infimum of a functional in $W^{1,p}((0,1))$ Consider the functional 
$$\mathcal{F}(u)=\int_{0}^{1}x^{\alpha}|u'(x)|^pdx,\ \ u\in W^{1,p}((0,1)),$$
where $\alpha\ge 0$ and $1<p<\infty$. Given $a<b$, find the value of 
$$\inf\{\mathcal{F}(u): u\in W^{1,p}((0,1)), u(0)=a, u(1)=b\}.$$
For the case $p>\alpha +1$, my proof look like this: 
Since $u\in W^{1,p}((0,1))$, we have 
$$u(1)-u(0)=\int_{0}^{1}u'(x)dx.$$
Base on the above equality, we have
\begin{align*}
(b-a)&=u(1)-u(0)\\
&=\int_{0}^{1}u'(x)dx\\
&=\int_{0}^{1}x^{\frac{\alpha}{p}}u'(x)x^{-\frac{\alpha}{p}}dx\\
&\le (\int_{0}^{1}x^{\alpha}|u(x)|^pdx)^{1/p}(\int_{0}^1x^{-\frac{\alpha q}{p}})^{1/q}\\
&=(\int_{0}^{1}x^{\alpha}|u(x)|^pdx)^{1/p}(\int_{0}^1x^{-\frac{\alpha}{p-1}})^{1/q}
\end{align*}
where we have $\frac{1}{p}+\frac{1}{q}=1$. Since $-\frac{\alpha}{p-1}>-1$, we have 
\begin{align*}
\int_{0}^1x^{-\frac{\alpha}{p-1}}&=\frac{1}{1-\frac{\alpha}{p-1}}\\
&=\frac{p-1}{p-\alpha-1}.
\end{align*}
Thus, we imply that 
$$(b-a)\le (\int_{0}^{1}x^{\alpha}|u(x)|^pdx)^{1/p}(\frac{p-1}{p-\alpha-1})^{1/q}.$$
We deduce that 
$$\int_{0}^{1}x^{\alpha}|u(x)|^pdx \ge (b-a)^p(\frac{p-\alpha-1}{p-1})^{p/q}$$
I am trying to prove that 
$$(b-a)^p(\frac{p-\alpha-1}{p-1})^{p/q}=\inf\{\mathcal{F}(u): u\in W^{1,p}((0,1)), u(0)=a, u(1)=b\}$$
by finding a function $u\in W^{1,p}((0,1)), u(0)=a, u(1)=b$ such that $\mathcal{F}(u)=(b-a)^p(\frac{p-\alpha-1}{p-1})^{p/q}$. However, I have not succeeded. Can any one help me with that ?
 A: You need to see when we have an equality in Holder's inequality $$\|fg\|_1\leq \|f\|_p\|g\|_q$$
(https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality). The equality is achieved iff $c_1|f|^p=c_2|g|^q,\,\,\mu-a.e$ for some constants $(c_1,c_2)\neq (0,0)$ where $\mu$ in our case is the Lebesgue measure on $\mathbb R$.
Applying this to $|f|^p=x^\alpha|u'|^p$ and $|g|^q=x^{-\frac{\alpha q}{p}}$ we get $$c_1 x^\alpha|u'|^p=c_2x^{-\frac{\alpha q}{p}}$$ 
Here, obviously $c_1\neq 0$, so WLOG  $c_1=1$ and we rename $c_2=:c>0$ a generic constant.
Because you also use the inequality 
$$\int\limits_{0}^{1}{x^{\frac{\alpha}{p}}u'x^{-\frac{\alpha}{p}}dx}\leq \int\limits_{0}^{1}{x^{\frac{\alpha}{p}}|u'|x^{-\frac{\alpha}{p}}dx}$$
we want $u'\ge 0,\,\,\mu-a.e$ in order to have equality here too. Finally 
$$(u')^p=c x^{-\frac{\alpha q}{p}-\alpha}=c x^{-\alpha-\frac{\alpha}{p-1}}=cx^{-\frac{\alpha p}{p-1}}$$
$$\Rightarrow u'=c x^{-\frac{\alpha}{p-1}}\in L^p(0,1)$$
$$\Rightarrow u(x)=a+c\int\limits_{0}^{x}{t^{-\frac{\alpha}{p-1}}}=a+c\frac{p-1}{p-1-\alpha}x^{\frac{p-1-\alpha}{p-1}}$$ where $c$ is determined from $$b-a=u(1)-u(0)=c\int\limits_{0}^{1}{t^{-\frac{\alpha}{p-1}}}=c\frac{p-1}{p-1-\alpha}$$
Therefore $u(x)=a+(b-a)x^{\frac{p-1-\alpha}{p-1}}\in L^p(0,1)$
