What is the norm of this linear functional?

Question

Let $X = C[0,1]$ be the space of continuous functions with the max norm. Then define the linear functional;

$$f(x) = \int_0^1 x(t)dt - x(1/2)$$

Then is the norm of this operator 1 or 0? I'm a little confused how to argue this?

Thanks!

Answer 1

Answer. The norm of this operator is equal to 2.

Explanation. Clearly the norm of this operator is less or equal to 2.

Set $$ x_n(t)=\left\{\begin{array}{llll} 1 & \text{if} & x\in[0,1/2-1/n], \\ n-1-2nt & \text{if} & x\in[1/2-1/n,1/2],\\ -n-1+2nt & \text{if} & x\in[1/2,1/2+1/n],\\ 1 & \text{if} & x\in[1/2+1/n,1]. \end{array}\right. $$ Then $$ \|x_n\|=1, $$ and $$ \Big|\int_0^1 x_n(t)\,dt-x(1/2)\,\Big|\ge\int_0^{1/2-1/n}dt+\int_{1/2+1/n}^1dt -\int_{1/2-1/n}^{1/2+1/n}+1=2-\frac{4}{n}. $$ Hence, the norm of this operator is greater or equal to $2-4/n$, for every $n\in\mathbb N$, and hence, it is greater or equal to $2$.