that for any ##\delta>0##, the image of ##(c-\delta,c+\delta)## contains both positive and negative points.

(3) By definition of continuity, there must be a real number ##L## that ##f(c)## converges to. Reason that ##L## cannot be negative or positive.

===(1)===

Suppose that there are no such intervals. Consider ##b##. Choose ##\epsilon\leq f(b)##. Hence, for any ##\delta>0##, if ##|b-x|<\delta##, then ##f(x)<0##.

\begin{align*}

|f(x)-f(b)|&=&f(b)+|f(x)|\\

&\geq&f(b)\\

&\geq&\epsilon

\end{align*}

This contradicts the fact that ##f## is continuous. Hence, there must be some ##\delta>0## such that ##f(x)>0## if ##b-\delta<x<b##.

===(2)===

Now choose ##c=b-\delta_b## where ##\delta_b>0## with the property that if ##x<b-\delta_b##, then ##f(x)\leq 0##.

Similarly, ##c=a+\delta_a## where ##\delta_a>0## with the property that if ##x>a+\delta_a##, then ##f(x)\geq 0##.

% Note: I feel like I should be invoking completeness here, but wasn't sure if it applies or if it does, how to invoke it.

Let ##\delta_c>0##.

If ##\delta_c+c>x>c=a+\delta_a##, then ##f(x)\geq 0##.

If ##c-\delta_c<x<c=b-\delta_b##, then ##f(x)\leq 0##.

===(3)===

Suppose ##f(c)>0## and choose ##\epsilon\leq f(c)##. Let ##\delta>0## and choose ##x\in (c-\delta,c)##.

Then ##f(x)\leq 0## and:

\begin{align*}

|f(c)-f(x)|&=&-f(x)+f(c)\\

&=&|f(x)|+f(c)\\

&\geq&f(c)\\

&\geq&\epsilon

\end{align*}

Hence, ##f## cannot be continuous at ##x=c## if ##f(c)>0##.

Now suppose ##f(c)<0## and choose ##\epsilon\leq -f(c)##. Let ##\delta>0## and choose ##x\in (c,c+\delta)##.

Then ##f(x)\geq 0##:

\begin{align*}

|f(x)-f(c)|&=&f(x)+(-f(c))\\

&=&|f(x)|-f(c)\\

&\geq&-f(c)\\

&\geq&\epsilon

\end{align*}

Hence, ##f## cannot be continuous at ##x=c## if ##f(c)<0##.

This leaves only the one possibility by the trichotomy of the real numbers.

(1) Prove the existence of open intervals centered around the end-points of the domain such that the image of the points in these intervals through ##f## has the same sign as the image of the end-point through ##f##. In other words, prove that there is a ##\delta>0## such that if ##b>x>b-\delta##, ##f(x)f(b)>0##; and similarly for ##a##.

(2) Conclude there is a point ##c\in (a,b)## such that for any ##\delta>0##, the image of ##(c-\delta,c+\delta)## contains both positive and negative points.

(3) By definition of continuity, there must be a real number ##L## that ##f(c)## converges to. Reason that ##L## cannot be negative or positive.

===(1)===

Suppose that there are no such intervals. Consider ##b##. Choose ##\epsilon\leq f(b)##. Hence, for any ##\delta>0##, if ##|b-x|<\delta##, then ##f(x)<0##.

\begin{align*}

|f(x)-f(b)|&=&f(b)+|f(x)|\\

&\geq&f(b)\\

&\geq&\epsilon

\end{align*}

This contradicts the fact that ##f## is continuous. Hence, there must be some ##\delta>0## such that ##f(x)>0## if ##b-\delta<x<b##.

===(2)===

Now choose ##c=b-\delta_b## where ##\delta_b>0## with the property that if ##x<b-\delta_b##, then ##f(x)\leq 0##.

Similarly, ##c=a+\delta_a## where ##\delta_a>0## with the property that if ##x>a+\delta_a##, then ##f(x)\geq 0##.

% Note: I feel like I should be invoking completeness here, but wasn't sure if it applies or if it does, how to invoke it.

Let ##\delta_c>0##.

If ##\delta_c+c>x>c=a+\delta_a##, then ##f(x)\geq 0##.

If ##c-\delta_c<x<c=b-\delta_b##, then ##f(x)\leq 0##.

===(3)===

Suppose ##f(c)>0## and choose ##\epsilon\leq f(c)##. Let ##\delta>0## and choose ##x\in (c-\delta,c)##.

Then ##f(x)\leq 0## and:

\begin{align*}

|f(c)-f(x)|&=&-f(x)+f(c)\\

&=&|f(x)|+f(c)\\

&\geq&f(c)\\

&\geq&\epsilon

\end{align*}

Hence, ##f## cannot be continuous at ##x=c## if ##f(c)>0##.

Now suppose ##f(c)<0## and choose ##\epsilon\leq -f(c)##. Let ##\delta>0## and choose ##x\in (c,c+\delta)##.

Then ##f(x)\geq 0##:

\begin{align*}

|f(x)-f(c)|&=&f(x)+(-f(c))\\

&=&|f(x)|-f(c)\\

&\geq&-f(c)\\

&\geq&\epsilon

\end{align*}

Hence, ##f## cannot be continuous at ##x=c## if ##f(c)<0##.

This leaves only the one possibility by the trichotomy of the real numbers.

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