# Showing that $\max\{f+g\} \leq \max f + \max g$

Given real-valued continuous functions $f, g$, is the following (and why?) inequality true?

$$\max \{f + g \} \leq \max f + \max g$$

Can someone give me a proof? I suspect the min is the reverse inequality

-
$(f+g)(x)=f(x)+g(x)$ and $f(x)\leq \max f$, and $g(x)\leq \max g$. –  Álvaro Lozano-Robledo Mar 10 '13 at 20:18
Assuming the maxima exist, then yes, that is true. In general, you would have to use the supremum. –  Zev Chonoles Mar 10 '13 at 20:22
Oh so $\max(f + g)(x) \leq \max \max (f + g) \leq \max ( \max f + \max g ) = \max f + \max g$ –  Hawk Mar 10 '13 at 20:22
@sizz That's an interesting way of looking at it. –  goblin Mar 10 '13 at 20:23
Remark: Use the following result. If $f$ has domain $X$, and if $a$ is a real number, and if for all $x \in X$ it holds that $f(x)\leq a$, then $\max f \leq a$. –  goblin Mar 10 '13 at 20:32

## 3 Answers

For all $x$, $$f(x)\le \max f(x)$$ and $$g(x)\le\max g(x).$$ Now add the two together.

-

More generally we have:$$\sup_{x\in A} ( f(x)+g(x))\le \sup_{x\in A} \left( f(x)+\sup_{y\in A} g(y)\right)=\sup_{x\in A} f(x)+\sup_{y\in A} g(y)$$

-

let $\ max g(z):=g(z^*)$,$\ max f(z):=f(z^*)$ then $$f(z)\le f(z^*)$$ $$g(z)\le g(z^*)$$ then $$\forall z : (f+g)(z)=f(z)+g(z)\le f(z^*)+g(z^*)$$ $$\max(f+g)(z)\le f(z^*)+g(z^*)=\ max f(z)+\ max g(z)$$

-
the only way to replace max $g(x)$ by $g(x^*)$ is that g is continuous and the domain should be compact, for example if the domain is $\Bbb R$ and g=x you cant replace max g by $g(x^*)$, that is the max is not necessarily an image. –  i.a.m Mar 10 '13 at 20:52