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alia
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awarded
 Supporter
comment
Show $v(x,t) \in L^2([0,T];H^2(\mathbb{R}))$ when $v(x,t)$ is a transformation of a $L^2([0,T];H^2(\mathbb{R}))$ function
you are right! it holds for any $T>0$, then I can take $T \rightarrow \infty$ and cover all the same values as with $\log(5T+1)$? I believe that is what you mean to say
awarded
 Scholar
accepted
Show $v(x,t) \in L^2([0,T];H^2(\mathbb{R}))$ when $v(x,t)$ is a transformation of a $L^2([0,T];H^2(\mathbb{R}))$ function
comment
Show $v(x,t) \in L^2([0,T];H^2(\mathbb{R}))$ when $v(x,t)$ is a transformation of a $L^2([0,T];H^2(\mathbb{R}))$ function
thank you for the details! how do i reconcile this with $v(x,t)$ being in $L^2([0,T];H^2(\mathbb{R}))$, since the interval is not always smaller than $[0,\log(5T+1)]$?
awarded
 Student
asked
Show $v(x,t) \in L^2([0,T];H^2(\mathbb{R}))$ when $v(x,t)$ is a transformation of a $L^2([0,T];H^2(\mathbb{R}))$ function
awarded
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