Find continuous functions on $[0,1]$ satisfying this inequality. For a real valued function $f \in C([0,1])$ define $s(f) =\max _{x \in [0,1]}\mid x \cdot f(x) \mid$. One can easily check that $s$ defines a norm on $C([0,1])$. Let $n \in \mathbb{N}$. I am trying to find  functions $f, g \in C([0,1])$ such that  $s(fg) &gt; n \cdot s(f)s(g)$.   

I am having trouble finding these $f,g$ explicitly. I messed around with $f,g$ beeing polynomials of degree $n$, but I am not making a lot of progress. Some help would be appreciated.
Take $f(x) = \frac{|\sin (\pi k x / 2)|}{x}$ on $[0,1]$, with the understanding that $f(0) = \frac{\pi k }{2}$, so that $f$ is continous on $[0,1]$, where we take $k &gt; n$. Then $s(f) = 1$ and $s(f^2) = \sup_{x \in [0,1]} |x f(x)^2| \geq |1/k \cdot f(1/k)^2| = k &gt; n$, so taking $g = f$ will give you examples for any $k &gt; n$.
Let's take $f(x) = g(x) = 1-x$. Then $s(f)=s(g)=max \left|x \cdot (1-x) \right| = \frac{1}{4}$ for $x=\frac{1}{2}$.

If we take $s(fg)=max \left|x \cdot (1-x)^2 \right| = \frac{4}{27}$ for $x=\frac{1}{3}$.

At last $\frac{4}{27} &gt; n \cdot \frac{1}{16}$ is correct for $n=1$ and $n=2$.

p.s. Feeling a certain guilt and not having an answer, I hurry to help, as far as I can. Once again, I apologize if something was/is wrong.