Let $\mathcal{C}[0,1] $ be the space of continuous functions on the interval $[0,1]$ and define $ \| f \|_1 = \int_{0}^{1} |f(x)| \ dx \quad \text{and} \quad \| f \|_0 = \int_{0}^{1} x |f(x)| \ dx. $ Assume that $ \| \cdot \|_1 $ is a norm on $ \mathcal{C}[0,1] $. I was having a difficult time showing the property that $ \| f \|_0 = 0 \implies f \equiv 0$. I can't help but feel there must be a slick way to do it since I didn't use the first norm $ \| \cdot \|_1 $ . Here is my proof (?) of the "separates points" property of a norm for $ \| \cdot \|_0 $ :
Suppose $ f \not\equiv 0$. Using integration by parts, we set $ u = x $ and $dv = |f(x)| \ dx $. Then, $ \int_{0}^{1} x |f(x)| \ dx =\left[ x \cdot \int_{0}^{x} |f(t)| \ dt \right]_{0}^{1} - \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx = \int_{0}^{1} |f(t)| \ dt - \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx. $
By the fundamental theorem of calculus, $ v(x) = \int_{0}^{x} |f(t)| \ dt, \quad \text{with} \ x \in [0,1] . $
Observe that for all $x \in [0,1]$, $ v(x) = \int_{0}^{x} |f(t)| \ dt \leq \int_{0}^{1} | f(t) | \ dt = M, $ which follows from the fact that our function is positive. Now, since we're assuming $ f(t) $ is not $0$, $ M > 0$. And as $v(0) = 0 $, the continuous function $ v(x) $ is ``strictly bounded above" by $ M$. Hence,
$ \int_{0}^{1} v(x) \ dx < \int_{0}^{1} M \ dx \iff \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx < \int_{0}^{1} \int_{0}^{1} | f(t) | \ dt \ dx, $ where the inequality here is strict. It follows that
$ \|f \|_0 = \int_{0}^{1} |f(t)| \ dt - \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx > \int_{0}^{1} |f(t)| \ dt - \int_{0}^{1} \int_{0}^{1} |f(t)| \ dt \ dx = 0. $
We conclude $ f \not\equiv 0 \implies \|f \|_0 > 0 $.