Prove that the linear Functional $\phi:C[0,1]\to\mathbb{F}$ such that $\phi(f)=\int\limits_0^{1/2}f-\int\limits_{1/2}^1f$ Doesn't achieve its norm.

Question

For the norm, I'm using the definition: $$||\phi||=\sup\{|\phi(f)|: f\in C[0,1], ||f||\leq1 \}$$

I was able to prove that
For $f\in C[0.1]$ With $||f||\leq1$,
$|\phi(f)|\leq\left|\int\limits_{0}^{1/2}f\right|+\left|\int\limits_{1/2}^1f\right|\leq\int\limits_{0}^{1/2}|f|+\int\limits_{1/2}^1|f|\leq 1$ Thus $||\phi||\leq 1$

And for the other side of the inequality, I chosed the functions $f_{\epsilon}$ where it takes $+1$ for $x\in[0,1/2-\epsilon]$ , $-1$ for $x\in[1/2+\epsilon,1]$ and for the $x\in[1/2-\epsilon,1/2+\epsilon]$, $f(x)$ will be the straight line joining $+1$ with $-1$.
Thus it gives $|\phi(f)|=1-\epsilon$
So $||\phi||\geq1-\epsilon$ for all $\epsilon>0$.
That is $||\phi||\geq1$
Therefore $||\phi||=1$

But how can I prove that there cannot be any function on $C[0,1]$ such that $||f||\leq 1$ and $|\phi(f)|=1$.
I think this has to do something with the function being continuous at $1/2$ but I don't really see how I should do it.

Answer 1

Suppose $\|f\| \leq 1$ and $\phi (f)=1$. Then $1=\int_0^{1/2} f(x)dx-\int_{1/2}^{1} f(x)dx \leq \int_0^{1/2} 1dx+\int_{1/2}^{1} 1dx =1$. This implies that $f(x)=1$ for all $x <\frac 1 2$ and $-f(x)=1$ for all $x >\frac 1 2$ because if strict inequality holds at some point (then it holds throughout some open interval around the point and ) you will get the contradiction $1<1$. But now $f(x)=1$ for all $x <\frac 1 2$ and $f(x)=-1$ for all $x>\frac 1 2$ contradicting the fact that $f$ is continuous.

Answer 2

Suppose such $f$ existed.

• What would $\int_0^{1/2} f$ and $\int_{1/2}^1$ have to be?

• What would that say about the values of $f$ on $(0,1/2)$ and on $(1/2,1)$?

• Why is that bad?

Answer 3

If $f\in C[0,1]$ and $|f(x_0)|<\|f\|$ for some $x_0\in [0,1|,$ let $r=\|f\|-|f(x_0)|.$

Then there exist $a,b$ with $0\le a<b\le 1$ and $x_0\in [a,b],$ such that $$\forall x\in [a,b]\,(\,|f(x)|<\|f\|-r/2\,).$$ Then

$$|\phi(f)|\le \int_0^1 |f(x)|dx=(\int_0^a+\int_a^b+\int_b^1)\,|f(x)|dx\le$$ $$\le a\|f\|+(b-a)(\|f\|-r/2)+(1-b)\|f\|<\|f\|.$$

So if $|\phi(f)|=\|f\|$ then $|f(x)|=\|f\|$ for all $x\in [0,1].$ But for a continuous $f:[0,1]\to \Bbb R$ that is not possible unless $f$ is constant. And for a constant $f$ we have $\phi(f)=0.$

So if $|\phi(f)|=\|f\|$ then $f=0.$