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

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

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

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

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

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

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

Hence:


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

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 $x$:


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




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

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

We have that:

Hence:

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:

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

Also see

 * Definition:P-Seminorm