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A family $\mathcal{A}\subset C([0,1])$ of functions is called uniformly bounded if $$\sup_{x\in\mathcal{A}}\sup_{t\in[0,1]}|x(t)|<\infty.$$ Can I interchange the order of the supremums in the definition, i.e. $$\sup_{x\in\mathcal{A}}\sup_{t\in[0,1]}|x(t)|=\sup_{t\in[0,1]}\sup_{x\in\mathcal{A}}|x(t)|?$$

Ѕᴀᴀᴅ
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RandomUser
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  • What have you tried? You should try to check this on your own, just use the definitions – Thomas Mar 07 '18 at 09:27
  • I think that $\sup_{i}\sup_{j}a_{ij}$ is the same as $sup_{i,j}a_{ij}$ and therefore I can interchange the supremums always. But is this enough to prove that they are equal? – RandomUser Mar 07 '18 at 09:39

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let $\infty > L=\sup_{x\in \cal{A}}\sup_{t\in [0,1]} |x(t)|$ and $R=\sup_{t\in [0,1]} \sup_{x\in \cal{A}}|x(t)| $

You have to show $L\le R $ and $R\le L$.

Let $\varepsilon >0$. By definition of $L$, there is $x_L\in A$ such that $$L<\sup_{[0,1]}|x_L(t)| + \varepsilon $$

Since $c$ is continuous, the $\sup$ in the last inequality is attained, so there is $t_0\in[0,1]$ (depending on $x_L$) such that $$L< x_L(t_0) + \varepsilon$$ Now the rhs is certainly less than or equal to $\sup_{x\in \cal{A}}|x(t_0)|+\varepsilon$ (since by adding choices you can only increase the sup), which in turn is certainly less than or equal to the $\sup_t$ of this last expression $+\varepsilon$, which is just $=R+ \varepsilon$. So $L< R +\varepsilon$ for any $\varepsilon >0$ which implies $L\le R$.

Can you do the reverse inequality on your own?

Thomas
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  • Thank you very much, it's clear to me how to prove the reverse inequality. Does this property hold for any supremums or are there examples where I cannot exchange the supremums? – RandomUser Mar 07 '18 at 12:14
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    @RandomUser have a look at this question: https://math.stackexchange.com/questions/53794/simple-question-the-double-supremum. Be careful, however, when it comes to $\inf\sup$ -- see e.g. here: https://math.stackexchange.com/questions/31030/exchangeability-between-inf-min-and-sup-max – Thomas Mar 07 '18 at 13:11