Elementary interpolation inequality between Lebesgue and Sobolev-Slobodeckij spaces Let $W^{s, 2}$ for $0 < s < 1$ denote the Sobolev-Slobodeckij spaces on the interval $(0, 1)$ and $L^2$ the Lebesgue space on the same interval.  I'm interested in an elementary proof that there exists $C > 0$ such that for any $f \in W^{s, 2}$ there holds
$$ \| f \|_{W^{s/2, 2}} \leq C \| f \|_{L^2}^{1/2} \| f \|^{1/2}_{W^{s, 2}}. $$
I'm not super comfortable with interpolation theory, but as far as I know one has $(L^2, W^{s, 2})_{1/2,2} = W^{s/2, 2}$ by the real interpolation method (and reiteration theorem), such that this estimate should be true. I wish to apply a very similar estimate in another context where I cannot use this abstract result directly, which is why I'm interested in an elementary proof of the inequality above.
Let for $0 < s < 1$ the Gagliardo semi-norm of $f \in W^{s, 2}$ be denoted by
$$| f |_{W^{s, 2}} = \left( \int_0^1 \int_0^1 \frac{|f(x) - f(y)|^2}{|x-y|^{2s+1}}\,\mathrm d x \mathrm dy \right)^{1/2}.$$
Then it is sufficient to prove that there is $C > 0$ such that for any $f \in W^{s,2}$ there holds
$$ | f |_{W^{s/2, 2}} \leq C \| f \|_{L^2}^{1/2} | f |^{1/2}_{W^{s, 2}}. $$
I tried to prove this estimate directly, but I can ultimately only prove
$$ | f |_{W^{(s+\mu)/2, 2}} \leq C_\mu \| f \|_{L^2}^{1/2} \| f |^{1/2}_{W^{s, 2}} $$
for $0 < \mu \ll 1$ with $C_\mu \to \infty$ as $\mu \to 0$. Using Hölder's inequality I estimate (omitting the bounds in the integrals for simplicity)
$$\begin{align}
| f |_{W^{(s+\mu)/2, 2}}^2 
&\leq \int \int \frac{| f(x) - f(y) |}{|x - y|^{s+\mu+1}} \left( |f(x)| + |f(y)| \right)\,\mathrm dx \mathrm dy\\
&\leq 2 \int |f(y)| \int \frac{| f(x) - f(y) |}{|x - y|^{s+\mu+1}}\,\mathrm dx \mathrm dy\\
&\leq 2 \left( \int |f(y)|^2 \,\mathrm dy \right)^{1/2} \left( \int \left( \int \frac{| f(x) - f(y) |}{|x - y|^{s+1/2}} \frac{1}{|x-y|^{1/2+\mu}} \,\mathrm dx \right)^2 \mathrm dy \right)^{1/2}\\
&\leq 2 \left( \int |f(y)|^2 \,\mathrm dy \right)^{1/2} \left( \int \int \frac{| f(x) - f(y) |^2}{|x - y|^{2s+1}} \,\mathrm dx \mathrm dy \right)^{1/2} \left( \int \int \frac{1}{|x-y|^{1+2\mu}} \,\mathrm dx\mathrm dy \right)^{1/2}
\end{align}$$
such that
$$ |f|_{W^{(s+\mu)/2,2}} \leq C_\mu \|f \|_{L^2}^{1/2} |f|_{W^{s,2}}^{1/2} \quad\text{with} \quad C_\mu = \sqrt 2 \left( \int \int \frac{1}{|x-y|^{1+2\mu}} \,\mathrm dx\mathrm dy \right)^{1/4}. $$
But there holds $C_0 = \infty$ as the integral in the constant is unbounded for $\mu = 0$.
As an alternative approach I again used only Hölder's inequality to show that for any $0 < \mu \ll 1$ there holds
$$ |f|_{W^{s/2,2}} \leq \operatorname{ess\,sup}_{(x, y) \in (0, 1)^2} | x - y|^{\mu/2} |f |_{W^{\mu, 2}}^{1/2} |f|_{W^{s,2}}^{1/2} $$
but I read in the Hitchhikers Guide to Fractional Sobolev Spaces that only $\lim_{\mu \to 0} \mu |f|^2_{W^{\mu, 2}} = C \| f \|_{L^2}^2$ holds and the essential supremum in the last inequality is actually $1$, so the constant again blows up as $\mu \to 0$.
 A: Using a Fourier approach the interpolation inequality follows easily. In contrast to the original question I consider the function spaces with domain $\mathbb R^n$ for any $n \in \mathbb N$, the estimate for the interval $(0, 1)$ follows for $n = 1$ since $(0, 1)$ is an extension domain.
According to the Hitchhicker's Guide to Fractional Sobolev Spaces (Prop 3.4), there holds
$$ | u |_{W^{s, 2}(\mathbb R^n)}^2 = C(s, n) \int_{\mathbb R^n} |\xi|^{2s} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi $$
with some factor $C(s, n)$, where $\mathcal F$ denotes the Fourier transform. Then
$$ \begin{align}
| u |_{W^{s/2, 2}(\mathbb R^n)}^2 
&= C(s/2, n) \int_{\mathbb R^n} |\xi|^{s} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi \\
& \leq C(s/2, n) \left( \int_{\mathbb R^n} |\xi|^{2s} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi \right)^{1/2} \left( \int_{\mathbb R^n} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi \right)^{1/2}\\
& \leq C(s/2, n) C(s, n)^{-1/2} |u|_{W^{s, 2}(\mathbb R^n)} \| u \|_{L^2(\mathbb R^n)}
\end{align}$$
using Hölder's inequality and Plancherel's theorem.
While this answers the original question I'm still very interested in how one can directly argue using the Gagliardo semi-norm.
