# Thread: Continuity of Function on Intervals

1. ## Continuity of Function on Intervals

If we let $\displaystyle I := [a,b]$ and let $\displaystyle f:I\rightarrow\mathbb{R}$ be a continuous function on $\displaystyle I$ such that for each $\displaystyle x$ in $\displaystyle I$ there exists $\displaystyle y$ in $\displaystyle I$ such that $\displaystyle |f(y)| \leq \frac{1}{2}|f(x)|$. Prove there exists a point $\displaystyle c$ in $\displaystyle I$ such that $\displaystyle f(c) = 0$.

2. Assume that $\displaystyle \forall x \in I , 0 < |f(x)|$. You know that f takes on a minimum on $\displaystyle I = [a,b]$. Say it takes on its minimum at $\displaystyle x = m$. Note that $\displaystyle \frac{|f(m)|}{2} < |f(m)|$ since $\displaystyle |f(m)| \neq 0$.

$\displaystyle \exists y \in I : |f(y)| \le \frac{|f(m)|}{2} < |f(m)|$, contradicting the fact that $\displaystyle f(m)$ is the minimum of f on I.

3. Additionally, you could suppose that $\displaystyle \sup_I|f|=1$ on the interval (if you can't do this then f is the zero function and there is nothing to prove).
Then choose a point $\displaystyle x_0$, so you have a new point $\displaystyle x_1$ such that $\displaystyle |f(x_1)|\le\frac{|f(x_0)|}{2}\le\frac{1}{2}$.
The by induction you have points $\displaystyle \{x_n\}_{n=0}^\infty$ which satisfy $\displaystyle |f(x_n)|\le\frac{1}{2^n}$

Now some subsequence $\displaystyle x_{n_k}$ converges to a point $\displaystyle x$. And since $\displaystyle f$ is continuous

$\displaystyle \lim_k|f(x_{n_k})|=\lim_k\frac{1}{2^{n_k}}=0=|f(x) |$.

Really, the $\displaystyle \frac{1}{2}$ wasn't important. You just needed a constant $\displaystyle 0<\alpha<1$ and the same proof will work.

4. Originally Posted by JG89
Assume that $\displaystyle \forall x \in I , 0 < |f(x)|$. You know that f takes on a minimum on $\displaystyle I = [a,b]$. Say it takes on its minimum at $\displaystyle x = m$. Note that $\displaystyle \frac{|f(m)|}{2} < |f(m)|$ since $\displaystyle |f(m)| \neq 0$.

$\displaystyle \exists y \in I : |f(y)| \le \frac{|f(m)|}{2} < |f(m)|$, contradicting the fact that $\displaystyle f(m)$ is the minimum of f on I.
Sorry but I need a little extra help on this one... I follow everything you're saying but can't see how to use that to show there is a c where f(c) = 0....

5. f has a root in [a,b] if and only if |f| has a root in [a,b], right?

If |f| doesn't have a root in [a,b], then 0 < |f(x)| for all x in [a,b], right? From that assumption, I derived a contradiction. Which must mean that there indeed does exist a point x = c such that 0 = |f(c)| = f(c)