Prove that $f(x) = 0$ for some $x$ under the assumption that there exists a continuous function $g$ such that $f + g$ is non decreasing. Let $f(0) > 0$, $f(1) < 0$. Prove that $f(x) = 0$ for some $x$ under the
assumption that there exists a continuous function $g$ such that $f + g$ is
non decreasing.
A hint for this problem says the following: bisect the interval $[0,1]$ choosing the right interval of the mid point if there's $x$ in the right interval for which $f(y)\ge 0$. Otherwise choose the left interval. As such you will form $[a_n, b_n]=I_n$ such that $a_n, b_n \rightarrow c$. It turns out $c$ is our desired point. The hint goes further and states that, 'notice that for all $n$, there's $y_n \in [a_n, c]$ such that $f(y_n)\ge 0$'. Here's where I am confused. (I tried very hard to prove it but I couldn't) I do not want a proof of this statement (if it exists). I am just looking for a sanity check. Could the $[a_n, c]$ actually be $[a_n,b_n]$? Was it just a typo?
 A: Let $n_m$ be the (possibly finite) subsequence so that the $[a_{n_m},b_{n_m}]$ are the left intervals. See that $f(y) < 0$ for all $y \in [b_{n_m}, b_{n_{m-1}}]$.  In this way, it must be the case that $f(y) < 0$ for all $y \in \bigcup_m [b_{n_m}, b_{n_{m-1}}]$.  And $(c,1] \subseteq \bigcup_m [b_{n_m}, b_{n_{m-1}}]$.
A: Let $g:[0,1]\rightarrow\mathbb{R}$ be a continuous function such
that $h:=f+g$ is monotonic increasing on $[0,1]$. Observe that if
$x\in(0,1)$, right-hand limit $f(x+)=\lim_{t\rightarrow x+}f(t)=\lim_{t\rightarrow x+}h(t)-g(x)$
exists. Similarly, left hand limit $f(x-)=\lim_{t\rightarrow x-}f(t)$
exists.
Let $A=\{x\in[0,1]\mid f(y)>0 \mbox{ for all } y\in[0,x]\}.$ Note
that $A\neq\emptyset$ (because $0\in A$) and bounded above by $1$,
so $\xi=\sup A$ exists. We go to show that $\xi\notin A$ by contradition.
Suppose the contrary that $\xi\in A$, then $f(\xi)>0$. Note that
$\xi<1$. For each $x\in(\xi,1]$, $x\notin A\Rightarrow\exists y\in(\xi,x]$
such that $f(y)\leq0$. From which we can choose a sequence $(y_{n})$
in $(\xi,1]$ such that $y_{1}>y_{2}>\ldots>\xi$ , $y_{n}\rightarrow\xi$,
and $f(y_{n})\leq0$. Letting $n\rightarrow\infty$, we have $f(\xi+)\leq0$.
Observe that $0\leq h(\xi+)-h(\xi)=f(\xi+)-f(\xi)<0,$ which is a
contradiction.
$\xi\notin A$ implies that there exists a sequence $(x_{n})$ in
$A$ such that $x_{1}<x_{2}<\ldots<\xi$ and $x_{n}\rightarrow\xi$.
Therefore $f(\xi-)=\lim_{n}f(x_{n})\geq0$. Now, $0\leq h(\xi)-h(\xi-)=f(\xi)-f(\xi-)$,
so $0\leq f(\xi-)\leq f(\xi)$. Finally, we go to prove that $f(\xi)=0$.
Prove by contradiction. Suppose the contrary that $f(\xi)>0$. Let
$t\in[0,\xi)$ be arbitrary, then $t$ is not an upper bound of $A$.
Therefore, there exists $x\in A$ and $x>t$. In particular $f(t)>0$
by the very definition of $A$ (because $x\in A$ and $t\in[0,x]$).
Hence, we prove that $f(t)>0$ for each $t\in[0,\xi)$. Together with
the fact that $f(\xi)>0$, we have $\xi\in A$, which is a contradiction.
