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This should be an easy question, but I am kind of unsure today.

I have a compact subset $I \subset \mathbb{R}$ and two real valued functions $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$, where $g(x) >0$.

Do I then have:

$$\frac{\sup_{t\in I}|f(t)|}{\sup_{t\in I}g(t)} \leq \sup_{t\in I} \frac{|f(t)|}{g(t)}?$$

I claim yes, since $\sup_{t\in I}g(t) \geq g(t)$ for all $t \in I$ from which the assertion then already should follow.

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  • $\begingroup$ It follows from the inequality $\sup(fg) \le \sup f\sup g$ for non-negative $f$, $g$. $\endgroup$ – Hanul Jeon Jan 30 '17 at 15:05
  • $\begingroup$ Coding $\sup$ as \sup with a backslash not only prevents italicization but also results in proper spacing in things like $a\sup A$ and $a\sup(A)$ and, in a displayed as opposed inline setting, affects the positions of subscripts, thus: $$ \sup_{t\in I} f(t). $$ It is standard usage and I edited accordinly. Similarly \max, \det, \log, \sin, etc. $\endgroup$ – Michael Hardy Jan 30 '17 at 17:28
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Hint: If $u$ and $v$ are nonnegative on $I$, then $\sup\limits_I (uv) \leq (\sup\limits_I u)( \sup\limits_I v)$.

Consider $u=g$ and $v=|f|/g$.

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