This is a problem in a homework assignment (which I turned in, and has been graded) from some time ago.
Let $w$ be a continuous, nonnegative, real function on $[0,1]$. Define the norm $\|.\|_w$ on $C([0,1])$ by: $$ \sup_{x\in [0,1]}w(x)|f(x)| $$
It is easy to see this is a norm and to check all the necessary properties. Moreover, this norm is equivalent to the usual $\|f\|_\infty$ norm when $w >0$. However, the question asks to show that this norm is not equivalent to $\|f\|_\infty$ when $w(x) = x$. That is, the norm: $$ \|f\|_w = \sup_{x\in [0,1]}x|f(x)| $$ is not equivalent to the infinity norm. I provided the following solution which was marked as incorrect:
Note that for $w(x) = x$, the equivalence cannot occur due to the second inequality (that is $m\|f\|_\infty < \|f\|_w$, $m>0$) not being fulfilled. Specifically, take the family of functions defined by $f_\epsilon (x) = \frac{1}{x + \epsilon}$. These functions are all continuous on $[0,1]$. Now, notice that for any of these functions he maximum possible value of $\|f_\epsilon\|_w = \sup_{x\in [0,1]} \frac{x}{x+\epsilon}$ is approximately $1$. However, $\|f_\epsilon\|_{\infty} = \frac{1}{\epsilon}$, which is unbounded. Thus, no choice of $m$ will work for all $\epsilon$, as the left hand side will grow unbounded.