Show that a set of linear functionals on $L^{\infty}$ is not weakly-* sequentially compact I have a homework question which is really giving me some difficulty.
The question:
Let 0<$\epsilon$<1, and define $T_\epsilon \in L^{\infty}((0,1))^*$ by
$$ T\epsilon f = \frac{1}{\epsilon} \int_0^\epsilon f(x)dx$$
Prove that $\{T_\epsilon : 0<\epsilon<1\}$ is bounded in $L^{\infty}((0,1))^*$ but that it is not weakly-* sequentially compact.
Hint Update (I had the wrong f before): 
Consider the function $f(x) = \sum_{n=1}^{\infty} (-1)^n\chi_{[\epsilon_{n+1},\epsilon_n)}(x)$.
My work:
I have shown that the set is bounded. For weak-* sequential compactness, by the hint, I have considered an arbitrary decreasing sequence $(\epsilon_n)_{n\in \mathbb{N}}$ values. To show that the set is not sequentially compact, I note that any subsequence of $(\epsilon_{n_k})_{k\in \mathbb{N}}$ is also a decreasing sequence of epsilon values, so it suffices to show that the sequence $(T_{\epsilon_n})_{n\in \mathbb{N}}$ does not converge in the weak-* topology.
It suffices to show that $(T_{\epsilon_n}(f))_{n\in \mathbb{N}}$ is not Cauchy in $\mathbb{R}$, where $f$ is given as in the hint. So it is enough to show that $|T_{\epsilon_n}(f) - T_{\epsilon_{n+1}}(f)|\ge c$ for some constant c>0 which is independent of n. 
However, going through the computation, I have not been able to find a bound for this given an arbitrary decreasing sequence of $\epsilon_n$ values. I can show I don't get weak-* convergence for some specific sequences of $\epsilon_n$ values, but then I have to show that I don't get weak-* convergence for ANY subsequence, which again has proven to be difficult.
 A: Consider $\epsilon_n= 4^{-n}$ and suppose $T_{\epsilon_n}$ has a weak* convergent sub-sequence $T_{\epsilon_{n_k}}$. As you can tell things are getting pretty illegible. Lets call the sub-sequence of $\epsilon$s $a(n)$ (now indexed by $n$), relable the $T_{a(n)}$ to $T_n$ and denote the characteristic function of $(a(n+1),a(n)]$ by $I(n)$.
Now the important part of choosing $\epsilon_n=4^{-n}$ is that we get $a({n+1}) ≤ \frac{a(n)}4$. Letting $f= \sum_{k=1}^\infty (-1)^k I(k)$ we have:
\begin{align}
T_n f &= \frac1{a(n)}\sum_{k=n}^\infty (-1)^k\, (a(k)-a(k+1))\\
&= (-1)^n\frac{a(n)-a(n+1)}{a(n)}+\frac1{a(n)}\sum_{k={n+1}}^\infty (-1)^k (a(k)-a(k+1))
\end{align}
Now the contribution of the second summand is bounded by $\frac1{a(n)}\sum_{k={1}}^\infty 4^{-k} a(n+1)≤\frac14$ (actually bounded by $\frac1{16}$). The first summand on the other hand will have magnitude $≥\frac12$ (actually $≥\frac34$). Hence $T_nf$ will have absolute value $≥\frac14$ with sign $(-1)^n$. As such $|T_n(f)-T_{n+1}(f)|≥\frac12$ for all $n$. Hence $T_n$ does not converge pointwise.
