Show that there is a continuous function $g$ over $\mathbb{R}$ such that $0 < g(x) \leq f(x)$ Suppose $f$ is a function over $\mathbb{R}$ such that $f(x) &gt; 0$ for all $x \in \mathbb{R}$ and that $f$ is strictly decreasing. Show that there is a continuous function $g$ over $\mathbb{R}$ such that $0 &lt; g(x) \leq f(x)$ and that $\displaystyle \lim_{x \to \infty} \frac{g(x)}{f(x)} = 0$.


Attempt:

We have $f(x) &lt; f(y)$ if $x &lt; y$ and $f(x) &gt; 0$ for all $x \in \mathbb{R}$. I am not sure how to take into account the condition that $0 &lt; g(x) \leq f(x)$, but the condition that $\displaystyle \lim_{x \to \infty} \frac{g(x)}{f(x)}$ means that $f(x)$ grows faster than $g(x)$, so that intuitively fits the previous definition. Maybe proof by contradiction will work here?
For $n\in\Bbb N$ let $g(n)=\frac1nf(n+1)$ and linearly interpolate between these points. That is, let
$$ g(x)=\begin{cases}f(2)&amp;x\le 1\\
(x-\lfloor x\rfloor)\frac1{\lceil x\rceil}f(\lceil x\rceil+1)+(\lfloor x\rfloor +1-x)\frac1{\lfloor x\rfloor}f(\lfloor x\rfloor +1)&amp;x&gt; 1\end{cases}$$
and verify that is is indeed continuous (the only "problematic" points being $x\in\Bbb N$).
Also, $g$ is clearly positive.
For $n\le x&lt;n+1$ we have $g(x)\le \frac 1nf(n+1)$ and thus $0&lt;\frac{g(x)}{f(x)}\le \frac 1n$. We conclude that $\lim_{x\to\infty}\frac{g(x)}{f(x)}=0$.



Alternatively: The function $h(x)=\frac{f(x)}{\max\{1,x\}}$ is also positive and strictly decreasing. It suffices to find continuous $g$ with $0&lt;g(x)&lt;h(x)$. Such $g$ can be found by "mollification", e.g. simply
$$g(x)=\int_0^1 h(x+t)\,\mathrm dt $$
The key point is that the monotonicity of $h$ implies integrability and that the integral is continuous as a function of $x$.
For simplicity, only consider $x \ge 0$. Since $f(x)$ is strictly decreasing, we know there are countable jump points $x_0 = 0 &lt; x_1 &lt; x_2 &lt; ..&lt;x_n &lt; ...$ for $f(x)$. $f(x)$ is continnous in $(x_i, x_{x+1})$, and f(x) has finite left limit $f(x_i^+)$ at $x_i$ and the finite right limit $f(x_{i+1}^-)$ at $x_{i + 1}$. 

We notice that $f(x_i^-) &gt; f(x_i^+)$

Now we can construst a continuouds function $G(x)$ from f(x) by connecting $f(x_i - \epsilon_i)$ to $f(x_i^+)$ with the straight line, otherwise, G(x) = f(x), where $0 &lt; \epsilon_i &lt; (x_i - x_{i - 1})$.

The final $g(x) = G(x)/( x + 1)$.  Clearly,
$$0 &lt;= g(x) \le f(x),$$ and $$limit_{x \to \infty} \frac{g(x)}{f(x)} \le limit_{x-&gt;\infty} \frac{1}{1 + x} = 0 $$