1-Seminorm on Continuous on Closed Interval Real-Valued Functions is Norm
Theorem
Let $C \closedint a b$ be the space of real-valued functions continuous on $\closedint a b$.
Let $x \in C \closedint a b$ 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 $C \closedint a b$.
Proof
Positive definiteness
Let $x \in C \closedint a b$.
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 $x \in C \closedint a b : \norm x_1 = 0$.
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 $\ds \epsilon = \frac {\size {\map x {t_0} } } 2$.
We have that:
\(\ds \size {\map x t}\) | \(=\) | \(\ds \size {\map x t + \map x {t_0} - \map x {t_0} }\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \size {\map x {t_0} } - \size {\map x t - \map x {t_0} }\) | Reverse triangle inequality | |||||||||||
\(\ds \) | \(>\) | \(\ds \size {\map x {t_0} } - \frac {\size {\map x {t_0} } } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\size {\map x {t_0} } } 2\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 0\) | $\map x {t_0} \ne 0$ |
Hence:
\(\ds 0\) | \(=\) | \(\ds \norm x_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \size {\map x t} \rd t\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \int_{t_0 - \delta}^{t_0 + \delta} \size {\map x t} \rd t\) | $t_0 + \delta < b$, $a < t_0 - \delta$ | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \int_{t_0 - \delta}^{t_0 + \delta} \frac {\size {\map x {t_0} } } 2 \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \delta \frac {\size {\map x {t_0} } } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \delta \size {\map x {t_0} }\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 0\) |
Hence, we reached a contradiction.
Therefore:
- $\paren {\norm x_1 = 0} \implies \paren {x = 0}$
Positive homogeneity
Let $x \in C \closedint a b$, $\alpha \in \R$.
Then:
\(\ds \size {\alpha x}_1\) | \(=\) | \(\ds \int_a^b \size {\alpha \map x t} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size \alpha \int_a^b \size {\map x t} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size \alpha \norm x_1\) |
Triangle inequality
Let $x, y \in C \closedint a b$
\(\ds \norm {x + y}_1\) | \(=\) | \(\ds \int_a^b \size {\map {\paren {x + y} } t} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \size {\map x t + \map y t} \rd t\) | Definition of Pointwise Addition of Real-Valued Functions | |||||||||||
\(\ds \) | \(\le\) | \(\ds \int_a^b \paren {\size x + \size y} \rd t\) | Triangle Inequality for Real Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \size x \rd t + \int_a^b \size y \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm x_1 + \norm y_1\) |
$\blacksquare$
Also see
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed Spaces