Show that
\begin{align} & \|y\|_M= \max_{a \leq x \leq b} |y(x)| \tag 1 \\[8pt] & \|y\|_1=\int_a^b |y(x)|\, dx \tag 2 \end{align}
satify the properties of a norm in $C[a,b]$.
That's what I have tried:
$$\|y\|_M=0 \Rightarrow \max_{a \leq x \leq b} |y(x)|=0.$$
But $|y(x)| \leq \max_{a \leq x \leq b} |y(x)|=0$ and $|y(x)| \geq 0$. Thus $|y(x)|=0 \Rightarrow y(x)=0$.
If $y(x)=0$ then $\|y\|_M=\max_{a \leq x \leq b} |y(x)|=0$.
Let $\lambda \in \mathbb{R}$. Then $\|\lambda y\|_M= \max_{a \leq x \leq b} |\lambda y(x)|=|\lambda| \max_{a \leq x \leq b} |y(x)|=|\lambda| \|y\|_M $
Let $y_1, y_2 \in C[a,b]$. Then $\|y_1+y_2\|_M= \max_{a \leq x \leq b} |y_1(x)+y_2(x)| \overset{\text{ triangle inequality}}{ \leq } \max_{a \leq x \leq b} (|y_1(x)|+|y_2(x)|)= \max_{a \leq x \leq b} |y_1(x)|+ \max_{a \leq x \leq b} |y_2(x)|=\|y_1\|_M+\|y_2\|_M$
So $(1)$ satisfies the properties of a norm in $C[a,b]$.
$$$$
$\|y\|_1=0 \Rightarrow \int_a^b |y(x)|dx=0 \overset{|y(x)| \text{continuous and non-negative}}{\Rightarrow } |y(x)|=0 \Rightarrow y(x)=0$.
If $y(x)=0$ then $\|y\|_1=\int_a^b |y(x)|dx=0$.
Let $\lambda \in \mathbb{R}$. Then $\|\lambda y\|_1= \int_a^b |\lambda y(x)|dx= |\lambda| \int_a^b |y(x)|dx=|\lambda| \|y\|_1$.
Let $y_1, y_2 \in C[a,b]$. Then $\|y_1+y_2\|_1=\int_a^b |y_1(x)+y_2(x)|\,dx \leq \int_a^b (|y_1(x)|+|y_2(x)|)\,dx= \int_a^b |y_1(x)|\,dx+ \int_a^b |y_2(x)|\,dx=\|y_1\|_1+\|y_2\|_1$
So $(2)$ satisfies the properties of a norm in $C[a,b]$.
Is that what I have tried right?
