# About the boundedness of the derivative of a function which is in special function space.

If $f \in C^1 ([0,T] , L^2) \cap C^0 ([0,T] , W^{1,2} )$, then how can I conclude that $$\left \| \frac{\partial f}{\partial t} \right \|_{L^\infty([0,T] \times \Bbb R^n )} < \infty ?$$ Here $f$ is defined on $[0,T] \times \Bbb R^n$ , and the notation $f \in C^1([0,T], L^2)$ means that $\| f(t) \|_{L^2 (\Bbb R^n)}$ is continuously differentiable on $[0,T]$, $W^{s,p}$ means the usual Sobolev space.

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Handwaving: $\partial f/\partial t = \mathrm df/\mathrm dt - \partial f/\partial x\ \partial x/\partial t$, thus the first inclusion gives (ess.) finiteness for the first term on the rhs, the second gives (ess.) finiteness for the second term, thus their difference is also essentially finite. – filmor Oct 21 '12 at 15:45
@filmor Thank you, but here $x$ is not depend on $t$. $f = f(t,x_1 , \cdots , x_n)$. – Ann Oct 21 '12 at 15:49
I don't think this is true: Take $f(t,x)=t\eta(x)|x|^{n-\alpha}$ with $\eta\in C_c^{\infty}(\mathbb{R}^n)$ a cut-off function, then for a suitable $\alpha>0$ we have $f\in W^{1,2}$, but $\eta(x)|x|^{n-\alpha}\notin L^{\infty}$. – Jose27 Oct 21 '12 at 17:52

Unless I am missing something, this does not seem to be correct. Let $n=3$, and define $$F(x) = \|x\|^{-1/3} \quad\text{ for \|x\|\le 1,}$$ and extend it smoothly to a function with compact support on $\mathbb{R}^3$. Then $$\|\nabla F(x)\| = \frac13 \|x\|^{-4/3} \quad\text{ for \|x\|\le 1},$$ so $F \in L^2(\mathbb{R}^3)$, $\nabla F \in L^2(\mathbb{R}^3)$ and thus $F\in W^{1,2}(\mathbb{R}^3)$. Now define $$f(t,x) = t F(x).$$ Then $$\frac{\partial f}{\partial t} (t,x) = F(x),$$ so this example seems to satisfy the even stronger assumption that $f \in C^1([0,T],W^{1,2})$, but it $\frac{\partial f}{\partial t} = F \notin L^\infty$.
The Sobolev embedding doesn't give that inclusion: Take $f(x)=\log\log(1+|x|^{-1})$. – Jose27 Oct 21 '12 at 18:06
Thank you very much. But how it becomes if we add the additional assumption: $f(t,x)$ is bounded on $[0,T] \times \Bbb R^n$ ? – Ann Oct 21 '12 at 19:42