# Math Help - Continuity of Function on Intervals

1. ## Continuity of Function on Intervals

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

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

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

3. Additionally, you could suppose that $\sup_I|f|=1$ 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 $x_0$, so you have a new point $x_1$ such that $|f(x_1)|\le\frac{|f(x_0)|}{2}\le\frac{1}{2}$.
The by induction you have points $\{x_n\}_{n=0}^\infty$ which satisfy $|f(x_n)|\le\frac{1}{2^n}$

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

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

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

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

$\exists y \in I : |f(y)| \le \frac{|f(m)|}{2} < |f(m)|$, contradicting the fact that $f(m)$ 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)