1-Seminorm on Continuous on Closed Interval Real-Valued Functions is Norm

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\CC \closedint a b$ be the space of continuous on closed interval real-valued functions.

Let $x \in \CC \closedint a b$ be a continuous real valued function.

Let $\displaystyle \norm x_1 := \int_a^b \size {\map x t} \rd t$ be the 1-seminorm.


Then $\norm {\, \cdot \,}_1$ is a norm on $\CC \closedint a b$.


Proof

Positive definiteness

Let $x \in \CC \closedint a b$.

Then $\forall t \in \closedint 0 1 : \size {\map x t} \ge 0 $.

Hence:

$\displaystyle \int_a^b \size {\map x t} \rd t = \norm x_1 \ge 0$.

Suppose $\forall t \in \closedint a b : \map x t = 0$.

Then $\norm x_1 = 0$.

Therefore:

$\paren {x = 0} \implies \paren {\norm x_1 = 0}$


Let $x \in \CC \closedint a b : \norm x_1 = 0$.

Suppose:

$\forall t \in \openint a b : \map x t = 0$.

By assumption of continuity of $x$:

$\forall t \in \closedint a b : \map x t = 0$.

Aiming for a contradiction, suppose:

$\exists t_0 \in \openint a b : \map x {t_0} \ne 0$.

By assumption, $x$ is continuous at $t_0$.

$\forall \epsilon \in \R_{> 0} : \exists \delta \in \R_{> 0} : \size {t - t_0} < \delta \implies \size {\map x t - \map x {t_0} } < \epsilon$

Furthermore:

$\exists \delta \in \R_{> 0} : \paren{ a < t_0 - \delta} \land \paren {t_0 + \delta < b}$

Let $\displaystyle \epsilon = \frac {\size {\map x {t_0}}} 2$.

We have that:

\(\displaystyle \size {\map x t}\) \(=\) \(\displaystyle \size {\map x t + \map x {t_0} - \map x {t_0} }\)
\(\displaystyle \) \(\ge\) \(\displaystyle \size {\map x {t_0} } - \size {\map x t - \map x {t_0} }\) Reverse triangle inequality
\(\displaystyle \) \(>\) \(\displaystyle \size {\map x {t_0} } - \frac {\size {\map x {t_0} } } 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\size {\map x {t_0} } } 2\)
\(\displaystyle \) \(>\) \(\displaystyle 0\) $\map x {t_0} \ne 0$

Hence:

\(\displaystyle 0\) \(=\) \(\displaystyle \norm x_1\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \size {\map x t} \rd t\)
\(\displaystyle \) \(\ge\) \(\displaystyle \int_{t_0 - \delta}^{t_0 + \delta} \size {\map x t} \rd t\) $t_0 + \delta < b$, $a < t_0 - \delta$
\(\displaystyle \) \(\ge\) \(\displaystyle \int_{t_0 - \delta}^{t_0 + \delta} \frac {\size {\map x {t_0} } } 2 \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \delta \frac {\size {\map x {t_0} } } 2\)
\(\displaystyle \) \(=\) \(\displaystyle \delta \size {\map x {t_0} }\)
\(\displaystyle \) \(>\) \(\displaystyle 0\)

Hence, we reached a contradiction.

Therefore:

$\paren {\norm x_1 = 0} \implies \paren {x = 0}$


Positive homogeneity

Let $x \in \CC \closedint a b$, $\alpha \in \R$.

Then:

\(\displaystyle \size {\alpha x}_1\) \(=\) \(\displaystyle \int_a^b \size {\alpha \map x t} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \size \alpha \int_a^b \size {\map x t} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \size \alpha \norm x_1\)


Triangle inequality

Let $x, y \in \CC \closedint a b$

\(\displaystyle \norm {x + y}_1\) \(=\) \(\displaystyle \int_a^b \size {\map {\paren {x + y} } t} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \size {\map x t + \map y t} \rd t\) Definition of Pointwise Addition of Real-Valued Functions
\(\displaystyle \) \(\le\) \(\displaystyle \int_a^b \paren {\size x + \size y} \rd t\) Triangle Inequality for Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \size x \rd t + \int_a^b \size y \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \norm x_1 + \norm y_1\)

$\blacksquare$


Also see


Sources