To find the norm of a linear functional $f(x)$

Question

Let $\displaystyle{f(x)=\int_{-1}^{1}x(t)dt-\frac{1}{2n+1}\sum_{k=-n}^{n}x(k/n)}$ for all $x\in C[-1,1]$. I want to show that $f$ is continuous and calculate its norm.

I have tested the continuity as follows:

Let $x\in C[-1,1]$ such that $\left\|x\right\|\leq 1\Longrightarrow\displaystyle{ \max_{t\in[-1,1]}|x(t)|\leq 1}$ \begin{eqnarray} |f(x)|&=&\left|\int_{-1}^{1}x(t)dt-\frac{1}{2n+1}\sum_{k=-n}^{n}x(k/n)\right|\\ &\leq&\left|\int_{-1}^{1}x(t)dt\right|+\left|\frac{1}{2n+1}\sum_{k=-n}^{n}x(k/n)\right|\\ &\leq&\int_{-1}^{1}\left|x(t)\right|dt+\frac{1}{2n+1}\sum_{k=-n}^{n}\left|x(k/n)\right|\\ &\leq&\left\|x\right\|\int_{-1}^{1}dt+\frac{1}{2n+1}\sum_{k=-n}^{n}\left\|x\right\|\\ &=&2\left\|x\right\|+\frac{2n+1}{2n+1}\left\|x\right\|\\ &=&3\left\|x\right\| \end{eqnarray}

I need to find the norm. Please help!

Answer 1

Your bound is tight. To find an $x$ that almost achieves the bound, you might take $x$ that is $-1$ at each of the points $k/n$ but $+1$ except in small intervals around those points.

Answer 2

$$\displaystyle{f(x)=\int_{-1}^{1}x(t)dt-\frac{1}{2n+1}\sum_{k=-n}^{n}x(k/n)}\quad \textrm{for all}\quad x\in C[-1,1]$$ taking $\lim_{n\rightarrow\infty}$ both sides we get, $$\displaystyle{f(x)=\int_{-1}^{1}x(t)dt-\lim_{n\rightarrow\infty}\frac{1}{2n+1}\sum_{k=-n}^{n}x(k/n)}\\\implies\displaystyle{f(x)=\int_{-1}^{1}x(t)dt-\lim_{n\rightarrow\infty}\frac{1}{2}.\frac{2}{2n+1}\sum_{k=-n}^{n}x(k/n)}\\ \implies \displaystyle{f(x)=\int_{-1}^{1}x(t)dt-\frac{1}{2}}\int_{-1}^{1}x(t)dt=\frac{1}{2}\int_{-1}^{1}x(t)dt $$ See if this does help.