Prove that the function $φ:C(I)\rightarrow\Bbb R$ defined as $φ(f):=f(1)$ for any $f\in C(I)$ is not continous with respect the $L_2$ topology.

Question

If $C(I)$ is the set of continous function on $I:=[0,1]$ then it is a well know result that the positions

1. $d_\infty(f,g):=\sup\{|f(x)-g(x)|:x\in I\}$
2. $\langle f,g\rangle:=\int_0^1fg$

for any $f,g\in C(I)$ define a distance and an inner product. So let be now $\varphi:C(I)\rightarrow\Bbb R$ the function defined as $$ \varphi(f):=f(1) $$ for any $f\in C(I)$. So given $f_0\in C(I)$ for any $a,b\in\Bbb R$ such that $a<b$ let be $$ \delta:=\min\{f_0(1)-a,b-f_0(1)\} $$ so that we observing that $$ \varphi[B(f_0,\delta)]\subseteq(a,b) $$ we conclude that $\varphi$ is continuous with repsect the distance $d_\infty$. Now my topology text claim that $\varphi$ is not continous with respect the topology on the inner product $\langle\,,\,\rangle$ but unfortunately I was not able to show this so that I thought to put a specific question where I ask to do this. Could someone help me, please?

Answer 1

$(C[0, 1],\|•\|_2)$

$\|f\|_2=[\int_{0}^{1} |f(t) |^2 dt]^{\frac{1}{2}}$

$\varphi:C(I)\rightarrow\Bbb R$ defined by $$\varphi(f):=f(1)$$

Claim :$\varphi$ is not continuous.

First of all, $\varphi \in {\scr{L}}(C[0, 1], R) $ i.e $\varphi$ is linear.

A linear map between two normed space is continuous iff continuous at $0$ .

To prove $\varphi$ is not continuous it is enough to show that it is not continuous at $0$ .

Choose, $(f_n) \subset C[0, 1]$ defined by $$f_n(t) =t^n$$

Then, $f_n \to 0 $ in the space $(C[0, 1], \|•\|_2) $(Verify)

But $\varphi(f_n) =f_n(1) =1 $ doesn't converges to $\varphi(\textbf{0}) =0$

Hence, $\varphi$ is not continuous at $0$ and hence discontinuous on the whole domain.

Hope it helps.