"$\epsilon-\delta$" for $\lim_{x\to \infty}\sup_{\alpha\in [0,1]}f_{\alpha}(x)$ is bounded

Question

Suppose that I proved that $$\lim_{x\to \infty}\sup_{\alpha\in [0,1]}f_{\alpha}(x)\le C$$

where $C$ is a constant.

Does it means that there exist a constant $A$ such that $f_{\alpha}(x)\le C$ holds for every $\alpha\in[0,1]$ and $\vert x\vert\ge A$ ?

Answer 1

Let $l$ be that limit and we fix $\epsilon>0$. Then there exists $A$ such that for each $x\geq A$ we have

$$ \sup_{\beta\in[0,1]}f_\beta(x)-l<\epsilon . $$

For an arbitrary $\alpha \in [0,1]$ we have $$ f_\alpha(x)\leq \sup_{\beta\in[0,1]}f_\beta(x)<l+\epsilon . $$ Thus for each $\epsilon>0$ we have $f_\alpha(x)\leq l+\epsilon\to_{\epsilon\to 0^+}l\leq C$, which means

$$ f_\alpha(x)\leq C \quad\text{for each}\quad x\geq A .\tag{1} $$ Since $\alpha$ was chosen arbitrarily in $[0,1]$ equation $(1)$ holds for every $\alpha \in [0,1]$.

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Concerning your question in the comment: It is not possible to choose $A=1$ in general because, for example, regard

$$ f_\alpha(x)=f(x)=\chi_{[0,2]} . $$

Then you have that $l\leq C=\frac{1}{2}$ but for $|x|\geq 1$ it is not true that $f(x)\geq 1$ because, for example, if $x=1$, you have

$$f(1)=1\not \leq \frac{1}{2}.$$

Answer 2

Well given some $\beta\in [0,1]$, we have that for any $x$, $f_\beta (x) \leq \sup_{\alpha\in [0,1]} f_\alpha (x)$. So in particular, it's true that $$\lim_{x\to\infty} f_\beta (x) \leq \lim_{x\to\infty} \sup_{\alpha\in [0,1]} f_\alpha (x) \leq C$$ for every $\beta\in [0,1]$. We can write this as $$ \lim_{x\to\infty} (C - f_\beta (x) ) \geq 0 $$

By definition, we can take arbitrary $\epsilon >0$ and rephrase this as:

There exists an $A >0$ such that for all $|x| \geq A$, $$ C - f_\beta (x) > -\epsilon $$ $$\implies f_\beta (x) < C + \epsilon $$ $$\implies f_\beta (x) \leq C $$