Functional Analysis, a question that needs clarification. Find the norm of the linear operator $A:C[-1,1]\to L^p[-1,1]; p\geq1$ that is defined as: $$A(x(t))=\int_{-1}^{1}{{x(s)\over (s-t)^{1 \over 3}}}ds$$
Can someone provide an answer with a little more detail, to see where I make a mistake? Thank you in advance.
 A: First of all, the operator is unbounded if $p\geq 3$.
From Minkowski's integral inequality, we see that
$$||A(x)(t)||_{L^p(dt)} = 
\lvert\lvert\int_{-1}^1 \dfrac{x(s)}{(s-t)^{1/3}} ds 
\rvert\rvert_{L^p(dt)}
\leq 
\int_{-1}^1 \lvert\lvert 
\dfrac{x(s)}{(s-t)^{1/3}} \rvert\rvert_{L^p(dt)} ds 
= \int_{-1}^1 \lvert x(s)\rvert
\lvert\lvert
(s-t)^{-1/3} \rvert\rvert_{L^p(dt)} ds$$
Obviously, 
$\lvert\lvert 
{(s-t)^{-1/3}} \rvert\rvert_{L^p(dt)}=\infty$ if  $p\geq 3$, where $s,t\in [-1,1]$.   
Therefore, in order for the operator to be bounded, we require  $1\leq p<3$.
We compute
$$\lvert\lvert 
{(s-t)^{-1/3}} \rvert\rvert_{L^p(dt)} =
\left(\int_{-1}^1 |s-t|^{-p/3}dt\right)^{1/p}=
\left(\int_{-1}^s (s-t)^{-p/3}dt + \int_{s}^1 (t-s)^{-p/3}dt\right)^{1/p}
= \alpha^{-1/p} [(1+s)^\alpha+(1-s)^\alpha]^{1/p},$$
where $\alpha=1-p/3>0, s\in [-1,1]$.
In view of the right-hand side in the first inequality, we now have 
$$||A(x)(t)||_{L^p(dt)} \leq \int_{-1}^1 \lvert x(s)\rvert
\lvert\lvert
(s-t)^{-1/3} \rvert\rvert_{L^p(dt)} ds
=\alpha^{-1/p} \int_{-1}^1 \lvert x(s)\rvert
[(1+s)^\alpha+(1-s)^\alpha]^{1/p} ds
\leq \alpha^{-1/p} \int_{-1}^1 
[(1+s)^\alpha+(1-s)^\alpha]^{1/p} ds ,$$
where we assume $||x(t)||_{\text{sup}}\leq 1.$
Thus the quantity on the right-hand side above gives an upper bound for the norm of $A$.
I cannot find the exact norm of $A$. The idea below might help or it might not.
Note that equality holds in Minkowski's inequality iff the integrand is in the form $f(x,y)=g(x)h(y)$. Hence in your case, maybe we can approximate $(s-t)^{-1/3}$ by functions in the form $g(s)h(t)$(this is not possible though since $s=t$ produces a singularity) or a sum of them, i.e., $\sum_{i=1}^r g_i(s)h_i(t) $, to see how close we can reach the upper bound. 
PS: Actually, one can show that $A(x)(t)$ is continuous using dominated convergence theorem.
This implies that $A:C[-1,1]\to C[-1,1].$
