Which of the following statements are true $(NBHM - 2015)$?

Question

Let $X =\{f \in C[-5, 5] : f(-5)= f(5) = 0 \} $

• There exist $ f \in X$ such that $ f \equiv 2$ on $[-1, 0]$ and $ f \equiv 3$ on $[1 , 2] \cup [3 , 4]$

• For every $ f \in X$, there exist distinct points $x_1$ and $x_2$ in $]-5,5[$ such that $f(x_1) = f(x_2)$

• For every $f \in X$ , therer exist $x \in ]-5,5[$ such that $f(x) = x$

• If $f :\mathbb R \rightarrow \mathbb R$ is differentiable on $\mathbb R$ and $| f'| \leq M$ for all $t \in \mathbb R$, then there exist $\epsilon_o >0$ such that for all $0 < \epsilon \leq \epsilon_o$, the function $g(x) = x + \epsilon f(x)$ is injective.

  I would like to explain what all i have tried and if there is any better way to do please let Me know.

• define $f(x) = \begin{cases} \frac{1}{2}(x+5) & x \in [-5,-1] \\ 2 & x \in [-1,0] \\ x+2 & x \in [0,1] \\ 3 & x \in [1,4] \\ -3x +15 & x \in [4,5] \end{cases}$

Easly we can prove that $f \in [-5,5]$

• I think this can be proved by sketching the graph of $f$, but if someone have any counter example , provide me.

• Define $g(x) = f(x) -x$ $\Rightarrow g(-5) = 5$ and $ g(5) = -5$, So by intermediate value theorem , there exist $x \in ]-5, 5[$ such that $g(x) = x$

$\Rightarrow f(x) =x$

• Let if possible there exist $\epsilon_o > 0$, for each $\epsilon \leq \epsilon_o$

define $g(x) = x + \epsilon f(x) $

$\Rightarrow g'(x) = 1 + \epsilon f'(x) $

$\Rightarrow f'(x) = \frac{g'(x)- 1}{\epsilon}$

as $f'(x) \rightarrow \infty$ for $\epsilon \rightarrow 0$ , which is a contradiction

So This is a false statement.

Please check my solution. Thank you

Answer 1

The second statement is true, and you can do better than "sketching the graph of $f$".

If $f$ is constantly zero in $]-5,0[$ the proof is obvious. Otherwise, there is a value $c$ in $]-5,0[$ where $f(c)\ne 0$. Then by continuity there are $x_1$ and $x_2$ such that $-5<x_1<c<x_2<5$ and $f(x_1)=f(x_2)=\frac{f(c)}2$.

Answer 2

About the last problem $g'(x)=1+\epsilon f'(x)>1-\epsilon M$ , so choosing $\epsilon_0\le\frac{1}{M}$ will ensure that for all $\epsilon\le\epsilon_0$ , $g'(x)>0$ thus $g$ is injective.