# Proof that $\liminf_{x\to\infty}f(x) \leq \limsup_{x\to\infty}f(x)$.

I can't understand this proof from my old lecture notes. $\liminf$ is defined as: \begin{align*} \liminf_{z\to\infty}f(z) = \inf_{x< y} \sup_{y< z}f(z) \end{align*} and $\limsup$ is defined similarly. We have already shown that if $f$ is an increasing (decreasing) function then $\sup f$ ($\inf f$) is unique, if it exists, hence the definition of liminf is safe. We have also shown that \begin{align*} \inf_{y<z} f(z) \leq \sup_{y<z} f(z). \end{align*} We assume that both $\liminf f$ and $\limsup f$ exist. The proof is presented as follows: \begin{align*} \inf_{y<z} f(z) &\leq \sup_{y<z} f(z)\\ \sup_{x<y} \inf_{y<z} f(z) &\leq \sup_{x<y} \sup_{y<z} f(z)\\ \inf_{x<y} \inf_{y<z} f(z) &\leq \inf_{x<y} \sup_{y<z} f(z) \end{align*} I don't see how this proves the theorem. Could somebody please explain the conclusion?

P.S. At this stage we do not have $\lim$ in our toolbox.

• Are you sure your definition is correct? Because it makes very little sense, with three variables where $x$ is only ever under the expression... – 5xum Apr 18 '16 at 10:16
• Yes, in this context it makes because $\sup_{y< z}f(z)$ is decreasing in $y$ and hence $\inf_{x< y} \sup_{y< z}f(z)$ is independent of $x$, if it exists. The theorem assumes that $\liminf_{x\to\infty} f(x)$ exists so we are safe. – lampishthing Apr 18 '16 at 10:21
• You are missing my point. The point is that $\sup_{y<z}f(z)$ is independent of $x$, therefore $\inf_{x<y}\sup_{y<z} f(z)$ is the infimum of a constant, which means $\inf_{x<y}\sup_{y<z} f(z)=\sup_{y<z} f(z)$, which is a function of $y$. – 5xum Apr 18 '16 at 10:40

## 1 Answer

The correct definitions should be $$\liminf_{z\to\infty} f(z)=\sup_y\inf_{z\ge y}f(z),\qquad\limsup_{z\to\infty} f(z)=\inf_{y}\sup_{z\ge y}f(z).$$ Here is two-line proof: $\;\inf_{z\ge y} f(z) \leq \sup_{z\ge y} f(z)$, hence \begin{align*} \inf_{z\ge y} f(z) &\leq \inf_y\sup_{z\ge y} f(z)&&\qquad \text{by definition of the g.l.b.}\\ \sup_y\inf_{z\ge y} f(z) &\leq \inf_y\sup_{z\ge y} f(z)&&\qquad \text{by definition of the l.u.b.} \end{align*}

• Thank you. Literally "by definition of the g.l.b." clicked it for me. My sticking point was thinking that the inf sup on the RHS doesn't necessarily have to be a sup and as such could slip below the inf on the LHS. The LHS is an l.b. but the inf on the RHS is the g.l.b. and thus fits between. Gah. Thanks again. – lampishthing Apr 18 '16 at 11:52