Suppose $\sum_{n=1}^{\infty} V(f_n - f;a,b) < \infty$, prove that $f_n^{'}$ converges almost everywhere to $f^{'}$

Question

Suppose $f$, $f_n$ $\in$ BV[a,b] and suppose $\sum_{n=1}^{\infty} V(f_n - f;a,b) < \infty$, prove that $f_n^{'}$ converges almost everywhere to $f^{'}$. I could prove that $f_n^{'}$ converges to $f^{'}$ in $L^p$, so that there is a subsequence of {$f_n^{'}$} which converges almost everywhere to $f^{'}$. Could someone help me to prove the stronger argument? Thanks a lot.

Answer 1

Notice (by replacing $f_n - f$ by a new function $g_n$) that it is enough to prove that for a sequence of functions $f_n\in BV[a,b]$ if $$ \sum\limits_{n=1}^\infty V(f_n; a, b) < \infty $$ then $f_n' \to 0$ a.e. To this end recall a well-known fact that for $f \in BV[a,b]$ one has $$\tag{1} \int\limits_a^b |f'(x)|dx \leq V(f; a,b).$$ To see $(1)$ define $V(f; a,x)=: T(x)$ where $a\leq x \leq b$ and observe that at points $x\in [a,b]$ where both $f'(x)$ and $T'(x)$ are defined we have $|f'(x)| \leq T'(x)$. The latter is due to the fact that for any $h>0$ small we have $$ \frac { T(x + h) - T(x) }{h} = \frac 1h V(f; x, x + h) \geq \frac{1}{h}|f(x + h) - f(x)|, $$ and the claim follows by taking $h\to 0$. Since derivatives of $f$ and $T$ exist almost everywhere in $[a,b]$ it follows that $|f'(x)| \leq T'(x)$ a.e. Integrating this over $[a,b]$ we recover $(1)$.

Using $(1)$ we have $$ \sum\limits_{n=1}^\infty \int\limits_{a}^b |f_n'(x)|dx \leq \sum\limits_{n=1}^\infty V(f_n; a,b) <\infty. $$ It follows that $\sum\limits_{n=1}^{\infty} |f_n'(x)| < \infty $ a.e. and hence $f_n'(x) \to 0$ a.e. completing the proof.