Let $f : [0,\infty)\to\mathbb{R}$ be a non-negative continuous function and $\limsup\limits_{t\to\infty}f(t)=M<\infty$. Can we deduce that $f$ is bounded from above?
Here is my attempt so far :
I take any random sequence $(t_{n})_{n\in\mathbb{N}}\subset[0,\infty)$ such that $t_n\to\infty$ as $n\to\infty$. Then, by definition of limit superior, I fix $\varepsilon=1>0$ so that I can have $N_{0}\in\mathbb{N}$ such that for any $n>N_{0},\,f(t_{n})<M+1$. Since $t_{N_{0}}<\infty$, then I can deduce $\max\limits_{t\in[0,t_{N_{0}}]}f(t)<\infty$ by continuity of $f$ and $\forall n>N_{0}, f(t_n)<M+1<\infty$. So, my question is how to show that $\sup\limits_{t\in(t_{N_{0}},\infty)}f(t)<\infty$? I know that I choose any arbitrary sequence in the beginning but I cannot see how it helps here.
Any help is much appreciated!
For each closed interval $[n,n+1]$, $n=0,1,...$, take $x_{n}\in[n,n+1]$ as $f(x_{n})$ being maximal on it, so, $f(x_{n})=\sup_{x\in[n,n+1]}f(x)$. Now it is easy to see that $x_{n}\rightarrow\infty$, and hence $\limsup_{n}f(x_{n})\leq M$. But then $\sup_{x\in{\mathbb{R}}}f(x)=\sup_{n}\sup_{x\in[n,n+1]}f(x)=\sup_{n}f(x_{n})$. If $\sup_{n}f(x_{n})$ is attained by some $f(x_{n_{0}})$, then we are done. If it is not attained, we can extract a subsequence $(f(x_{n_{k}}))$ that converges to $\sup_{n}f(x_{n})$, for then $\sup_{n}f(x_{n})\leq\limsup_{n}f(x_{n})\leq M$.