# Dual space isometry

1. May 26, 2010

### complexnumber

1. The problem statement, all variables and given/known data

For the Banach space $$X = C[0,1]$$ with the supremum norm, fix
an element $$g \in X$$ and define a map $$\varphi_g : X \to \mathbb{C}$$
by
\begin{align*} \varphi_g(h) := \int^1_0 g(t) h(t) dt, \qquad h \in X \end{align*}
Define $$W := \{ \varphi_g | g \in X \}$$.

Prove that $$\varphi_g \in X^*$$ and calculate
$$||\varphi_g||_{X^*}$$.

2. Relevant equations

3. The attempt at a solution

The supremum norm for a function $$f \in C[0,1]$$ is $$\displaystyle ||f|| = \sup_{x \in [0,1]} |f(x)|$$.

For $$h_1,h_2 \in X$$ and $$\lambda \in \mathbb{C}$$
\begin{align*} \varphi_g(h_1 + h_2) =& \int^1_0 g(t) (h_1(t) + h_2(t)) dt = \int^1_0 g(t) h_1(t) dt + \int^1_0 g(t) h_2(t) dt = \varphi_g(h_1) + \varphi_g(h_2) \\ \varphi_g(\lambda h_1) =& \int^1_0 \lambda g(t) h_1(t) dt = \lambda \int^1_0 g(t) h_1(t) dt \end{align*}
So $$\varphi_g$$ is linear functional.

For any $$h \in X$$, $$\varphi_g$$ is continuous if $$\forall \varepsilon > 0$$ $$\exists \delta > 0$$ such that any $$h' \in X$$ satisfies
\begin{align*} d(h,h') < \delta \implies d(\varphi_g(h), \varphi_g(h')) < \varepsilon \text{.} \end{align*}
The distance function is $$\displaystyle d(h,h') = ||h - h'|| = \sup_{x \in [0,1]} |h(t) - h'(t)|$$.
\begin{align*} d(\varphi_g(h), \varphi_g(h')) =& |\int^1_0 g(t) h(t) dt - \int^1_0 g(t) h'(t) dt | = |\int^1_0 g(t) h(t) - g(t) h'(t) dt | \\ =& |\int^1_0 g(t) (h(t) - h'(t)) dt| \leq \int^1_0 |g(t) (h(t) - h'(t))| dt \\ \leq & \int^1_0 |g(t)| |(h(t) - h'(t))| dt \leq \sup_{t \in [0,1]} |(h(t) - h'(t))| \int^1_0 |g(t)| dt \\ =& d(h,h') \int^1_0 |g(t)| dt \end{align*}
For any $$\varepsilon > 0$$, we get $$\displaystyle d(h,h') \int^1_0 |g(t)| dt < \varepsilon$$ when $$\displaystyle d(h,h') < \frac{\varepsilon}{\displaystyle \int^1_0 |g(t)| dt} = \delta$$.
Thus $$\varphi_g$$ is continuous. Are the above proofs correct?

The formula for $$||\varphi_g||_{X^*}$$ is
\begin{align*} ||\varphi_g||_{X^*} :=& \sup \{ |\varphi_g(h)| | \norm{h} = 1 \} \\ =& \sup \left\{ |\int^1_0 g(t)h(t) dt| \bigg| \sup_{t \in [0,1]} |h| = 1 \right\} \end{align*}
How can I find the value of this?

Last edited: May 26, 2010
2. May 26, 2010

### Office_Shredder

Staff Emeritus
For the supremum, maybe some sort of triangle inequality application can give you a starting upper bound

3. May 27, 2010

### kompik

As far as I can say, they are.

You have
$$\left|\int^1_0 g(t)h(t) dt\right| \le \int^1_0 |g(t)| |h(t)| dt \le \int^1_0 |g(t)| dt$$
therefore
$$||\varphi_g|| \le \int^1_0 |g(t)| dt$$.

I think that $$||\varphi_g|| = \int^1_0 |g(t)| dt$$ holds. A naive approach to show this would be trying to approximate the step function $$h(t)=g(t)/|g(t)|$$ by continuous functions; but I guess there is a simpler solution.

BTW the same question was asked here: http://www.mathhelpforum.com/math-h...tial-geometry/146627-dual-space-isometry.html

4. May 27, 2010

### complexnumber

Is this correct?

[PLAIN]http://9ya7ng.blu.livefilestore.com/y1pFP94kdkanTL7oWHQXuecgsG7MYfNfM3fHYVt7AE01cgDtbQY8VkjQk94V8H5WceDMp8kOlh-X1WSs79GZtIUTEWTFMCBtceU/q3.jpg [Broken]

Last edited by a moderator: May 4, 2017