Intermediate value property for f(x)= sin (1/x) when x is not zero and $a$ when x is zero

Question

If $f:[-1,1]\to \mathbb{R} $ with $f(x)= \begin{cases}\sin(1/x), & x\neq 0\\ a,& x=0\end{cases} $ , how can I prove that $f$ has intermediate value property if and only if $a$ is between $[-1,1]$?

Answer 1

Let $a\in[-1,1]$. Then:

• for $-1\leq b
• for $-1\leq b \leq 0 < c\leq 1$ (analogously for $-1\leq b < 0 \leq c\leq 1$ ) :
  • if $f(b)f(c)$), lets take $w=\frac{1}{(2k-\frac{1}{2})\pi}$, where $k$ is positive integer such that $k>\frac{1}{2\pi c}+\frac{1}{4}$. Then we have $0
  • if $f(b)=f(c)$, then take $w=\frac{c}{2\pi c+1}$. We have then $0

We see then, that if $a\in [-1,1]$, then function $f$ has intermediate value property.

On the other hand, if $a\not \in [-1,1]$, let's take interval $[b,0]$ ($b\in [-1,0)$)

• if $a>1$ (analogously for $a<-1$), let's take $p=\frac{a+1}{2}$. Of course $f(b)\leq 1 < p < f(0)$. Because $1

We see then that if $a\not \in [-1,1]$, then $f$ doesn't have intermediate value property.