Show that $f(c)= 0$ for some $c \in (0, 1)$

Question

Let $f : [0,1] \to \Bbb R$ satisfy $f(0) < 0$ and $f(1) > 0$ and $g$ continuous on $[0, 1]$ such that $f+g$ is decreasing. Show that $f(c)= 0$ for some $c \in (0, 1)$

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I just want to show that $f+g$ follows the intermediate value property on $[0, 1]$.

As $f+g$ is decreasing function on $[0, 1]$ and if it follows the intermediate value property then $f+g$ will be continuous on $[0, 1]$ & as $g$ is continuous on $[0, 1]$ then $f$ must have to be continuous on $[0, 1]$.

As $f(0) < 0$ and $f(1) > 0$ then there exists $c$ in $(0, 1)$ such that $f(c) = 0$.

How can I show $f+g$ follows intermediate value property on $[0, 1]$?

Thank you!

Answer 1

Here is an approch (need not to prove $f+g$ has the intermediate value property)

Let $$A=\left\{x\in [0,1]:f(t)<0,\forall t\in [0,x]\right\}$$ then $A\ne\varnothing$ since $0\in A$ and $A$ is bounded above by $1,$ so $\sup A$ exists, we denote it by $\xi=\sup A.$ We claim

[1] $\xi<1:$ Otherwise $\xi=1,$ then $f(t)<0,\forall t\in [0,1),$ since $f+g$ is decreasing, we have $$f(t)+g(t)\geq f(1)+g(1),\forall t\in [0,1)$$ let $t\to 1^-,$ since $\displaystyle\lim_{t\to 1^-}g(t)=g(1)$ and $\displaystyle\liminf_{t\to 1^-}f(t)\leq 0,$ we have $0+g(1)\geq f(1)+g(1)\Rightarrow f(1)\leq 0,$ which is absurd!

[2] $f(\xi)\geq 0:$ Suppose by contradiction that $f(\xi)<0,$ we claim there is some $\delta>0,$ so that $f(t)<0,\forall t\in (\xi,\xi+\delta)$ .Otherwise $\exists \left\{t_n\right\}\subset [0,1],t_n\downarrow \xi$ and $f(t_n)\geq 0,$ use again $f+g$ is decreasing, we obtain: $f(t_n)+g(t_n)\leq f(\xi)+g(\xi),$ similarly let $n\to\infty$ we have $f(\xi)\geq 0,$ which is absurd! This shows that $f(t)<0$ also holds for $t\in [\xi,\xi+\delta),$ but it already holds for $t\in [0,\xi),$ hence $\xi+\frac{\delta}{2}\in A,$ which contradicts to $\xi=\sup A.$ It also implies that $\xi>0.$

[3] $f(\xi)=0:$ If not, $f(\xi)>0,$ since $\xi=\sup A,$ for any $\left\{t_n\right\}\subset [0,\xi),t_n\uparrow \xi,$ we have $f(t_n)<0.$ Use $f+g$ is decreasing, we have $f(t_n)+g(t_n)\geq f(\xi)+g(\xi),$ similarly let $n\to\infty$ we derive that $f(\xi)\leq 0,$ which is absurd.

So $\xi$ is what you want.

Answer 2

Consider $S:=\{x\in[0,1]\mid f(x)\leq 0\}$. You may check that this set has a supremum $c:=\sup S$.

Let $\epsilon>0$. By continuity of $g$, there is a $\delta>0$ such that for any $x\in(c-\delta,c+\delta)$, we have $|g(c)-g(x)|<\epsilon$.

By supremum property, there is an $a_1\in(c-\delta,c]$ such that $a_1\in S$, meaning that $f(a_1)\leq 0$. By monotonicity of $f+g$, we can get $f(c)\leq g(a_1)-g(c)$. As $g(a_1)-g(c)<\epsilon$, we have $f(c)<\epsilon$.

Take any $a_2\in(c,c+\delta)$, which must be outside of $S$. Hence, $f(a_2)>0$. By monotonicity of $f+g$, we can get $f(c)> g(a_2)-g(c)$. As $g(a_2)-g(c)>-\epsilon$, we have $f(c)>-\epsilon$.

Hence, $|f(c)|<\epsilon$ for any $\epsilon>0$, and thus $f(c)=0$.