Supremum of continuous function on an unbounded interval

Question

Let f(x) be a continuous function on an unbounded interval $[a,\infty)$. If $\lim_{x\to\infty}f(x)=c<1$ and $f(x)<1$ $\forall x\geq a$, can we conclude that $\sup\{f(x)\}<1$? Why or why not?

Answer 1

I think yes.

Since $\lim_{x \to \infty}f(x)=c<1$, for $\epsilon$ small enough there exists $M >a$ such that $f(x)<c+\epsilon<1$ for all $x>M$.

By contradiction, assume that $\sup_{x\geq a} f(x)=1$. Since $f(x)<1$ for all $x>M$, we must have $\sup_{a\leq x \leq M} f(x)=1$. Since $f(x)$ is continuous and $[a,M]$ is compact, $f$ achieves supremum on $[a.M]$, i.e., there exists $x_0 \in [a,M]$ such that $f(x_0)=1$, contradiction to assumption $f(x)<1$ for all $x \geq a$.

Answer 2

By contradiction : if the supremum is equal to $1$ there is an increasing sequence $u_n$ such that $f(u_n) \rightarrow 1$.

Since $f(\infty) = c < 1$, $u_n$ must converge to some $s \in [a, \infty)$, but by continuity of $f$, $f(u_n) \rightarrow f(s) = 1$ which contradicts the fact that $f<1$

Answer 3

yes: we know that: $\lim _{x \rightarrow \infty} f(x)=c<1$. we can infer from this that there exists a b, such that for all x>b: $f\left(x\right)<1-\dfrac{1-c}{2}$. now lets look at the values of f on [a,b]. the function is continues, and therefore : $$sup_{x\in\left[a,b\right]}\left(f\left(x\right)\right)=max{}_{x\in\left[a,b\right]}\left(f\left(x\right)\right)$$ but we know that: $max_{x\in\left[a,b\right]}\left(f\left(x\right)\right)<1$. All together we get that for every x>a: $$f\left(x\right)\leq max\left(max_{x\in\left[a,b\right]}\left(f\left(x\right)\right),1-\dfrac{1-c}{2}\right)<1$$ which finishes the proof