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If $U_{n} \to U$ weakly* in $BV[0,1]$ is it true that $U_{n}\mid_{[0,\frac{1}{2}]} \to U\mid_{[0,\frac{1}{2}]}$ weakly* in $BV[0,\frac{1}{2}]$?


I'll give some explication. Let $U_{n} \in BV[0,1]$ and $U_n \to U$ weakly*, i.e. for any $y \in C[0,1]$
$$ \int\limits_{0}^{1} y(t)dU_{n}(t) \to \int\limits_{0}^{1}y(t)dU(t) $$ Restrictions $U_{n} \mid _{[0,\frac{1}{2}]}, U \mid_{[0,\frac{1}{2}]} \in BV[0,\frac{1}{2}]$. For any $y \in C[0,\frac{1}{2}]$ such that $y(\frac{1}{2}) = 0$ we have $$ \int\limits_{0}^{\frac{1}{2}} y(t)dU_{n}(t) \to \int\limits_{0}^{\frac{1}{2}}y(t)dU(t) $$ But is it true for any $y \in C[0,\frac{1}{2}]$? In other words, it it true, that $U_{n}\mid_{[0,\frac{1}{2}]} \to U\mid_{[0,\frac{1}{2}]}$ weakly*?

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1 Answer 1

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Not if I understand your conventions correctly. Consider $$U_n=\chi_{(1/2+1/n,1]},\quad U=\chi_{(1/2,1]}.$$ In general, you have to be very careful about jump discontinuities at interval endpoints.

**Edit:** More than that, you have to worry about concentrations of variation. Here is another example: Let $U_n(x)=0$ for $x\in\{0,\frac12-\frac1n,\frac12+\frac1n,1\}$, $U_n(\frac12)=1$, and interpolate linearly between the points named. Then $U_n\stackrel*\rightharpoonup0$, but $U_n|_{[0,1/2]}\stackrel*\rightharpoonup\chi_{\{1/2\}}$ (where $\stackrel*\rightharpoonup$ is weak* convergence).

If you wish for a positive result, you have to make more stringent assumptions, such as $$ \lim_{\delta\downarrow0}\sup_n \operatorname{TV}(U_n|_{[1/2-\delta,1/2+\delta]})=0. $$ (I think that should be sufficient.)

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Dear Harald, is it true that if we require additionally the continuity of $U$ in point $\frac{1}{2}$ then the statement will be correct? –  Nimza Jun 9 '12 at 21:14
    
@Nimza: No. See my expanded answer. –  Harald Hanche-Olsen Jun 11 '12 at 8:13

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