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How do limits and infimums work with each other? For instance, let $C[0,T]$ be the space of real valued functions on $[0,T]$, how does one justify

$$ \lim_{r\to \infty}\inf_{\{z\in \mathbb{R} : z\geq r\}}\inf_{\{f\in C[0,T]:\sup_{s\in[0,T]}f(s)\geq z\}}\int_0^T f(s)ds=-\infty~~?$$

Theoretically I can see if the limit was taken inside the infimum we would have the infimum over an empty set which is $-\infty$, but im not $100$% sure about the details here.

$\textbf{EDIT :}$ Mistake in my question, really I should of made $f$ positive, I will only add as an edit since people have given answers : So what about the case :

$$ \lim_{r\to \infty}\inf_{\{z\in \mathbb{R} : z\geq r\}}\inf_{\{f\in C[0,T]:\sup_{s\in[0,T]}f(s)\geq z\}}\int_0^T |f(s)|ds=-\infty~~?$$

amWhy
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joedondonjoe
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1 Answers1

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It's a strange thing to be looking at to me because the outer expressions here don't really do anything. The inner infimum is just $-\infty$.

To see this, let $z\in \mathbb{R}$ be given and let $$ f_n(x)= \begin{cases} -n+ \frac{2(z+n)}{T}x & x\in[0,T/2] \\ z-\frac{2(z+n)}{T}(x-\frac{T}{2}) & x\in [T/2,T] \end{cases}, $$ that is the function with graph the upper part of the equilateral triangle with vertices $(0,-n)$, $(T,-n)$ and $(T/2,z)$.

Then, clearly, $\sup_{s\in [0,T]} f_n(s)=z$ and $f_n$ is continuous. However,

$$ \int_0^T f_n(t)\textrm{d}t=2\int_0^{\frac{T}{2}} f_n(t)\textrm{d}t=2 \left(\frac{(n+z)T}{2}-n\right), $$ which goes to $-\infty$ as $n\to\infty$. Accordingly, your expression has the form $\lim_{r\to\infty} \inf_{z\geq r} \{-\infty\}$, which is $-\infty$.

WoolierThanThou
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