Let endow $\mathcal{C}([0;1], \mathbb{R})$ with the $L^1$-norm.
Let $L$ be a mapping from $\mathcal{C}([0;1], \mathbb{R})$ to $\mathbb{R}$ such that:
$\forall f \in \mathcal{C}([0;1], \mathbb{R}), L(f)=f(a)$ where $a \in [0;1]$.
Show that $L$ is a discontinuous mapping.
I tried to prove it by contradiction but I ended up with an equality that is true. I can't manage to find a contradiction. I was also thinking about finding a sequence of function that violates the sequential characterization of continuity, but I could not find one. Any help would be appreciated!
Note that $L$ depends on $a$: it’s the evaluation function at $a$, so I’ll call it $L_a$. So you’re asked to show that all maps $L_a$ where $a \in [0,1]$ are discontinuous (they are linear by definition of the operations on $C([0,1])$).
For $L_0$ consider maps like $f \in C([0,1])$ that are defined by
$$f_n(x) = \begin{cases} -nx + 1 & 0 \le x \le \frac1n\\ 0 & \frac1n \le x \le 1\\ \end{cases}$$
We have $\|f_n\|_1 = \frac{1}{2n} \to 0$ so that $f_n \to 0$ in the $L^1$-norm while $L_0(f_n) = 1$ all the time and $L_0$ is $0$ at the limit... Try to apply this idea for any $a$ to prove all of them discontinuous...