# Continuous, Bounded Normed Linear Spaces

For $f$ in $C[a, b]$, define $||f||_1 = \int_a^b |f|$.

1. Show that this is a norm on $C[a, b]$.

2. Also, show that there is no number $c\geq0$ for which $||f||_{\max}\leq c||f||_1$ for all $f$ in $C[a,b]$, but there is a $c\geq0$ for which $||f||_1\leq c||f||_{\max}$ for all $f$ in $C[a,b]$.

Solution:

Part 1:

Let $f\in C[a,b]$, $g\in C[a,b]$, and $\alpha \in \Re$.

a) (Triangle- Inequality) $||f+g||_1 = \int_a^b |f+g| \leq \int_a^b |f| + \int_a^b |g| = ||f||_1 + ||g||_1$

b) (Positive Homogeneity) $||\alpha f||_1 = \int_a^b |\alpha f| = |\alpha| \int_a^b |f| = |\alpha| ||f||_1$

c) From my book (Royden "Real Analysis" 4th):

(Nonnegativity) $||f||\geq 0$ and $||f|| = 0$ if and only if $f=0$.

So, does this mean $||f||\geq 0$ and $||f|| = 0$ which implies just $||f|| = 0$? Or $||f||\geq 0$ or $||f|| = 0$? I took it to mean the former. So,

Let $||f||_1||\geq 0$ and $=0$. $0=||f||_1 = \int_a^b |f|$

$0=||f||_1 = \int_a^b |f|$ = $F(b)-F(a) = 0$, so $F(b) = F(a)$ which either means that $a=b$ or $f=0$. Since $a\neq b$, $f=0$.

Conversely, let $f=0$. Then, $||f||_1 = \int_a^b |f| =\int_a^b |0| = 0$.

Part 2:

I understand the second part: "but there is a $c\geq0$ for which $||f||_1\leq c||f||_{\max}$ for all $f$ in $C[a,b]$". Intuitively, we can say that this is true because if you make a box with base $|b-a|$ and height $f_{\max}$, the area will always be at least as much as the area under the curve.

$||f||_1 = \int_a^b |f| = |F(b)| - |F(a)| \leq |F(b) - F(a)|$. Let $c=|b-a|$, then $c||f||_{\max} \geq ||f||_1$ by definition of the integral. How can I say this more mathematically?

The first part I am having more trouble comprehending. Since $f$ are continuous, there is no infinite value of any $f$. So, let $||f||_{\max} = N$, then couldn't $c=N$ and $||f||_{\max}\leq c||f||_1$ for all $f$ in $C[a,b]$?

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For the nonnegativity part, that just means the norm acts like an absolute value. So what does it mean to say $|x|\geq0$ with equality if and only if $x=0$? It means if $x\neq0$, then $|x|>0$. For the first part of ($2$), try to build a tent map with a decreasing base but increasing height. This will show that there is no fixed $N$ that works for all $f\in C[a,b]$. For the second part of ($2$), just say $$\int_a^b|f|\leq\int_a^b\Vert f\Vert_{max}=\Vert f\Vert_{max}|b-a|.$$ –  Clayton Jan 25 '13 at 20:48
Thanks, Clayton. –  Jake Casey Jan 25 '13 at 23:16
So, you're saying: 1.(Nonnegativity) Do the proof "$||f|| \geq 0$ if and only if $f=0$ as well? Also, was my proof of $||f|| = 0$ satisfactory? –  Jake Casey Jan 25 '13 at 23:33
Jake, it seems okay, but I don't think it's the standard proof (at least not that I've seen). Mostly I've seen the contrapositive. Suppose $|f|$ is not identically $0$. Then there exists an $x_0$ such that $|f(x_0)|>0$, thus there is a $\delta>0$ so that for all $x\in(x_0-\delta,x_0+\delta)$ $|f(x)|>0$. Take the minimum of $|f|$ in $[x_0-\delta/2,x_0+\delta/2]$, and call it $A$ then $\int_a^b|f|\geq\int_{x_0-\delta/2}^{x_0+\delta/2}|f|\geq 2A\delta>0$. –  Clayton Jan 26 '13 at 0:26