# Suprema proof: prove $\sup(f+g) \le \sup f + \sup g$

Suppose D is a nonempty bounded subset of reals. let $f:D \to R$ and $g:D \to R$. Define $(f+g)(x)=f(x)+g(x)$. Prove $\sup(f+g)(D) \le \sup f(D) + \sup g(D)$ (also prove that $\sup (f+g)$ exists).

I understand why this is the case, just not how to prove it. Left side is pretty much $\sup (f(x)+g(x))$ and right side is $\sup (f(x) + g(y))$ for $x,\,y \in D$. Basically $f+g$ has to use the same variable and $f(D)+g(D)$ use different ones. But I don't know how to go about proving this.

The second part of the question is to find a specific example where strict inequality holds. Let $D=\{a,b\}$ and $f: a \to 1,\, b\to 0, \,g: a \to 0,\, b\to 1$. $\sup f(D) = 1,\, \sup g(D) = 1,\, \sup f(D) + \sup g(D) = 2.$ $\sup (f+g)(D) = 1$ (if we choose a, $f+g = 1+0,\, b,\, f+g=0+1$).

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Welcome to Math.SE. You can use $\LaTeX$ markup to build mathematical expressions. I've fixed some of your question. Click on edit to see how it's done (and fix the rest if you like). :) –  Ayman Hourieh Oct 4 '12 at 19:26

Consider $x\in D$. Then $f(x)\le \sup f$ and $g(x)\le \sup g$, hence $(f+g)(x)=f(x)+g(x)\le \sup f+\sup g$. Therefore, $\sup f + \sup g$ is some upper bound, but the least upper bound may be smaller.

Your example for strictness is fine.

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Wow, that was obnoxiously obvious. Thank you –  fhyve Oct 4 '12 at 18:51
Or, in the last step, the way I like to think about it is to take the sup of both sides. The right hand side is a constant so it stays the same. This could also work for things like limits. –  Graphth Oct 4 '12 at 19:28
So $\sup (f(x)+g(x)) \le \sup (\sup f + \sup g) = \sup f + \sup g$? –  fhyve Oct 4 '12 at 19:35

Another example for < from Abbott

The two sides are usually not equal because the functions f and g could easily take on their larger values in different places of each subinterval. For example, consider f(x) = x and g(x) = 1 − x on the interval [0, 1]. Then $\sup\{f(x) : x ∈ [x_{k−1}, x_k]\} = 1$,

$\sup\{g(x) : x ∈ [x_{k−1}, x_k]\} = 1$,

but $\sup\{\color{seagreen}{f(x) + g(x) = 1} : x ∈ [x_{k−1}, x_k]\} = \sup\{\color{seagreen}{1}\} = 1 \neq 2$

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The pdf doesn't expatiate on this, but $\sup f(x) + g(x) = 1$ at both $x = -1, 1$.
pdf simply chagrins about $x = 1$. But at $x = -1$, $f(-1) + g(1) = 1 + 0 = 1$.