 Solution

The finite-energy comparison $I_{\epsilon }(u_{\epsilon })\le I_{\epsilon }(x)<\infty$ gives $\epsilon \int |u_{\epsilon }^{\prime }{|}^{4/3}<\infty$. Together with the endpoint value and absolute continuity, this proves $u_{\epsilon }\in W^{1,4/3}(0,1)$.

For $\alpha >1/4$, the power competitor $u_{\alpha }=x^{\alpha }$ is admissible and has finite derivative energy. Direct integration givesChoose $\delta ={\epsilon }^{1/3}$ and $\alpha =1/4+\delta$. The first three terms tend to zero as $\alpha \downarrow 1/4$, while the last equals $\frac{3}{4}{\epsilon }^{2/3}(1/4+{\epsilon }^{1/3}{)}^{4/3}\to 0$. By minimality and nonnegativity,One also obtains the quantitative bound $I_{\epsilon }(u_{\epsilon })=O({\epsilon }^{2/3})$: for $\alpha =1/4+\delta$, the mean value theorem gives $|x^{4\alpha }-x|\le 4\delta x|\log x|$, and ${\int }_0^1{x}^2{\log }^{2}xdx=2/27$. Thus the squared mismatch is at most $32{\delta }^2/27$, matching the order of the derivative penalty with this choice. This is a power-law comparison for vanishing derivative regularization.