# Operator norm on $C([-1,1])$ w.r.t. $\Vert \cdot \Vert_{\infty}$-norm

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 $$\varphi(f ) =\int_{-1}^0 f (x) dx −\int_0^1 f (x) dx$$ 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?

• – Kelenner Aug 5 '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 5 '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 5 '16 at 11:50
• Ok thank you @Mathematician42 ! Sorry for my mistake and for the duplication – MorganeMaPh Aug 5 '16 at 12:02
• 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 5 '16 at 12:25

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 $[-1/n,1/n]$: $$\varphi(f_n) = 2(1-1/n),\;\;\; \|f_n\|=1.$$
$|\psi (f)|\leq \int_{-1}^0|f(x)|\;dx+\int_0^1|f(x)|\;dx=$ $\int_{-1}^1 |f(x)|\;dx\leq \int_{-1}^1 \|f\|\;dx=2\|f\|.$ 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 $|\psi(f)|=\| \psi \|\cdot \|f\|.$