Is it possible to prove that $|\sup (f+g)| \le |\sup f| + |\sup g|$?

Question

Let C the space of continuous functions $f: [0,1] \to \mathbb R$. How to prove that $|\sup (f+g)| \le |\sup f| + |\sup g|$? I thought that it was a simple consequence of the well-known fact that $\sup (f+g) \le \sup f + \sup g$. But I don't know if the inequality holds when we apply the modulus function on both sides.

Answer 1

The inequality need not hold.

Take $f(x)=-x-1$ and $g(x)=x-2$, so $(f+g)(x) = -3$. Then $|\sup(f+g)|=|-3|=3$. On the other hand, $f$ is decreasing, so $\sup f=f(0)=-1$, and $g$ is increasing, so $\sup(g)=g(1)=-1$, so $|\sup f|+|\sup g|=|-1|+|-1|=2$.

Thus, $|\sup(f+g)|\gt |\sup f|+|\sup g|$ in this instance.

Though of course it is possible it holds, per your subject line. It will hold in many cases, e.g., when the functions are nonnegative, or when $\sup(f+g)\geq 0$, among others.

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In the comments you ask about the inequality $$|\sup(f+g)|+|\inf(f+g)|\leq |\sup f|+|\sup g| + |\inf f| + |\inf g|.$$ This inequality also need not hold; it's a bit easier to find counterexamples, because all we need is a pair of nonnegative functions with $\sup(f+g)=\sup(f)+\sup(g)$ and $\inf(f+g)\gt \inf(f)+\inf(g)$, and those are plentiful

One example is to take $$f(x) =\left\{\begin{array}{ll} 1&\text{if }1\leq x\leq \frac{1}{2}\\ 2-2x&\text{if }\frac{1}{2}\leq x\leq 1 \end{array}\right.\qquad\text{and}\qquad g(x) = \left\{\begin{array}{ll} 1-2x&\text{if }0\leq x\leq\frac{1}{2}\\ 2x-1 &\text{if }\frac{1}{2}\leq x \leq 1. \end{array}\right.$$ It is easy to verify that $\sup(f)=\sup(g)=1$, $\inf(f)=\inf(g)=0$. Now, $$(f+g)(x) = \left\{\begin{array}{ll} 2 - 2x&\text{if }0\leq x\leq\frac{1}{2}\\ 1&\text{if }\frac{1}{2}\leq x \leq 1. \end{array}\right.$$ So $\sup(f+g) = 2$ (achieved at $x=0$), and $\inf(f+g)=1$. So we have $$\sup(f+g) + \inf(f+g) = 2+1 = 3 \gt 2 = \sup(f)+\inf(f)+\sup(g)+\inf(g).$$

Answer 2

The inequality fails already in two dimensional space, or equivalently functions on the two point set $\{0,1\}.$ Consider $(0,-1)$ and $(-1,0).$ Then the left hand side of the inequality is equal $1$ while the right hand side is equal $0.$ This immediately can be extended to $C[0,1]$ by taking strictly monotonic functions $f$ and $g$ such that $(f(0),f(1)) =(0,-1)$ and $(g(0),g(1))=(-1,0).$ Then $f+g$ is negative, so RHS is positive while LHS is equal $0.$ For example $f(x)=-x,$ $g(x)=x-1.$