Let $\mathcal Y=C([a,b])$. Show that the function $N:\mathcal Y \rightarrow \mathbb R _+$ defined by $$N(y)= \left\lvert \int_a^b y(x) dx \right\rvert$$ does not define a norm on $\mathcal Y$.
I was able to find a similar question which was answered here, where it is clearly shown how to prove that
$$\|f(x)\|:= \int_a^b \vert y(x) \vert dx$$
is a norm on $C([a,b])$.
I am basically confused as to how my question differs from the question that can be found at the aforementioned link. Obviously, the placement of the absolute value signs must have a critical impact on either Positive Definiteness, the Triangle Inequality, or Homogeneity, but I can't see how that will play out. It seems to me like the two functions are identical.
My only guess here is that the mapping of the function $N:\mathcal Y \rightarrow \mathbb R _+$ might have an important impact, but I'm not quite sure how...
Any help would be highly appreciated! Thank you in advance for your time.
Say, $[a,b]=[-1,1]$ and $y(x)=x$, then $y$ is not the zero function but $N(y)=\left|\displaystyle\int_{-1}^{1}xdx\right|=0$.
Triangle inequality: \begin{align*} N(y+z)&=\left|\int_{a}^{b}(y+z)(x)dx\right|\\ &=\left|\int_{a}^{b}y(x)dx+\int_{a}^{b}z(x)dx\right|\\ &\leq\left|\int_{a}^{b}y(x)dx\right|+\left|\int_{a}^{b}z(x)dx\right|\\ &=N(y)+N(z). \end{align*}