TITLE: Application of Baire theorem for proof of non-existence of a universal function
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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.