Let $P$ be the space of all real-valued polynomials, defined on $\mathbb{R}$. For a polynomial $p\in P$, such that $p(t)= \sum_{k=0}^na_kt^k$, set $||p||:= \sum_{k=0}^n|a_k|$. Consider the linear mappings $f_i : P \to \mathbb{R}, i = 1, ..., 3$, which are given by
$f_1(p)= \int_0^1p(t)dt $ , $f_2(p)= p'(0) $, $f_3(p)= p'(1) $
Examine for $i=1, ..., 3$ whether $f_i$ is continuous and determine the operator norm $||f_i||$
Since this is a homework, I would love to solve it by myself but am really having trouble with how to start since I'm not used to this type of exercise. What I have been thinking of so far is: if $f_i$ is countinous, I should show that it is bounded (but don't really know how), else I should find some point where I can show a counterexample. Is this the right way to solve this exercise? Any hint would be really appreciated!
Alright, I'll show an example with $f_1$. Let $p=\sum_{k=0}^n a_kx^k$. Then:
$|f_1(p)|=|\int_0^1\sum_{k=0}^n a_kx^kdx|=|\sum_{k=0}^n a_k\int_0^1 x^kdx|=|\sum_{k=0}^n \frac{a_k}{k+1}|\leq\sum_{k=0}^n |a_k|=||p||$
Since this is true for all $p\in P$ we proved that the functional is bounded (which is equivalent to continuous) and its norm is at most $1$. It makes sense to believe that the norm is equal to $1$. And this is true, because if we take $p=1$ (the constant polynomial $1$) then it is easy to check that $\frac{|f_1(p)|}{||p||}=1$. So the norm of $f_1$ is at least $1$.
Now try to do the same thing with $f_2$ and $f_3$.