2
$\begingroup$

Consider a continuous function $f: \mathbb R \rightarrow \mathbb R$ and define $f_n: \mathbb R \rightarrow \mathbb R$ by $f_n(x) = n(f(x+1/n)-f(x))$.

I want to show that for $a < b \in \mathbb R$ we have $$\lim_{n \rightarrow \infty} \int^b_a f_n(x) \lambda(dx) = f(b)-f(a),$$ where $\lambda$ denote the Lebesgue measure.

I've tried to use dominated convergence, but it didn't work. Evaluating the integral I get $$\lim_{n \rightarrow \infty} \int^b_a f_n(x) \lambda(dx) = \lim_{n \rightarrow \infty} n\int^b_a f(x+1/n) dx - n\int^b_a f(x) dx,$$ and I can't really continue from here.

$\endgroup$
  • $\begingroup$ $f_n(x) \to f^\prime (x)$ pointwise. $\endgroup$ – Mustafa Said Oct 29 '14 at 8:51
  • $\begingroup$ Suggestion: Consider transforming the first integral on the right hand side by $x\mapsto x-1/n$. $\endgroup$ – Jonas Dahlbæk Oct 29 '14 at 8:51
  • 1
    $\begingroup$ @MustafaSaid $f$ is not assumed differentiable. $\endgroup$ – Jonas Dahlbæk Oct 29 '14 at 8:51
2
$\begingroup$

Note that, for every $n$, $$\int^b_a f_n(x) \lambda(dx)=n\int^{b+1/n}_b f(x) \lambda(dx)-n\int^{a+1/n}_a f(x) \lambda(dx).$$ Can you identify the limit of each term on the RHS? Hint: only the continuity of $f$ is needed.

$\endgroup$
  • $\begingroup$ Have you made the transformation suggested above to get this equality ? And do you use the fact that the Riemann integral is continuous ? On the right side, I get a limit on the form $\infty \cdot (f(b)-f(a))$ ? $\endgroup$ – Shuzheng Oct 29 '14 at 9:16
  • $\begingroup$ Also, how do you get your integral boundaries ? $\endgroup$ – Shuzheng Oct 29 '14 at 9:23
  • $\begingroup$ "Have you made the transformation suggested above to get this equality ?" Which transformation? "And do you use the fact that the Riemann integral is continuous ?" Nope. "On the right side, I get a limit on the form ∞⋅(f(b)−f(a)) ?" This is not the limit. How did you get that? As I said, only the continuity of $f$ (at $a$ and at $b$) is involved. $\endgroup$ – Did Oct 29 '14 at 9:28
  • $\begingroup$ I get that limit, since we I let $n \rightarrow \infty$ on R.H.S, both scalars ($n$) approach $\infty$. $\endgroup$ – Shuzheng Oct 29 '14 at 9:37
  • $\begingroup$ Sure---but $n$ is multiplied by an integral$$\int_{c}^{c+1/n}f(x)\lambda(dx),$$ for $c=b$ or for $c=a$, hence if these integrals converge to $0$ (and they do), the limit $\infty\times0$ is indeterminate. As a first step, can you show that each of these integrals converges to $0$? $\endgroup$ – Did Oct 29 '14 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.