We have $$ DF(g)=\frac{\|\nabla g \|_{L^{5/3}(X)}}{ \|g\|_{L^{10}(X)} -\|g\|_{L^1(X)}},$$

which is a homogeneous expression. Hence we can drop the condition $\int_X g^{10}dx=1$ and replace it by $\|g\|_{L^1(X)}=\int_X g dx=1$. If $dim(X)=2$ we have by the Sobolev and the triangle inequality $$ DF(g)=\frac{\|\nabla g \|_{L^{5/3}(X)}}{ \|g\|_{L^{10}(X)} -\|g\|_{L^1(X)}}\geq \frac{C\|g-1 \|_{L^{10}(X)}}{ \|g\|_{L^{10}(X)} -1}\geq C. $$

For $dim(X)>2$ the Sobolev space $W^{1,5/3}(X)$ can not be embedded into $L^{10}(X)$, i.e we could find $g\in W^{1,5/3}(X)$ with $g \notin L^{10}(X)$ or with $\|g\|_{L^{10}(X)}$ arbitrarily large. Consider for example that $B_0(1)(\mathbb{R}^3) \subset X$ and $g(x)=|x|^{-3/10+\epsilon}$ in $B_0(1)(\mathbb{R}^3)$.

We have $$ DF(g)=\frac{\|\nabla g \|_{L^{5/3}(X)}}{ \|g\|_{L^{10}(X)} -\|g\|_{L^1(X)}},$$

which is a homogeneous expression. Hence we can drop the condition $\int_X g^{10}dx=1$ and replace it by $\|g\|_{L^1(X)}=\int_X g dx=1$. If $dim(X)=2$ we have by the Sobolev and the triangle inequality $$ DF(g)=\frac{\|\nabla g \|_{L^{5/3}(X)}}{ \|g\|_{L^{10}(X)} -\|g\|_{L^1(X)}}\geq \frac{C\|g-1 \|_{L^{10}(X)}}{ \|g\|_{L^{10}(X)} -1}\geq C. $$

For $dim(X)>2$ the Sobolev space $W^{1,5/3}(X)$ can not be embedded into $L^{10}(X)$, i.e we could find $g\in W^{1,5/3}(X)$ with $g \notin L^{10}(X)$ or with $\|g\|_{L^{10}(X)}$ arbitrarily large. Consider for example for $d\geq 3$ that $B_0(1)(\mathbb{R}^d) \subset X$ and $g(x)=|x|^{-d/10+\epsilon}$ in $B_0(1)(\mathbb{R}^d)$.