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Let $f$ and $g$ be functions from $\mathbb{R}$ to $\mathbb{R}$. Prove that:

a.) $\sup\{f(x)+g(x):x\in\mathbb{R}\}\leq \sup\{f(x):x\in\mathbb{R}\}+\sup\{g(x):x\in\mathbb{R}\}$

b.) $\sup\{f(x)+g(y):x,y\in\mathbb{R}\} = \sup\{f(x):x\in\mathbb{R}\}+\sup\{g(y):y\in\mathbb{R}\}$

Attempted proof a.) - Since $f(x)\leq \sup\{f(x):x\in\mathbb{R}$ and $g(x)\leq\sup\{g(x):x\in\mathbb{R}\}$ then we have $$f(x) + g(x) \leq \sup\{f(x):x\in\mathbb{R}\} + \sup\{g(x):x\in\mathbb{R}\}$$ Thus, $f(x) + g(x)$ are bounded from above by $\sup\{f(x):x\in\mathbb{R}\} + \sup\{g(x):x\in\mathbb{R}\}$. So, $$\sup\{f(x)+g(x):x\in\mathbb{R}\}\leq \sup\{f(x):x\in\mathbb{R}\}+\sup\{g(x):x\in\mathbb{R}\}$$

I am not exactly sure how to show b.) any suggestions is greatly appreciated.

Community
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Wolfy
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2 Answers2

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b) Let $x, y \in \mathbb{R}$

Then $f(x)+g(y) \leq \sup f +\sup g$. Taking the supremum for all $x,y$ we get that: $\sup \{f(x)+g(y)| x,y \in \mathbb{R}\} \leq \sup f +\sup g$.

Besides, is obvious that for $x,y \in \mathbb{R}, f(x)+g(y) \leq \sup \{f(x)+g(y)| x,y \in \mathbb{R}\} $

Taking the supremum in $x$, we get $\sup f +g(y) \leq \sup \{f(x)+g(y)| x,y \in \mathbb{R}\} $.

Then taking the supremum in $y$, we get $\sup f + \sup g \leq \sup \{f(x)+g(y)| x,y \in \mathbb{R}\} $

Finally $\sup \{f(x)+g(y)| x,y \in \mathbb{R}\} = \sup f + \sup g $

Dark
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  • I see so you just said that since $$f(x) + g(y)\leq \sup{f(x) + g(y):x,y\in\mathbb{R}}$$ then taking the supremum of $x$ and $y$ we get that $$\sup{f(x):x\in\mathbb{R}} + \sup{g(y):y\in\mathbb{R}} \leq \sup{f(x) + g(y):x,y\in\mathbb{R}}$$ thus from a.) we have equality? – Wolfy Jun 03 '16 at 11:29
  • No it is different from a). Basically to show the equality $a=b$ I show that $a \leq b$ and $b \leq a$ – Dark Jun 03 '16 at 11:31
  • Ok then I don't understand your last two lines where you say we finally have equality – Wolfy Jun 03 '16 at 11:37
  • We have equality because we have shown that $\sup {f(x)+g(y)| x,y \in \mathbb{R}} \leq \sup f + \sup g $ AND $ \sup f + \sup g \leq \sup {f(x)+g(y)| x,y \in \mathbb{R}}$ – Dark Jun 03 '16 at 11:38
  • right the first line is the same as what we wanted to prove for a.) so I dont see why I cant refer to a.) to show we have equality once I have shown the second to last line of your answer – Wolfy Jun 03 '16 at 11:39
  • In a) you refer to the same variable for $f$ and $g$ ($f(x)+g(x))$ but in b) you have two different variables $f(x)+g(y)$ – Dark Jun 03 '16 at 11:41
  • ahh ok, fair enough – Wolfy Jun 03 '16 at 11:42
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Note that in order to prove part b) it is suffices to prove that $$ \sup(A+B)=\sup A+\sup B $$ where $A+B=\{x+y:x\in A,y\in B\}$. Set $\sup A=a$ and $\sup B=b$. We shall prove that $\sup(A+B)=a+b$. We need to prove that $$ \begin{array}{ll} i) & \hbox{$\sup(A+B)\leqslant a+b$} \\ ii) & \hbox{for all $\epsilon>0$, there exist $z_0\in A+B$ such that $a+b-\epsilon<z_0$} \end{array} $$

Proof: i) Since $\sup A=a$ and $\sup B=b$, then $x\leqslant a$ and $y\leqslant b$ for every $x\in A$ and $y\in B$. Therefore $x+y\leqslant a+b$ for every $x+y \in A+B$, as required.

ii) Let $\epsilon>0$. Since $\sup A=a$ and $\sup B=b$, there exist $x_0\in A$ and $y_0\in B$ such that $a-\epsilon/2<x_0$ and $b-\epsilon/2<y_0$. Therefore, $a+b-\epsilon<x_0+y_0$. The assertion then follows.

To prove part a) note that $$ \{f(x)+g(x):x\in R\}\subseteq\{f(x):x\in R\}+\{g(x):x\in R\}\} $$ Therefore $$ \sup\{f(x)+g(x):x\in R\}\leqslant\sup(\{f(x):x\in R\}+\{g(x):x\in R\}\}) $$ The assertion then follows using the identity $\sup(A+B)=\sup A+\sup B$

boaz
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