An oscillatory L1-bounded sequence with unbounded spatial TV

The intervals are disjoint and have Lebesgue measure $2^{-(k+1)}$ by [F4], so $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_1={\sum }_{k=1}^N{2}^{-(k+1)}<1\mathrm{/}2$. Integration by parts on each interval, justified by [F5] and [F6], shows that $D{u}_N$ is an atomic signed measure: its jump at $1\mathrm{/}2$ has magnitude $1$, its $N-1$ interior jumps at $2^{-2},\dots ,2^{-N}$ have magnitude $2$, and its jump at $2^{-(N+1)}$ has magnitude $1$. These are all jumps in $(0,1)$, so $u_N\in BV((0,1))$ and $TV(u_N;(0,1))=1+2(N-1)+1=2N$.

The sequence converges in $L^1$ to $u=(-1{)}^k$ on each $I_k$ and $u=0$ on $(1\mathrm{/}2,1)$, since $\mathrm{\mid }{u}_N-u\mathrm{\mid }\le 1$ and its support lies in $(0,2^{-(N+1)})$, whose measure tends to zero by [F4]. For each $M$, choose disjoint small intervals around $x_j=2^{-j}$, $1\le j\le M$, avoiding every other dyadic jump, and smooth bumps equal to one at these points, with signs chosen opposite to the jumps of $u$. Their sum is a compactly supported smooth test function ${\varphi }_M$ with $\mathrm{\mid }{\varphi }_M\mathrm{\mid }\le 1$. Its support is bounded away from zero, so only finitely many intervals $I_k$ meet it; integration by parts on those intervals, using [F5] and [F6], gives ${\int }_0^{1}u{\varphi }_M^{\mathrm{\prime }}dx=1+2(M-1)=2M-1.$ Thus the test-field supremum for $u$ is infinite. If $u$ were in $BV((0,1))$, [F2] would make that supremum equal its finite total variation, a contradiction. This proves the refuted assertion fails despite the uniform $L^1$ bound.