Consider the fonction : $N(x,y)=\int_{0}^{1}\mid x+ty\mid dt $ where $x,y$ are real numbers. I have to proove that this function defines a norm $R^2$. First of all, I said that, by linearity : $$N(\lambda(x,y))=\int_{0}^{1}\mid x\lambda + t\lambda y \mid =\mid \lambda\mid N(x,y)$$ Then, i said that because $\mid x+ ty \mid $ is positive for all $x,y$ by definition of the absolute value, then if we take the integral, it remains positive. The problem is for the triangle inequality. I don't know if the following lines are true :
We have to proove that : $$N(x_1+x_2,y_1+y_2)\leq N(x_1,y_1)+N(x_2,y_2)\quad \forall (x_1,y_1), (x_2,y_2) \in R^2$$ We have : $$\mid x_1+x_2+t(y_1+y_2)\mid \leq \mid x_1+ty_1\mid +\mid x_2+ty_2\mid$$ Then, $$\int_{0}^{1}\mid x_1+x_2+t(y_1+y_2)\mid dt \leq \int_{0}^{1}\mid x_1+ty_1\mid +\mid x_2+ty_2\mid dt$$ $$ = \int_{0}^{1}\mid x_1+ty_1\mid dt + \int_{0}^{1}\mid x_2+ty_2\mid dt$$ Is that all ? Thank you for your help :)