Discontinuity of a linear functional

Question

Consider the space $C[0,1]$ of complex valued continuous functions on $[0,1]$ endowed with the norm $||f||_1=\int_0^1|f(t)|dt$. Is the linear functional, say $\lambda:C[0,1]\rightarrow \mathbf{C}, \lambda(f)=f(\frac{1}{2})$, discontinuous? Why?

Answer 1

Let the graph of $f_n$ be a little triangle: $f_n(1/2)=n$, linear on $[1/2-1/n, 1/2]$ and on $[1/2, 1/2+n]$, and $0$ otherwise. It's clear that $\lVert f_n \rVert=1$ and $\lambda(f_n)=n$. So we can bound the norm of $\lambda$ from below: $$\lVert \lambda \rVert = \sup_{\lVert v \rVert = 1} |\lambda(v)| \geqslant \sup_{n \in \mathbb{N}} |\lambda(f_n)| = +\infty$$ So it is not bounded, hence not continuous.