Let $V:=\{f:[0,1]\to\mathbb{R}|f\quad\text{continuous}\}$ provided with the norm $\|f\|_1:=\int_0^1 |f(x)|\, dx$ for $f\in V$.
Show that $\varphi : V\to\mathbb{R}$, $\varphi(f):=f(0)$ is linear but not continuous.
It is easy to show, that $\varphi$ is linear. Let $a\in\mathbb{R}$ and $f,g\in V$. Then $\varphi(af+g)=(af+g)(0)=af(0)+g(0)=a\varphi(f)+\varphi(g)$.
Now I want to show, that it is not continuous. I know some equivalent statements for a linear function $f:V\to W$ between two normed spaces, which I tried to disprove.
First, that there exists a $c>0$ such that $\|f(v)\|\leq c\cdot \|v\|$ for every $v\in V$
I tried several sequences of functions, but they did not work out.
What I want to do is to give a sequence such that $f_n(0)=\text{constant}>0$ but $\|f_n\|_1\to 0$ or that $f_n(0)\to \infty$ while $\|f_n\|_1$ stays bounded.
But nothing really worked. I tried a lot with $\cos(nx)$ but the absolute value makes it hard to inegrate. $\cos$ or $\sin$ should be involved I guess.
Hints are appreciated.
Thanks in advance.
Try with $f_n(t)=(1-t)^n$. The sequence $(f_n)_{n\in\mathbb N}$ converges to the null function, but you don't have $\lim_{n\in\mathbb N}\varphi(f_n)=\varphi(0)=0$.