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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?

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About the triangle inequality for the $\max$ norm, it is better to take the $\max$ in the "right order": $$|y_1(x)+y_2(x)| \leq |y_1(x)| + |y_2(x)| \leq \|y_1\|_M+\|y_2\|_M.$$So $\|y_1\|_M+\|y_2\|_M$ is an upper bound for $\{ |y_1(x)+y_2(x)| \mid x \in [a,b]\}$, and now you take the $\max$ to obtain $$\|y_1+y_2\|_M \leq \|y_1\|_M+\|y_2\|_M.$$

About the triangle inequality for $\|\cdot \|_1$, what you did is fine, but I would have justified a bit differently: $$\begin{align} &|y_1(x)+y_2(x)| \leq |y_1(x)|+|y_2(x)|, \quad \forall\,x\in[a,b] \\ &\implies\int_a^b|y_1(x)+y_2(x)|\,{\rm d}x \leq \int_a^b |y_1(x)|+|y_2(x)|\,{\rm d}x \\ &\implies \|y_1+y_2\|_1 \leq \int_a^b |y_1(x)|\,{\rm d}x+\int_a^b|y_2(x)|\,{\rm d}x = \|y_1\|_1+\|y_2\|_1. \end{align}$$

Everything else is fine.

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  • $\begingroup$ I see... Thank you!!! :) $\endgroup$ – evinda May 22 '15 at 18:48

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