1
$\begingroup$

Let $X = C([−1, 1], \mathbb{R})$ be the Banach space of continuous functions on $[−1, 1]$ equipped with $\Vert \cdot \Vert_{\infty}$-norm . Prove that the functional \begin{equation} \varphi(f ) =\int_{-1}^0 f (x) dx −\int_0^1 f (x) dx \end{equation} belongs to $X^{\ast}$ and compute its norm. Show that there is no $f \in C([−1,1],\mathbb{R})$ with $\Vert f \Vert_{\infty}$ ≤ 1 such that $\vert \varphi(f)\vert = \Vert f \Vert_{\infty}$.

I have some difficulties with this exercise. I get that $\Vert \varphi \Vert =0$... but that seams weird. And also I don't know how to do for the second part. Can someone help me please?

$\endgroup$
  • 3
    $\begingroup$ See here math.stackexchange.com/questions/1435937/… $\endgroup$ – Kelenner Aug 5 '16 at 11:45
  • $\begingroup$ You cannot have that $\|\phi\|=0$ as this would imply that $\phi=0$ which clearly is not the case. $\endgroup$ – Mathematician 42 Aug 5 '16 at 11:47
  • $\begingroup$ Yes of course Mathematician 42 I was too quick. @Kelenner Great! Thank you what about the second part? Is following Idea correct: Assume the contrary: there exists $f \in X$ with $\Vert f \Vert_{\infty}\leq 1$ such that $\vert \varphi (f)\vert= \Vert f \Vert_{\infty}$, then we have that $\Vert \varphi \Vert<1$... but this is a contradiction since $\Vert \varphi \Vert=2$ $\endgroup$ – MorganeMaPh Aug 5 '16 at 11:50
  • $\begingroup$ Ok thank you @Mathematician42 ! Sorry for my mistake and for the duplication $\endgroup$ – MorganeMaPh Aug 5 '16 at 12:02
  • 2
    $\begingroup$ In fact, the function $f$ such that $f(x)=-x$ on $[-1,1]$ is such that $\|f\|_{\infty}=1$ and $\varphi(f)=1$. Perhaps your problem is about functions $f$ such that $|\varphi(f)|=2\|f\|_{\infty}$. $\endgroup$ – Kelenner Aug 5 '16 at 12:25
1
$\begingroup$

Every continuous linear functional $\Phi$ on $C[-1,1]$ with the max norm has the form $$ \Phi(f) = \int_{-1}^{1}f(t)d\mu(t) $$ for a finite signed Borel measure on $[-1,1]$. And $\|\Phi\|=\|\mu\|$ where $\|\mu\|$ is the variation of $\mu$. In your case the measure associated with your functional $\varphi$ is positiveLebesgue measure on $[-1,0]$ and negative Lebesgue measure on $[0,1]$. The associated variation measure is ordinary Lebesgue measure, which gives $$ \|\varphi\| = m[-1,1] = 2. $$ You can prove this for the special case at hand by choosing functions $f_n$ that are $1$ on $[-1,-1/n]$, are $-1$ on $[1/n,1]$, and are extended to be continuous on $[-1,1]$ and linear on $[-1/n,1/n]$: $$ \varphi(f_n) = 2(1-1/n),\;\;\; \|f_n\|=1. $$

$\endgroup$
1
$\begingroup$

$|\psi (f)|\leq \int_{-1}^0|f(x)|\;dx+\int_0^1|f(x)|\;dx=$ $\int_{-1}^1 |f(x)|\;dx\leq \int_{-1}^1 \|f\|\;dx=2\|f\|.$ Therefore $$\|\psi\| \leq 2.$$ For $r\in (0,1)$ let $f_r(x)=1$ for $x\in [-1,-r],$ let $f_r(x)=-x/r$ for $x\in [-r,r],$ let $f_r(x)=-1$ for $x\in [r,1].$ Then $\|f_r\|=1$ and $\psi (f_r)>2-2r.$ Therefore $$\| \psi \|\geq \sup_{r\in (0,1)}(2-2r)=2.$$

For the second Q,$$|\psi (f)|=2\|f\|\implies 2\|f\|=|\int_{-1}^0f(x)\;dx-\int_0^1f(x)\;dx\;|\leq$$ $$\leq |\int_{-1}^0f(x)\;dx\;|+|\int_0^1f(x)\;dx\;|\leq \int_{-1}^0|f(x)|\;dx+\int_0^1|f(x)|\;dx\leq$$ $$\leq \int_{-1}^0\|f\|\;dx+\int_0^1\|f\|\;dx =2\|f\|.$$ We can only have equality from one end of the above to the other end if $\int_{-1}^0f(x)\; dx=-\int_0^1f(x)\;dx=\pm\|f\|,$ which, since $f$ is continuous, requires $f(x)=\|f\|$ for $x\in [-1,0]$ and $f(x)=-\|f\|$ (or vice-versa) for $x\in [0,1].$ But then $f(0)=\|f\|=-\|f\|$ so $\|f\|=0.$

The second part should say there is no $f\ne 0$ such that $|\psi(f)|=\| \psi \|\cdot \|f\|.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.