Prove Dual Space Isometry: X = C[0,1] Sup Norm

[complexnumber]

Homework Statement

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
<br /> \begin{align*}<br /> \varphi_g(h) := \int^1_0 g(t) h(t) dt, \qquad h \in X<br /> \end{align*}<br />
Define W := \{ \varphi_g | g \in X \}.


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

Homework Equations

The Attempt at a Solution

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

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

For any h \in X, \varphi_g is continuous if \forall \varepsilon<br /> &gt; 0 \exists \delta &gt; 0 such that any h&#039; \in X satisfies
<br /> \begin{align*}<br /> d(h,h&#039;) &lt; \delta \implies d(\varphi_g(h), \varphi_g(h&#039;)) &lt;<br /> \varepsilon \text{.}<br /> \end{align*}<br />
The distance function is \displaystyle d(h,h&#039;) = ||h - h&#039;|| =<br /> \sup_{x \in [0,1]} |h(t) - h&#039;(t)|.
<br /> \begin{align*}<br /> d(\varphi_g(h), \varphi_g(h&#039;)) =&amp; |\int^1_0 g(t) h(t) dt -<br /> \int^1_0 g(t) h&#039;(t) dt | = |\int^1_0 g(t) h(t) - g(t) h&#039;(t) dt | \\<br /> =&amp; |\int^1_0 g(t) (h(t) - h&#039;(t)) dt| \leq \int^1_0 |g(t) (h(t) - h&#039;(t))| dt \\<br /> \leq &amp; \int^1_0 |g(t)| |(h(t) - h&#039;(t))| dt \leq \sup_{t \in<br /> [0,1]} |(h(t) - h&#039;(t))| \int^1_0 |g(t)| dt \\<br /> =&amp; d(h,h&#039;) \int^1_0 |g(t)| dt<br /> \end{align*}<br />
For any \varepsilon &gt; 0, we get \displaystyle d(h,h&#039;) \int^1_0<br /> |g(t)| dt &lt; \varepsilon when \displaystyle d(h,h&#039;) &lt;<br /> \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
<br /> \begin{align*}<br /> ||\varphi_g||_{X^*} :=&amp; \sup \{ |\varphi_g(h)| | \norm{h} = 1<br /> \} \\<br /> =&amp; \sup \left\{ |\int^1_0 g(t)h(t) dt| \bigg| \sup_{t \in [0,1]}<br /> |h| = 1 \right\}<br /> \end{align*}<br />
How can I find the value of this?

 

[Office_Shredder]

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

 

[kompik]

Are the above proofs correct?

As far as I can say, they are.

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

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

 

[complexnumber]

Is this correct?

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