1. ## Dual space isometry

Is the above proof correct?

2. Originally Posted by forumbug

Is the above proof correct?
As far as I can say, yes.
Originally Posted by forumbug

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 posted here Dual space isometry

3. Is this correct?

4. Originally Posted by forumbug
Is this correct?
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 $W=\{\varphi_g; g\in X\}$.
W.l.o.g. you can assume $f_2=0$. (Do you know why?)
Now if a continuous function is non-zero, then there exists x such that $f(x)\ne 0$.
But then there exists $\varepsilon>0$ such that f does not change sign on $(x-\varepsilon,x+\varepsilon)$. Can you use this facts (or similar ideas) to find a function g such that $\varphi_g(f)\ne0$?

I think that your answer to the 3rd problem is correct. Another possibility is evaluation map like $f\mapsto f(1/2)$.

You get $\varphi_g(x_n)\to 0$ whenever $\int_0^1 x_n(t) dt$ converges to 0. So you can for instance choose
$
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

5. Originally Posted by kompik
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?
Now, when I think about it, you perhaps meant $\exists t\in[0,1] |h(t)|=1$ instead of $\forall t\in[0,1] h(t)=1$

6. 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
Code:
\begin{align*}
\end{align*}
For part 1, I meant that the second last step is true because the function $h(t) = 1 \forall
t \in [0,1]$
is in $X = C[0,1]$ and since all other $h$ satisfying
$\displaystyle \sup_{t \in [0,1]} |h| = 1$ are bounded above by
$h=1$, $\displaystyle \int^1_0 g(t) h(t) dt$ where $\displaystyle
\sup_{t \in [0,1]} \abs{h} = 1$
and $h(t) \ne 1$ is smaller than
$\displaystyle \int^1_0 g(t) h(t) dt$ where $\displaystyle \sup_{t
\in [0,1]} |h| = 1$
and $h(t) = 1$. Hence the supremum is
unchanged by limiting the set to having just $h(t) = 1$.

Is this correct?

For part 2, thanks for your clue, I put down the following.

Since $f_1$ and $f_2$ are both continuous, if $f_1 \ne f_2$ at
some point $t_0 \in [0,1]$, then there exists $\delta > 0$ such that
$f_1(t) - f_2(t) > 0 \forall t \in [t_0 - \delta, t_0 + \delta]$.
Then any $g \in X$ such that $g(t) \ne 0 \forall t \in [t_0 -
\delta, t_0 + \delta]$
will ensure $\varphi_g(f_1) \ne
\varphi_g(f_2)$
.

For part 4, I am not sure if the question meant $||\cdot||_\infty$ 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.

7. Originally Posted by forumbug
For part 1, I meant that the second last step is true because the function $h(t) = 1 \forall
t \in [0,1]$
is in $X = C[0,1]$ and since all other $h$ satisfying
$\sup_{t \in [0,1]} |h| = 1$ are bounded above by
$h=1$, $\displaystyle \int^1_0 g(t) h(t) dt$ where $\displaystyle
\sup_{t \in [0,1]} \abs{h} = 1$
and $h(t) \ne 1$ is smaller than
$\displaystyle \int^1_0 g(t) h(t) dt$ where $\displaystyle \sup_{t
\in [0,1]} |h| = 1$
and $h(t) = 1$. Hence the supremum is
unchanged by limiting the set to having just $h(t) = 1$.

Is this correct?
Well, you're right in claiming that $\int^1_0 g(t) h(t) dt \ le \int^1_0 g(t) dt$.
But you need to estimate $\left|\int^1_0 g(t) h(t) dt\right|$, which might be completely different if g(t).h(t) is not non-negative.

Originally Posted by forumbug
For part 2, thanks for your clue, I put down the following.

Since $f_1$ and $f_2$ are both continuous, if $f_1 \ne f_2$ at
some point $t_0 \in [0,1]$, then there exists $\delta > 0$ such that
$f_1(t) - f_2(t) > 0 \forall t \in [t_0 - \delta, t_0 + \delta]$.
Then any $g \in X$ such that $g(t) \ne 0 \forall t \in [t_0 -
\delta, t_0 + \delta]$
will ensure $\varphi_g(f_1) \ne
\varphi_g(f_2)$
.
Seems ok to me. You should also notice that you have used the fact that g does not change the sign on that interval.

8. Originally Posted by kompik
Well, you're right in claiming that $\int^1_0 g(t) h(t) dt \ le \int^1_0 g(t) dt$.
But you need to estimate $\left|\int^1_0 g(t) h(t) dt\right|$, which might be completely different if g(t).h(t) is not non-negative.
Should I consider $h(t) = \pm 1$ and hence get $\sup \left\{|\int^1_0 g(t) dt|, |\int^1_0 - g(t) dt|
\right\}$
?

Originally Posted by kompik
Seems ok to me. You should also notice that you have used the fact that g does not change the sign on that interval.
Thanks, got it.

9. Originally Posted by forumbug
Should I consider $h(t) = \pm 1$ and hence get $\sup \left\{|\int^1_0 g(t) dt|, |\int^1_0 - g(t) dt|
\right\}$
?
I do not think this is correct.
Notice that for $g(t)=\sin2\pi t$ then $\int_0^1 g(t)h(t) dt=0$ for $h(t)=\pm1$, but there exist function such that $\int_0^1 g(t)h(t) dt \ne 0$.

10. Originally Posted by kompik
I do not think this is correct.
Notice that for $g(t)=\sin2\pi t$ then $\int_0^1 g(t)h(t) dt=0$ for $h(t)=\pm1$, but there exist function such that $\int_0^1 g(t)h(t) dt \ne 0$.
I don't quite understand this. I will come back to this post when I find out the solution.