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Here is the statement: Suppose $f$ and $g$ are real-valued functions with a common domain $D\subset R$, such that $f(x)\le g(x)$ for each $x \in D$. Show that $inf_{x\in D}f(x) \le inf_{x \in D} g(x)$.

Here is my attempt at the proof: Given $f$ and $g$ are real-valued it follows that $f$ and $g$ are bounded, then let $u = inf_{x \in D} f(x)$ and $v = inf_{x \in D} g(x)$ be defined. By definition $u\le f(x)$ for each $x\in D$, and likewise $v\le g(x)$ for each $x \in D$. Since $f(x)\le g(x)$ for all $x \in D$ we can guarantee that $u$ is a lower bound for $g$ so that $u\le g(x)$. Since $v$ is the greatest lower bound of $g$ it follows that $u\le v$. Q.E.D.

It feels incomplete, I can't help but think that my logic concerning $u$ as a lower bound for $g$ is insufficient to conclude that $u \le v$. Any suggestions?

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    The statement, "it follows that $f$ and $g$ are bounded" is not true, but also not necessary for the proof, as long as we're ok with infimums being $-\infty$. Taking that out, your proof is just fine. – 2017-02-09
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    Please explain why "it follows that f and g are bounded" is not true. Is it not the case that real-valued functions are not bounded? – 2017-02-09
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    The function $f:\mathbb{R} \to \mathbb{R}$ defined by $f(x) = x$ is not bounded. – 2017-02-09
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    Ah, I've got mistaken logic. I assumed the reals are bounded which is clearly not the case. Thanks. – 2017-02-09

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