Boundedness of a linear functional with a parameter I'm working on this question:

Let consider the subspace $Z=\{x\in C[0,1];x(0)=0\}$. For which values of $r>0$ the linear functional
$$
f_r(x):=\int_0^1\frac{x(t)}{t^r}dt
$$
is bounded? For those $r$, find $\|f_r\|$.

Well, unitl now, I get
$$
|f_r(x)|\leqslant\|x\|_\infty\int_0^1t^{-r}dr,
$$
and then the values of $r$ that makes sense in the integral are $0<r<1$. So, we obtain
$$
|f_r(x)|\leqslant\frac{1}{1-r}\|x\|_\infty\quad\Longrightarrow\quad\|f_r\|\leqslant\frac{1}{1-r}.
$$
If my idea is right, then candidate to be the norm of $f_r$ is $1/(1-r)$. But I can't find the $x\in Z$ that does the work to the other inequality.
However, there is a big chance that what I'm doing is not correct...
Thanks for any help!

*

*to get norm equality you can test your functional by $\phi_n(t)=t^{1/n}$, and $n$ large enough. By doing the computations you get a maximizing sequence: $$||f_r||=\sup_{||x||_\infty=1, x \in Z} |f_r(x)| \geq |f_r(\phi_n)|$$ for all $n$, and the latter term tends exactly to $1/(1-r)$.


*the same sequence shows that for $r=1$ the functional is unbounded
