Is the above proof correct?
As far as I can say, yes.You have
$\displaystyle \left|\int^1_0 g(t)h(t) dt\right| \le \int^1_0 |g(t)| |h(t)| dt \le \int^1_0 |g(t)| dt$
therefore
$\displaystyle ||\varphi_g|| \le \int^1_0 |g(t)| dt $.
I think that $\displaystyle ||\varphi_g|| = \int^1_0 |g(t)| dt $ holds. A naive approach to show this would be trying to approximate the step function $\displaystyle h(t)=g(t)/|g(t)|$ by continuous functions; but I guess there is a simpler solution.
BTW the same question was posted here Dual space isometry
It would be much easier to answer, if you posted everything in latex format instead of images. (You obviously latexed this.)
Do you see that in your answer to question 1 you replaced supremum of some set with supremum of one-element set. Doesn't this seem suspicious to you?
You did not write what W means, but I guess $\displaystyle W=\{\varphi_g; g\in X\}$.
W.l.o.g. you can assume $\displaystyle f_2=0$. (Do you know why?)
Now if a continuous function is non-zero, then there exists x such that $\displaystyle f(x)\ne 0$.
But then there exists $\displaystyle \varepsilon>0$ such that f does not change sign on $\displaystyle (x-\varepsilon,x+\varepsilon)$. Can you use this facts (or similar ideas) to find a function g such that $\displaystyle \varphi_g(f)\ne0$?
I think that your answer to the 3rd problem is correct. Another possibility is evaluation map like $\displaystyle f\mapsto f(1/2)$.
You get $\displaystyle \varphi_g(x_n)\to 0$ whenever $\displaystyle \int_0^1 x_n(t) dt$ converges to 0. So you can for instance choose
$\displaystyle
x_n(t)=\begin{cases}
1-nx,&x\le1/n,\\
0,&\text{otherwise}
\end{cases}
$
Again, the same question was posted here: Dual space isometry
Thank you for your replies. I have been trying to finish off the question.
I tried latex the question in the forum but couldn't write multi-line equations using
For part 1, I meant that the second last step is true because the function $\displaystyle h(t) = 1 \forallCode:\begin{align*} \end{align*}
t \in [0,1]$ is in $\displaystyle X = C[0,1]$ and since all other $\displaystyle h$ satisfying
$\displaystyle \displaystyle \sup_{t \in [0,1]} |h| = 1$ are bounded above by
$\displaystyle h=1$, $\displaystyle \displaystyle \int^1_0 g(t) h(t) dt$ where $\displaystyle \displaystyle
\sup_{t \in [0,1]} \abs{h} = 1$ and $\displaystyle h(t) \ne 1$ is smaller than
$\displaystyle \displaystyle \int^1_0 g(t) h(t) dt$ where $\displaystyle \displaystyle \sup_{t
\in [0,1]} |h| = 1$ and $\displaystyle h(t) = 1$. Hence the supremum is
unchanged by limiting the set to having just $\displaystyle h(t) = 1$.
Is this correct?
For part 2, thanks for your clue, I put down the following.
Since $\displaystyle f_1$ and $\displaystyle f_2$ are both continuous, if $\displaystyle f_1 \ne f_2$ at
some point $\displaystyle t_0 \in [0,1]$, then there exists $\displaystyle \delta > 0$ such that
$\displaystyle f_1(t) - f_2(t) > 0 \forall t \in [t_0 - \delta, t_0 + \delta]$.
Then any $\displaystyle g \in X$ such that $\displaystyle g(t) \ne 0 \forall t \in [t_0 -
\delta, t_0 + \delta]$ will ensure $\displaystyle \varphi_g(f_1) \ne
\varphi_g(f_2)$.
For part 4, I am not sure if the question meant $\displaystyle ||\cdot||_\infty$ to be supremum norm or essential supremum norm. For essential supremum norm, I don't think such an example exists.
Thanks a lot for helping me get through this question.
Well, you're right in claiming that $\displaystyle \int^1_0 g(t) h(t) dt \ le \int^1_0 g(t) dt$.
But you need to estimate $\displaystyle \left|\int^1_0 g(t) h(t) dt\right|$, which might be completely different if g(t).h(t) is not non-negative.
Seems ok to me. You should also notice that you have used the fact that g does not change the sign on that interval.