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I have encountered the following problem:

Prove that there is no continuous function $U\colon [\,0;1\,]\times[\,0;1\,]\rightarrow\mathbb{R}$, so that $\forall f\in C[\,0;1\,], \quad|f(x)|\leq1\quad \forall x\in[\,0;1\,]$, exists such $y_f\in[\,0;1\,]$ so that $f(x)\equiv U(x,y_f)$.

It ia advised to use Baire theorem, but I have no idea which sets to choose. What sould I try?

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  • $\begingroup$ Are you sure the order of quantifiers in "for all x in [0,1] there exists y in [0,1]" shouldn't be swapped? Because y would depend on x and I'm not sure that is desired. $\endgroup$
    – abnry
    Jan 24, 2017 at 18:58
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    $\begingroup$ @pasta The statement is rather informal, but I believe what he means is $$\forall f\in C([0,1])((\forall x\ |f(x)|\le1)\implies\exists\ y_f\in[0,1]\dots)$$ or in other words, every continuous function $f:[0,1]\to[-1,1]$ is identical to the function $x\mapsto U(x,y_f)$ for some $y_f.$ $\endgroup$
    – bof
    Jan 24, 2017 at 23:16
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    $\begingroup$ @pasta I mean, that absolute value of $f$ is not greater than $1$ for every $x$. Statement about exsistence of $y_f$ is the next one. $\endgroup$ Jan 26, 2017 at 14:30
  • $\begingroup$ Okay, got you. To be precise, quantifiers always precede the property claimed. $\forall x: P(x)$ is correct but $P(x) \; \forall x$ is informal and not technically precise. $\endgroup$
    – abnry
    Jan 26, 2017 at 21:24

3 Answers 3

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Assume that such a $U$ exists. Since $U$ is continuous and $[0,1]\times[0,1]$ is compact, $U$ is uniformly continuous. Therefore we can choose $\varepsilon$ so that $0\lt\varepsilon\lt1$ and, for all $x_1,y_1,x_2,y_2\in[0,1],$ $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\le\varepsilon\implies|U(x_1,y_1)-U(x_2,y_2)|\le1;$$ in particular, $$0\le x\le\varepsilon,\ 0\le y\le1\implies|U(0,y)-U(\varepsilon,y)|\le1.$$ Define a continuous function $f:[0,1]\to[-1,1]$ with $f(0)=1$ and $f(\varepsilon)=-1.$ Then there is no $y_f\in[0,1]$ such that $f(x)\equiv U(x,y_f),$ contradicting the assumed universality of $U.$

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  • $\begingroup$ Nice, I can only award the bounty tomorrow however (having posted it today). Do you see a method to do with Baire? $\endgroup$
    – s.harp
    Jan 24, 2017 at 14:45
  • $\begingroup$ Yes, Baire theorem application will be very much appreciated. $\endgroup$ Jan 24, 2017 at 18:41
  • $\begingroup$ You mean the Baire category theorem? No, I'm sorry, I don't see how to use that. Is the use of BCT mandatory? $\endgroup$
    – bof
    Jan 24, 2017 at 23:10
  • $\begingroup$ @s.harp No, leave the bounty open until someone posts a solution using Baire's theorem. $\endgroup$
    – bof
    Jan 24, 2017 at 23:22
  • $\begingroup$ Ok, I think your answer is fine as an answer, but if somebody does it with Baire 'll give them the bounty :). If it doesn't look like anybody will answer before the bounty expires I'll give it to you so you get the $100$ and not half. $\endgroup$
    – s.harp
    Jan 25, 2017 at 19:17
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Consider the functions $f_n(x)=x^n$ which are in $C[0,1]$ and have $|f_n(x)| \leq 1$. Assuming the conclusion, there exists a sequence of $y_n$ such that $U(x,y_n)=x^n$. Because $y_n \in [0,1]$, there exists a convergent subsequence $y_{n_k} \to y \in [0,1]$. By continuity of $U$ we must have $U(x,y) = \chi_{\{1\}}(x)$. As pointed out in the comments, this violates the continuity of $U$.

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    $\begingroup$ $U(x,y')=\chi_{\{1\}}(x)$ immediately contradicts continuity of $U$. This proof makes the result seem very easy and it seems very strange that the Baire Category Theorem is suggested as useful! $\endgroup$
    – s.harp
    Jan 27, 2017 at 15:47
  • $\begingroup$ You are right. Thanks for making that point. I agree, using Baire Category seems strange without any additional context. However, I bet there is a way to approach it that would generalize to other cases. If the problem prompted a way to start a BCT argument, that'd be better. $\endgroup$
    – abnry
    Jan 28, 2017 at 1:54
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Write $$ g :[0,1]\rightarrow C([0,1])\\ y\rightarrow g(y)=U(\cdot,y) $$ It's not difficult to see that $g$ is continuous ($C([0,1])$ has the $\left \| \cdot \right \|_{\infty}$ topology). Since $[0,1]$ is compact, $g([0,1])$ is compact. The hypotesis asserts that $g([0,1]) = \{ h\in C([0,1]):\left \| h \right \|_{\infty}\leq 1 \}=:B$. But $B$ is not compact, because $C([0,1])$ has infinite dimension.

pd. This doesn't use the Baire's theorem, but I think we can use it by showing that compact subsets of $C([0,1])$ has empty interior.

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  • $\begingroup$ I think you mean $g:[0,1]\to C([0,1])$, and that $g$ is continuous (not $U$). I don't directly see how $g$ is continuous though, can't quite combine the uniform continuity of $U$ and the compactness of $[0,1]$ in the right way. $\endgroup$
    – s.harp
    Jan 30, 2017 at 10:12
  • $\begingroup$ Let $\epsilon >0,\ x\in [0,1],\ f=g(x)$. Take $\delta>0$ such that $\left \| (a,b)-(c,d) \right \| < \delta \Rightarrow |U(a,b)-U(c,d)|<\epsilon$. If $|y-x|<\delta$, we have, for all $z\in [0,1]$: $$ \| (z,y)-(z,x) \| = |y-x| < \delta $$ Then, $| g(y)(z) - g(x)(z) | = | U(z,y) - U(z,x) |< \delta$. So, $\| g(y)-g(x)\|_{\infty} < \epsilon$ $\endgroup$ Jan 30, 2017 at 10:58
  • $\begingroup$ Oh right. Was thinking in the wrong direction. $\endgroup$
    – s.harp
    Jan 30, 2017 at 12:37

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