I have been trying to solve an exercise in Normed Vector Spaces, and I'm stuck in the 2nd question.
My answer to the 1st question:
We have $\varphi$ linear.
Let
$$||\varphi(f)||_{1} = \int_0^1 |\varphi(f)(t)|dt$$
$$ = \int_0^1 |\int_0^t f(x)dx|dt$$
$$ \leq \int_0^1 \int_0^t |f(x)|dxdt$$
$$ \leq \int_0^1 \int_0^t |f(x)|dxdt + \int_0^1 \int_t^1 |f(x)|dxdt$$
$$ = \int_0^1 \int_0^1 |f(x)|dxdt$$
$$ = \int_0^1 |f(x)|dx\quad. \int_0^1 dt$$
$$ = \int_0^1 |f(x)|dx$$
$$ = ||f||_{1}$$
$$ \implies |||\varphi||| \leq 1, and \ \varphi \ is \ continuous $$
For the 2nd question, I'm trying to find a certain function in E that Verifies $=$ instead of $\leq$ in my answer to the previous question, so I can get the second inequality and hence the equality, using the sup in the definition of $|||\varphi|||$.
Finally,
I found the function $f_{0}(x)=e^{-nx}$, which verifies the conditions.
Where,
$$\frac{||\varphi(f_{0})||_{1} }{||f_{0}||_{1}} = \frac{1+\frac{e^{-n}}{n}-\frac{1}{n}}{1-\frac{e^{-n}}{n}}\rightarrow 1,\quad as \ n \rightarrow +\infty$$
And since $|||\varphi|||$ is bounded by $1$, $sup_{f \in E}(\frac{||\varphi(f)||_{1} }{||f||_{1}}) = 1$.
i.e, $|||\varphi||| = 1$.