$ \min_{x} [g(x) + h(x)] = \min_x g(x) + \min_{x} h(x)$ proof?

Question

So given $$\min_ \left.x\in[a,b \right] \left[g(x) + h(x) \right]= \min_{x\in[a,b]} g(x) + \min_{x\in[a,b]}h(x),$$

I am asked to prove or provide a case where this doesn't hold. I should also mention that x $\in$ [a,b] and [a,b] -> $\mathbb{R}$.

I don't really know how to approach this. If I wanted to find a contradiction to this I would simply find some functions g and h constrained to [a, b] and show that their respective minimum values equate. Is this a correct method? If this is in fact true then obviously I have to find a proof and I don't know how to do that either. So yeah, is the above correct and do I have to prove it?

Answer 1

It doesn't hold in general. Consider, for example, $g(x) = x$ and $h(x) = -x$ where $[a,b] = [0,1]$. Thus we obtain the counter-example \begin{align*} \min_{x\in[0,1]}\{g(x) + h(x)\} = \min_{x\in[0,1]}\{0\} = 0 \neq -1 = \min_{x\in[0,1]}\{g(x)\} + \min_{x\in[0,1]}\{h(x)\} \end{align*}

Answer 2

Hint: consider two functions $f,g$ where the minimum of each one appears at different points $x_1,x_2\in{}[a,b]$. For example take the simplest case $f(x)=1/2$ at $x_1$ and $f(x)=1$ elsewhere...

Answer 3

the left hand side is larger or equal to the right-hand side but equality may not hold. For instance, $g(x)=x^2$ was minimum $0$ in $[-1,1]$ and $h(x)=-x^2$ has minimum $-1$ in $[-1, 1]$. However, the minimum of $g+h$ is $0$.