$\newcommand{\al}{\alpha}$In [leo monsaingeon's answer][1], for the the value $J(\al)$ of the infimum it was shown that 
\begin{equation*}
	J(\al)\le9
\end{equation*}
and conjectured that 
\begin{equation*}
	J(\al)\to9 
\end{equation*}
as $\al\downarrow0$. 

---

This post is to confirm the conjecture and, moreover, to provide an explicit lower bound on $J(\al)$:  
\begin{equation*}
	J(\al)\ge9(1-\al-3^{1/3} (2 \pi)^{2/3}\al^{1/3}) \tag{$*$}\label{*}  
\end{equation*}
for $\al\in(0,1]$. 

Let 
\begin{equation*}
	\text{$a:=\pi/3$ and $b:=2\pi/3$.}
\end{equation*}

>**Lemma 1:** If $f$ is absolutely continuous on $[a,b]$ with $\int_a^b f^2\le\al$, then 
\begin{equation*}
	\int_a^b f'^2\ge\frac{(f_a+f_b)^4}{64\al} \tag{$\clubsuit$}\label{-2}
\end{equation*}
or 
\begin{equation*}
	f_a+f_b\le\sqrt{\frac{16\al}{b-a}}. \tag{$\clubsuit\clubsuit$}\label{-1}
\end{equation*}
Here and in what follows,
\begin{equation*}
	\text{$f_a:=|f(a)|$ and $f_b:=|f(b)|$.}
\end{equation*}

>**Lemma 2:** If $f$ is absolutely continuous on $[0,a]$ with ($f(0)=0$ and) $\int_0^a f^2\ge c^2$ for some $c\in[0,1]$, then 
\begin{equation*}
	\int_0^a f'^2\ge9(c-f_a\sqrt{a/3})_+^2\ge9(c^2-2f_a\sqrt{a/3}), \tag{$\heartsuit$}\label{-0.5}
\end{equation*}
where $u_+:=\max(0,u)$. 


>**Lemma 2':** If $f$ is absolutely continuous on $[b,\pi]$ with ($f(\pi)=0$ and) $\int_b^\pi f^2\ge 1-\al-c^2$ for some $c\in[0,\sqrt{1-\al}]$, then 
\begin{equation*}
	\int_b^\pi f'^2\ge9(1-\al-c^2-2f_b\sqrt{(\pi-b)/3}).  
\end{equation*}

These lemmas will be proved at the end of this answer. 

In the case when \eqref{-2} holds, it follows from Lemmas 2 and 2' and the definitions $a:=\pi/3$ and $b:=2\pi/3$ that 
\begin{equation*}
	\int_0^\pi f'^2=\int_a^b f'^2+\int_0^a f'^2+\int_b^\pi f'^2
	\ge B_1(u):=9(1-\al)+\frac{u^4}{64\al}-6\sqrt\pi\, u,  
\end{equation*}
where $u:=f_a+f_b$. Minimizing $B_1(u)$ in $u$, we get \eqref{*}. 

In the remaining case, by Lemma 1, \eqref{-1} must hold. So, it then follows from Lemmas 2 and 2' 
that 
\begin{equation*}
	\int_0^\pi f'^2\ge\int_0^a f'^2+\int_b^\pi f'^2
	\ge 9(1-\al-2(f_a+f_b)\sqrt\pi/3) 
	\ge 9(1-\al-8\sqrt{\al/3}).  
\end{equation*}
For $\al\in(0,1]$, the latter expression is no less than the expression on the right-hand side of \eqref{*}. 

Thus, \eqref{*} follows from Lemmas 1, 2, and 2'. 

---

It remains to prove these lemmas. 

*Proof of Lemma 1:* One can see that a minimizer $f$ of $\int_a^b f'^2$ subject to the conditions on $\int_a^b f^2$, $|f(a)|$, and $|f(b)|$ exists and is a real-analytic function on $[a,b]$. Replacing such a minimizer $f$ by $|f|$ does not change the values of $\int_a^b f'^2$, $\int_a^b f^2$, $|f(a)|$, and $|f(b)|$. So, without loss of generality (wlog) we may assume that $f$ is nonnegative minimizer. Let $c\in[a,b]$ be a point of minimum of $f$ on $[a,b]$, so that $0\le f(c)\le f(x)$ for all $x\in[a,b]$. 

Note that no point $\xi$ in the interval $(a,c)$ can be the point of strict local maximum of $f$: Otherwise, if $m:=\max(f(\xi_1),f(\xi_2))<f(\xi)$ for some $\xi_1$ and $\xi_2$ such that $a<\xi_1<\xi<\xi_2<c$, take any $l\in(m,f(\xi))$ and get a new function $g$ by replacing $f(x)$ by $g(x):=\min(l,f(x))$ for all $x\in[\xi_1,\xi_2]$, with $g=f$ on $[a,b]\setminus[\xi_1,\xi_2]$. Then the conditions on $\int_a^b f^2$, $|f(a)|$, and $|f(b)|$ will hold with $g$ in place of $f$, whereas $\int_a^b g'^2$ will be strictly less than $\int_a^b f'^2$ -- which will contradict the assumption that $f$ is a minimizer. 

Therefore and because $c$ is a point of (global) minimum of $f$ on $[a,b]$, 
it now follows that no point $\eta$ in the interval $(a,c)$ can be the point of strict local minimum of $f$: otherwise, there would exist a point in $(\eta,c)\subseteq(a,c)$ of strict local maximum of $f$. 

So, $f$ is decreasing on $[a,c]$ and, similarly, increasing on $[c,b]$. Hence, for any real 
\begin{equation*}
	h\ge f_c:=f(c), \tag{0}\label{0}
\end{equation*}
the set 
\begin{equation*}
	\{x\in[a,b]\colon f(x)>h\}
\end{equation*}
is the complement to $[a,b]$ of an interval $[a_h,b_h]\subseteq[a,b]$ such that  
\begin{equation*}
	\text{(i) $f(a_h)=h$ or $a_h=a\quad$ and $\quad$ (ii) $f(b_h)=h$ or $b_h=b$.}
\end{equation*}
Next, 
\begin{equation*}
	\al\ge\int_a^b f^2\ge\int_a^{a_h} f^2+\int_{b_h}^b f^2
	\ge(a_h-a+b-b_h)h^2,
\end{equation*}
so that 
\begin{equation*}
	(a_h-a)+(b-b_h)\le\frac\al{h^2}. \tag{1}\label{1} 
\end{equation*}
By Cauchy--Schwarz inequality, 
\begin{equation*}
	(a_h-a)\int_a^{a_h} f'^2\ge\Big(\int_a^{a_h} f'\Big)^2=(f_a-h)_+^2 
\end{equation*}
and similarly $(b-b_h)\int_{b_h}^b f'^2\ge(f_b-h)_+^2$. So, in view of \eqref{1},
\begin{equation*}
	\int_a^b f'^2\ge\int_a^{a_h} f'^2+\int_{b_h}^b f'^2
	\ge\frac{(f_a-h)_+^2}u+\frac{(f_b-h)_+^2}{\al/h^2-u}, \tag{2}\label{2}
\end{equation*}
where $u:=a_h-a$; if the denominator of either one of the ratios in \eqref{2} is $0$, then the corresponding ratio is understood as $0$. Minimizing the sum of the ratios in \eqref{2} in $u\in[0,\al/h^2]$, we get 
\begin{equation*}
	\int_a^b f'^2\ge\frac{h^2}\al\,((f_a-h)_+ + (f_b-h)_+)^2
	\ge\frac{h^2}\al\,((f_a+f_b-2h)_+)^2.  \tag{3}\label{3}
\end{equation*}

Now we need to distinguish between two cases: 

*Case 1: $f_a+f_b\ge4f_c$* and 

*Case 2: $f_a+f_b<4f_c$*. 
 
In Case 1, condition \eqref{0} will be satisfied with 
$h_*:=(f_a+f_b)/4$ in place of $h$. So, substituting $h_*$ for $h$ in \eqref{3}, we get \eqref{-2}. 

In Case 2, write $f_c^2(b-a)\le\int_a^b f^2\le\al$. So, \eqref{-1} follows by the case condition. 

This completes the proof of Lemma 1. $\quad\Box$

*Proof of Lemma 2:* Wlog, $f(a)\ge0$, so that $f(a)=f_a$. Let 
\begin{equation*}
	g(t):=f(t)-kt,
\end{equation*}
where 
\begin{equation*}
	k:=\frac{f_a}a,
\end{equation*}
so that $g(0)=0=g(a)$ and hence, by [Wirtinger's inequality][2], 
\begin{equation*}
	\int_0^a g'^2\ge\frac{\pi^2}{a^2}\,\int_0^a g^2. 
\end{equation*}
Also, in view of the fact that $\int_0^a g'=g(a)-g(0)=0$, we have 
\begin{equation*}
	\int_0^a f'^2\ge\int_0^a (g'+k)^2=\int_0^a g'^2+k^2 a\ge\int_0^a g'^2. 
\end{equation*}
So, 
\begin{equation*}
	\int_0^a f'^2\ge\frac{\pi^2}{a^2}\,\int_0^a g^2=9\int_0^a g^2. \tag{4}\label{4}
\end{equation*}
By Minkowski's inequality, 
\begin{equation*}
	c\le\sqrt{\int_0^a f^2}\le\sqrt{\int_0^a g^2}+k\sqrt{\int_0^a t^2\,dt},
\end{equation*}
whence 
\begin{equation*}
	\int_0^a g^2\ge(c-f_a\sqrt{a/3})_+^2. 
\end{equation*}
Now the first inequality in \eqref{-0.5} follows from \eqref{4}, and the second inequality in \eqref{-0.5} is trivial. 

This completes the proof of Lemma 2. $\quad\Box$

*Proof of Lemma 2':* Lemma 2' is quite similar to Lemma 2: it is obtained from Lemma 2 by the reflexions $t\leftrightarrow\pi-t$ and $c\leftrightarrow\sqrt{1-\al-c^2}$. $\quad\Box$

  [1]: https://mathoverflow.net/a/414597/36721 
  
  [2]: https://en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions#Second_version