In this question, you may assume that, if a continuous function
takes both positive and negative values in an interval, then
it takes the value \(0\) at some point in that interval.
The function \(\f\) is continuous and \(\f(x)\) is non-zero for some value
of \(x\) in the interval \(0\le x \le 1\). Prove by contradiction, or otherwise,
that if
\[
\int_0^1 \f(x) \d x = 0\,,
\]
then
\(\f(x)\) takes both positive and negative values
in the interval \(0\le x\le 1\).
The function \(\g\) is continuous and
\[
\int_0^1 \g(x) \, \d x = 1\,,
\quad
\int_0^1 x\g(x) \, \d x = \alpha\, ,
\quad
\int_0^1 x^2\g(x) \, \d x = \alpha^2\,.
\tag{\(*\)}
\]
Show, by considering
\[
\int_0^1 (x - \alpha)^2 \g(x) \, \d x
\,,
\]
that
\(\g(x)=0\) for some value of \(x\) in the interval \(0\le x\le 1\).
Find a function of the form \(\g(x) = a+bx\) that satisfies the conditions
\((*)\) and verify that \(\g(x)=0\) for some value of \(x\) in the interval
\(0\le x \le 1\).
The function \(\h\) has a continuous derivative \(\h'\) and
\[
\h(0) = 0\,,
\quad
\h(1) = 1\,,
\quad
\int_0^1 \h(x) \, \d x = \beta\,,
\quad
\int_0^1 x \h(x) \, \d x = \tfrac{1}{2}\beta (2 - \beta)
\,.
\]
Use the result in part (ii) to show that \(\h^\prime(x)=0\)
for some value of \(x\) in the interval \(0\le x\le 1\).
Solution:
Claim: If \(f(x)\) non-zero for some \(x \in [0,1]\) and \(\int_0^1 f(x) \d x =0\) then \(f\) takes both positive and negative values in the interval \([0,1]\).
Proof: Suppose not, then WLOG suppose \(f(x) > 0\) for some \(x \in [0,1]\). Then notice (since \(f\) is continuous) that there is some interval where \(f(x) > 0\) around the \(x\) we have already shown exists. But then \(\int_{\text{interval}} f(x) \d x > 0\) and since \(f(x) \geq 0\) everywhere \(\int_0^1 f(x) \d x > 0\), which is a contradiction.
\(\,\) \begin{align*}
&& \int_0^1 (x - \alpha)^2 g(x) \d x &= \int_0^1 x^2g(x) \d x - 2 \alpha \int_0^1 x g(x) \d x + \alpha^2 \int_0^1 g(x) \d x \\
&&&= \alpha^2 - 2\alpha \cdot \alpha + \alpha^2 \cdot 1 \\
&&&= 0
\end{align*} Therefore \(g(x)(x-\alpha)^2\) is a continuous function which is either exactly \(0\) (in which case we've already found our \(0\)) or it is a continuous function which is both positive somewhere and has \(0\) integral, and therefore by part (i) must take both positive and negative values (and therefore takes \(0\) in between those points by continuity).
\begin{align*}
&&1 &= \int_0^1 a+bx \d x \\
&&&= a + \frac12 b \\
&& \alpha &= \int_0^1 ax + bx^2 \d x \\
&&&= \frac12 a + \frac13 b \\
&& \alpha^2 &= \int_0^1 ax^2 + bx^3 \d x\\
&&&= \frac13 a + \frac14 b \\
\Rightarrow && \frac1{36}(3a+2b)^2 &= \frac1{12}(4a+3b) \\
\Rightarrow && \frac1{36}(3a+2(2-2a))^2 &= \frac1{12}(4a+3(2-2a)) \\
\Rightarrow && (4-a)^2 &= 3(6-2a) \\
\Rightarrow && 16-8a+a^2 &= 18-6a \\
\Rightarrow && a^2-2a-2 &= 0 \\
\Rightarrow && (a-1)^2 - 3 &= 0 \\
\Rightarrow && a &= \pm \sqrt{3}+1 \\
&& b &= \mp 2\sqrt{3}
\end{align*}
So say \(a = \sqrt{3}+1, b = -2\sqrt{3}\)
This has a root at \(-\frac{a}{b} = \frac{1+\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3}+3}{6} < 1\) so we have met our condition.
Consider \(h'\), we must have
\begin{align*}
&& \int_0^1 h'(x)\d x &= h(1)-h(0) =1\\
&& \int_0^1 xh'(x) \d x &= \left [x h(x) \right]_0^1 - \int_0^1 h(x) \d x \\
&&&= 1 - \beta \\
&& \int_0^1 x^2 h'(x) \d x &= \left [ x^2h(x) \right]_0^1 - \int_0^1 2xh(x) \d x \\
&&&= 1 - 2\tfrac12 \beta(2-\beta) \\
&&&= (1-\beta)^2
\end{align*}
Therefore \(h'\) satisfies the conditions with \(\alpha = 1-\beta\), so \(h'\) must have a root in our interval.