> [!NOTE] Lemma > Let $I$ be a [[Real intervals|real interval]]. Let $f,g:I\to \mathbb{R}$ be [[Real Function|real functions]]. Then $\sup_{I}(f+g) \leq \sup_{I} f + \sup_{I}(g)$but the two may not be equal, where $\sup_{I}f$ denotes the [[Supremum of Set of Real Numbers|supremum]] of $f(I),$ the [[Image of a set under a function|image]] of $I$ under $f.$ **Proof**: For all $x\in I,$ $f(x)\leq \sup_{I}f$ and $g(x)\leq \sup_{I}g.$ Thus $f(x)+g(x)\leq \sup_{I} f + \sup_{I} g$Therefore $\sup_{I}(f+g) \leq \sup_{I} f + \sup_{I}(g)$ To see that the two sides may not be equal, consider $f:x\mapsto x$ and $g: x \mapsto 1-x$ on the interval $[0,1].$ In this case $\sup_{I} f=\sup_{I} g = \sup_{I} (f+g)=1.$