Prove or disprove.
Let a arbtrary polynomial function $f:[a,b]\to \mathbb{R}$ such that $f(a)<0<f(b)$. There are functions $u:[a,b]\to\mathbb{R}$ and $v:[a,b]\to \mathbb{R}$ satisfying the conditions
- $u$ is continuous,
- $v$ is continuous,
- $u$ is strictly monotone (increasing or decreasing),
- $v$ is strictly monotone (increasing or decreasing,
and that fulfills equality $$ f=u+v \qquad (\ast) $$
My attempt. Set $f(x)=a_0+a_1\cdot x+a_2\cdot x^2+\ldots +a_n\cdot x^n$. Let's look for polynomial functions $$ u(x)=\alpha_0+\alpha_1 x+ \alpha_2 x^2+\ldots +\alpha_n x^n \\ v(x)=\beta_0+\beta_1 x+ \beta_2 x^2+\ldots +\beta_n x^n $$ that satisfy the equality $(\;\ast\;)$. So we have to solve the system $$ \alpha_0+\beta_0=a_0\\ \alpha_1+\beta_1=a_1\\ \vdots \\ \quad\\ \vdots \\ \alpha_n+\beta_n=a_n\\ $$ with the restriction that for all $x\in [a,b]$ we have $$ u^\prime(x)\neq 0, \mbox{ that is, } \alpha_1+ 2\cdot\alpha_2 x^{1}+\ldots+ i\cdot\alpha_i x^{i-1} +\ldots +n\alpha_n x^{n-1}\neq 0\\ v^\prime(x)\neq 0 \mbox{ that is, } \beta_1+ 2\cdot\beta_2 x^{1}+\ldots+ j\cdot\beta_j x^{j-1} +\ldots +n\beta_n x^{n-1}\neq 0\\ $$ In other words, we have to show that the set $$ C[x]= \left\{ (\beta_0, \beta_1, \ldots, \beta_n)\in\mathbb{R}^n \left| \begin{array}{l} \beta_1+ 2\cdot\beta_2 x^{1}+\ldots+ i\cdot\beta_i x^{i-1} +\ldots +n\beta_n x^{n-1}\neq 0 \\ \\ (a_1-\beta_1)+ 2\cdot(a_2-\beta_2) x^{1}+\ldots+ j\cdot(a_j-\beta_j) x^{j-1} +\ldots +n(a_n-\beta_n) x^{n-1}\neq 0 \end{array} \right. \right\} $$ is not empty for all $x\in[a,b]$.
But this approach to the problem is not looking very promising. Is there a smarter way to tackle this problem?
Your approach is much too complicated.
Hint: take $u(x) = c x$ where $c > \max_{a \le x \le b} f'(x)$.