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

If $f \in C^1 ([0,T] , L^2) \cap C^0 ([0,T] , W^{1,2} )$, $f (t,x) : bounded\; on \; [0,T] \times \Bbb R^n$ then how can I conclude that $$\left \| \frac{\partial f}{\partial t} \right \|_{L^\infty([0,T] \times \mathbb R^n )} < \infty ?$$

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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Why would you even expect this to be true? Or is something misstated here? –  Michael Renardy Oct 21 '12 at 21:17
You definitely cannot. You can easily make a counterexample by taking any unbounded function g in $W^{1,2}$ and making $f(t,x) = t g(x)$ –  Luis Silvestre Oct 27 '12 at 23:37