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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

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    $\begingroup$ $(f+g)(x)=f(x)+g(x)$ and $f(x)\leq \max f$, and $g(x)\leq \max g$. $\endgroup$ – Álvaro Lozano-Robledo Mar 10 '13 at 20:18
  • $\begingroup$ Assuming the maxima exist, then yes, that is true. In general, you would have to use the supremum. $\endgroup$ – Zev Chonoles Mar 10 '13 at 20:22
  • $\begingroup$ Oh so $\max(f + g)(x) \leq \max \max (f + g) \leq \max ( \max f + \max g ) = \max f + \max g$ $\endgroup$ – Hawk Mar 10 '13 at 20:22
  • $\begingroup$ @sizz That's an interesting way of looking at it. $\endgroup$ – goblin Mar 10 '13 at 20:23
  • $\begingroup$ 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 $\max f \leq a$. $\endgroup$ – goblin Mar 10 '13 at 20:32
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For all $x$, $$f(x)\le \max f(x)$$ and $$g(x)\le\max g(x).$$ Now add the two together.

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

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

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    $\begingroup$ 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. $\endgroup$ – i.a.m Mar 10 '13 at 20:52

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