$\ker(f)$ of linear functional with operator in it.

Question

Let $(X,\| \cdot\|)$ be finite dimension normed space, $\dim X=3$, where $X:=\{x(t)=a+bt+ct^{2}:t\in[-1,1],a,b,c\in \mathbb{R}\}\subset C([-1,1], \mathbb{R})$ with norm from $C([-1,1], \mathbb{R})$. In the space $X$ we have the operator

$$Tx(t)=t^{2}\frac{d^{2}x(t)}{d^{2}t}-2t\frac{dx(t)}{dt}+4x(t)$$

and the linear functional $f(x):=\int_{-1}^{1}\sin(\pi t)Tx(t)dt,\ x\in X$.

Find:

1) $\ker(f)\in X$?

2) $d(t,\ker(f))=$?

3) $||f||=$?

So I have started with computing the kernel of $f$.

First let's compute $Tx(t)=2ct^{2}+2bt+4a$.

So my $f(x)=0\iff \int_{-1}^{1}\sin(\pi t)(2ct^{2}+2bt+4a)=0$.

So I get $\frac{4b}{\pi}=0$, because inner and outer integral is equal to 0.

Then should I write $\ker(f)=\{x\in C([-1,1], \mathbb{R}): x(t)=a+ct^{2}\}$? If I'm correct, then what is the next step 2), 3)?

Answer 1

As you noticed, $Tx(t)=2ct^2+2bt+4a$, so we can rewrite functional as $f(x)=\dfrac{4b}{\pi}$. Thus we have $|f(x)|=\dfrac{4}{\pi}|b|$. Considering $x(t)=t$ we have $\|f\|\geq\dfrac{4}{\pi}$. Show that $\|f\|\leq\dfrac{4}{\pi}.$

Considering any $x(t)=a+bt+ct^2$ we can notice that $\|x\|=\max\{|a+b+c|,|a-b+c|,|a-\frac{b}{4c}|\}$ (here obviously $b\ne0$, $c\ne0$). Next, we'll be interested only the first two expressions under the sign of max. Consider all possible cases for the signs of $b$.

1. $b>0$: $|b|<|a+b+c|\leq\|x\|$ if $a+c>0$ and $|b|<|a-b+c|\leq\|x\|$ if $a+c<0$;

2. $b<0$: $|b|<|-a-b-c|\leq\|x\|$ if $-c-a>0$ and $|b|<|a-b+c|\leq\|x\|$ if $-c-a<0$;

Cases when $a = 0$ or $c = 0$ (or both$=0$) are considered even more simply. Finally, $|f(x)|\leq\dfrac{4}{\pi}\|x\|$ and $\|f\|\leq\dfrac{4}{\pi}$.

Value $d(t,\ker f)$ can be calculated by the formula $d(t,\ker f)=\dfrac{|f(t)|}{\|f\|}$ (for $x(t)=t$ we take $b=1$, $a=c=0$).