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Let $C_{b}[0, \infty)$ the space of all continuous and bounded functions $f:[0, \infty) \to \mathbb{R}$ with the supreme norm. Let $I$ the space of all $f \in C_{b}[0, \infty)$ such the $\lim_{x \to \infty}f(x)$ exist. Is this subspace of $I \subset C_{b}[0, \infty)$ closed? For being honest Im really struggling solving this problem. I got the intuition I is not closed, so my idea is using the equivalence of a closed set as set containing all its limit points, this way showing and proving a sequence in $\lbrace f_{n} \rbrace \subset C_{b}[0, \infty)$ such the limit exist for every $f_{n} \in C_{b}[0, \infty) $ but $lim_{n \to \infty} f_{n} \notin I$ im done. The problem is I cannot show a sequence satisfying what I mentioned.

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  • $\begingroup$ I think you should try to prove that it is closed. Think about why you couldn't find a counterexample. $\endgroup$ – Stan Tendijck Feb 27 at 19:42
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The subspace $I$ is indeed closed. To prove so, suppose that $\{ f_n \} \subset I$. If $f_n \to f$ in $C_b([0, \infty))$, then $\sup_{x \in [0, \infty)}|f_n(x) -f(x)| \to 0$. In particular, for any $\varepsilon >0$, there exists $N \in \mathbb{N}$ so that if $n \geq N$, then $|f_n(x) -f(x)| \leq \frac{\varepsilon}{2}$. Since $f_N \in I$, there exist $L \in \mathbb{R}$ and $R > 0$ so that if $x \geq R$, then $|f_N(x) - L| \leq \frac{\varepsilon}{2}$. Thus, by triangle inequality, it follows that if $x \geq R$, then $$ |f(x) - L | \leq |f(x) - f_N(x)| + |f_N(x) - L| \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. $$ That is, $\lim_{x \to \infty} f(x)$ is $L$.

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  • $\begingroup$ Very clear and concise! Thanks a lot! $\endgroup$ – Cos Feb 28 at 19:38
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Hint: if $f \notin I$ there is $\epsilon > 0$ such that for every $N$ there exist $x, y > N$ with $|f(x) - f(y)| > \epsilon$. What does this tell you about $g$ where $\|g - f\|$ is small?

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Usually, when we are stuck at a problem, we can try to investigate a simpler one. Here, maybe the fact that the limit at infinity of each $f_n$ depends on $n$ can be annoying. Therefore, we can have look at the space of the functions which vanish at infinity and show that it is indeed closed. The idea with sequences is good: let $\left(f_n\right)_{n\geqslant 1}$ be a sequence of functions such that $\sup_{x\geqslant 0}\left\lvert f_n(x)-f(x)\right\vert\to 0$ for a function $f\in C_b[0,\infty)$ and that for all $n$, $\lim_{x\to +\infty}f_n(x)=0$. We have to prove that $f$ vanishes at infinity. We can write for all $n\geqslant 1$ and all $x\gt 0$,
$$ \left\lvert f(x)\right\rvert\leqslant \left\lvert f(x)-f_n(x)\right\rvert+\left\lvert f_n(x)\right\rvert\leqslant \sup_{y\geqslant 0}\left\lvert f(y)-f_n(y)\right\rvert+\left\lvert f_n(x)\right\rvert. $$ For a fixed $\varepsilon$, choose $n$ such that the first term in the last expression is smaller than $\varepsilon$. Then use the fact that for this fixed $n$, the function $f_n$ vanishes at infinity.

Now, in the general case, the assumption become $\sup_{x\geqslant 0}\left\lvert f_n(x)-f(x)\right\vert\to 0$ for a function $f\in C_b[0,\infty)$ and that for all $n$, $\lim_{x\to +\infty}f_n(x)=c_n$.

  • First show that the sequence $(c_n)$, by writing for a fixed $x$ and fixed $m,n$, $$ \left\lvert f_n(x)-f_m(x)\right\vert\leqslant \sup_{y\geqslant 0}\left\lvert f_n(y)-f(y)\right\vert+ \sup_{y\geqslant 0}\left\lvert f_m(y)-f(y)\right\vert $$ and letting $x$ going to infinity.

  • Apply the result for the function vanishing at infinity to $g_n:=f_n-c_n$ instead of $f_n$.

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  • $\begingroup$ Thanks for the tips and guidance! Muito Obrigado! $\endgroup$ – Cos Feb 28 at 19:40

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