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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?

MorganeMaPh
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    See here http://math.stackexchange.com/questions/1435937/how-do-i-prove-the-norm-of-this-linear-functional-is-2 – Kelenner Aug 05 '16 at 11:45
  • You cannot have that $|\phi|=0$ as this would imply that $\phi=0$ which clearly is not the case. – Mathematician 42 Aug 05 '16 at 11:47
  • 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 $\Vert \varphi \Vert=2$ – MorganeMaPh Aug 05 '16 at 11:50
  • Ok thank you @Mathematician42 ! Sorry for my mistake and for the duplication – MorganeMaPh Aug 05 '16 at 12:02
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    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 $|\varphi(f)|=2|f|{\infty}$. – Kelenner Aug 05 '16 at 12:25
  • @Kelenner: You're absolutely right, I made a silly mistake as well then. – Mathematician 42 Aug 05 '16 at 12:32
  • @MorganeMaPh: The reasoning is false, if $|\phi(f)|=|f|{\infty}$ for some $f$ with $|f|{\infty}\leq 1$, then $|\phi|\geq 1$, since the norm is defined by a supremum. But Kelenner's latest comment is very interesting! – Mathematician 42 Aug 05 '16 at 12:36
  • Suppose that there exists such a (non zero) function (ie $|f|{\infty}>0$) with $|\varphi(f)|=2|f|{\infty}$. We may suppose (eventually we replace $f$ by $-f$) that $\int_{-1}^0 (f(t)dt-\int_0^1 f(t)dt=2|f|{\infty}$. Hence we have $\int{-1}^0 (f(t)-|f|{\infty})dt=\int_0^1(f(t)+|f|{\infty})dt$. The first integral is $\leq 0$, the second $\geq 0$. Hence both are zero. As the functions are continuous, we get $f(t)-|f|{\infty}=0$ on $[-1,0]$ and $f(t)+|f|{\infty}=0$ on $[0,1]$. It is easy to finish. – Kelenner Aug 05 '16 at 12:41
  • @Kelenner but then $f$ is not continuous ... – MorganeMaPh Aug 05 '16 at 12:52
  • Yes, and this show that a continuous function such that $|\varphi(f)|=2|f|_{\infty}$ does not exists – Kelenner Aug 05 '16 at 12:53
  • Ok ! All right, I get it thank you @Kelenner – MorganeMaPh Aug 05 '16 at 12:55

2 Answers2

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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 $[-1/n,1/n]$: $$ \varphi(f_n) = 2(1-1/n),\;\;\; \|f_n\|=1. $$

Disintegrating By Parts
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$|\psi (f)|\leq \int_{-1}^0|f(x)|\;dx+\int_0^1|f(x)|\;dx=$ $\int_{-1}^1 |f(x)|\;dx\leq \int_{-1}^1 \|f\|\;dx=2\|f\|.$ 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 $|\psi(f)|=\| \psi \|\cdot \|f\|.$

DanielWainfleet
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