Function $f$ is continuous on $[0, 1]$. Function $g$ is defined on $[0, 1]$ and $g(0)=1, g(1)=0$. Function $f+g$ is monotonic increasing. Prove: $g$ can take any value between $[0, 1]$.
My attempt:
Since $f+g$ is monotonic, then $f(0)+g(0)\le f(1)+g(1)\Rightarrow f(0)+1\le f(1)$. We also have:
$$f(0)+1\le f(x)+g(x)\le f(1)~~~~~\forall x\in [0,1]$$
move terms, $$f(0)+1-f(x)\le g(x)\le f(1)-f(x)$$
since $f$ is continuous, it can take any value between $[f(0), f(1)]$, the LHS can take any value between $[f(0)+1-f(1), 1]$, and $[0,1]\subseteq [f(0)+1-f(1), 1]$. The RHS can take any value between $[0, f(1)-f(0)]$, and $[0,1]\subseteq [0, f(1)-f(0)]$.
So for any value $a\in [0,1]$, there exists $x_1$, such that $a\le g(x_1)$.Also there exists a point $x_2$ such that $g(x_2)\le a$. I am stuck here. I can't conclude $x_1=x_2$. How to proceed next? Thank you a lot.
$\def\lim{\mathop{\mathrm{lim}}\limits}\def\sup{\mathop{\mathrm{sup}}\limits}\def\inf{\mathop{\mathrm{inf}}\limits}\let\leq=\leqslant\let\geq=\geqslant$Let $h := f + g$. First, note that $h$ has lateral limits: if $x_0 \in [0, 1]$ then $\lim_{x\to x_0^-} h(x) = \sup_{x<x_0} h(x)$ and $\lim_{x\to x_0^+} h(x) = \inf_{x>x_0} h(x)$. Since $f$ is continuous, it has lateral limits, hence $g = h - f$ has lateral limits. Second, note that for every $x_0 \in (0, 1]$ $$g(x_0) - \lim_{x\to x_0^-} g(x) = h(x_0) - \lim_{x\to x_0^-} h(x) - f(x_0) + \lim_{x\to x_0^-} f(x) = h(x_0) - \sup_{x<x_0} h(x) \geq 0$$ and similarly $\lim_{x\to x_0^+} g(x) - g(x_0) \geq 0$ for every $x \in [0, 1)$.
Suppose that $g$ doesn't take the value $y \in [0, 1]$. Then $[0, 1] = A \cup B$ where $A := \{x : g(x) < y\}$ and $B := \{x : g(x) > y\}$.
Since $g(0) = 1$ we have $y < 1$ and $0 \in B$. Let $\bar x := \sup\{x \in [0, 1] : [0, x) \subset B\}$. It's easy to check that $[0, \bar x) \subset B$. (Ask me if it's not clear.) We have $$g(\bar x) \geq \lim_{x\to \bar x^-} g(x) \geq \inf_{x < \bar x} g(x) \geq y$$ since $g(x) > y$ if $x < \bar x$. Hence $\bar x \not\in A$, hence $\bar x \in B$, and $\bar x < 1$ since $g(1) = 0 < y$.
Now $\lim_{x\to\bar x^+} g(x) \geq g(\bar x) > y$, hence $g(x) > y$ if $x \in [\bar x, \bar x + \epsilon)$ for some $\epsilon > 0$, so $[0, \bar x + \epsilon) \subset B$, which contradicts that $\bar x$ was the supremum of $x$ such that $[0, x) \subset B$. This contradiction proves that $g$ takes every value in $[0, 1]$, as desired.