Norm of the functional $f(x)=\int_0^2 x(t).(t^2-1)dt$ in the space $X=L_1(0,2)$.

Question

I am trying to calculate the norm of the functional $f(x)=\int_0^2 x(t).(t^2-1)dt$ in the space $X=L_1(0,2)$.

What we have is: $||f(x)||\le ||x||_1.||t^2-1||_{\infty}$ by extended Hölder's inequality.

Then we have $||f(x)||\le 3.||x||_1$.

Now let $y_n=n.\mathcal{X}_{[2-\frac{1}{n},2]}$ which has norm $||y_n||_1=1$

But the problem is the source I'm looking at says $f(y_n)\rightarrow 3$ as $n\rightarrow \infty$ thus $||f||=3$, and I calculate $f(y_n)$ different.

Can anyone tell me explicitly why $\int_0^2 y_n.(t^2-1)dt \rightarrow 3$

Answer 1

You mean you are unable to compute $$ \int_{2-1/n}^2n(t^2-1)\mathrm dt=\left.n\left(\frac{t^3}3-t\right)\right|_{2-1/n}^2=3-\frac2n+\frac1{3n^2},$$ how so?