Why is this set open? How to prove it?

Question

Let $f\in C_b(\mathbb{R})$ and $g$ any continuous function on $\mathbb{R}$ (or maybe there has to be a restiction on $g$?). Now let K be a compact subset of $\mathbb{R}$ and $U$ an open subset of $\mathbb{R}$. Then I want to show that the following set $$\Phi:=\{t\in\mathbb{R}_+:(e^{t\cdot g}f)(K)\subseteq U\}$$ is a open set. Hence I have to find for every $t\in\Phi$ an $\varepsilon>0$ such that $s\in\Phi$ whenever $|s-t|<\varepsilon$. Why is this true. Maybe the argument is simple but I don't see it. Already thank you.

Answer 1

Let $t\in\Phi$ and write $h=e^{tg}f$. Note $h$ is continuous so bounded on $K$. For $x\in K$:

$$|e^{(t+a) g(x)}f(x)-e^{tg(x)}f(x)|=|e^{ag(x)}h(x)-h(x)|≤\|h\|_K\cdot |1-e^{ag(x)}|≤\|h\|_K |1-e^{|a|\cdot \|g\|_K }|\tag{1}$$ So it is clear, for every $\epsilon>0$ one can choose a $\delta$ independent of $x$ so that if $a\in(-\delta,\delta)$ that then the expression in $(1)$ is smaller than $\epsilon$.

What we will do now is that we will show there exists and $\epsilon>0$ so that if $y\in h(K)$ that then $B_\epsilon(y)\subset U$. This implies from before that there exists a $\delta$ so that $(t-\delta,t+\delta)\subset\Phi$ and we would be finished.

This follows from $\mathbb R-U$ being closed and disjoint from the compact $h(K)$. Namely the infimum: $$\inf\{d(x,y)\mid x\in h(K), y\in \mathbb R-U\}$$ is realised and actually a minimum. The minimum cannot be zero, since the sets are disjoint.

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To see the minimum is realised, consider an $A$ so that $[-A,A]\supset h(K)$ and that contains points $\mathbb R -U$. Then every point outside of $[-2A,2A]$ has a distance from $h(K)$ that is greater than the distance from any point in $[-A,A]$ from $h(K)$. So we can restrict to the minimal distance to $[-2A,2A]\cap(\mathbb R-U)$ being realised, but since that set is compact it must be realised.

Answer 2

For a compact set $C$ which is contained in an open set $U$ there is open ball around $1$ such that $B_{\varepsilon}(1) \cdot C \subset U$.

Then you have $$ e^{B_{\delta}(t)\cdot g(K) }f(K) = e^{B_{\delta}(0)}e^{t\cdot g(K)}f(K)=B_{\varepsilon}(1) \cdot C \subset U $$ where $\varepsilon = \log \delta$ and $e^{t\cdot g(K)}f(K) =C$.

How to prove the first assertion? Assume $(\Bbb R \setminus U)\cap (\overline{ B_{\varepsilon}(1)} \cdot C)\ne \emptyset$, these sets are bounded and closed, thus compact, so their intersection(take $\varepsilon = \frac{1}{n}$ for $n>0$) contains an element which has distance zero to $C$ which is a contradiction.