1

I'm trying to solve some real analysis question, and I have no clue how to solve this one.

$$ \text{Let: }f \in L^1 (\mathbb{R}), f_n(x):=f(x-\frac{1}{n})$$ $$ \text{Prove that } f_n \text{ converges to } f \text{ in } L^1(\mathbb{R}) \text{ when } n\rightarrow\infty $$

Do you have any ideas? Thanks!

Jose Avilez
  • 12,710
Itai Cohen
  • 31
  • 5
    Interesting question, but please show some effort. – Hyperbolic PDE friend Jan 31 '22 at 15:39
  • Do you know that Lebesgue measure is invariant under translations? That means that $\varphi \in \mathscr{L}^1$ if and only if $\varphi(t - \cdot) \in \mathscr{L}^1$ for all $t \in \mathbf{R}$ (furthermore, the two functions integrate to the same number). – William M. Jan 31 '22 at 15:40
  • You mean that $$\mathbb{R} + \frac{1}{n} = \mathbb{R}?$$ – Itai Cohen Jan 31 '22 at 15:41
  • Is the fact that $$n\rightarrow\infty$$ redundant? @WilliamM. – Itai Cohen Jan 31 '22 at 15:42
  • No, I meant that $\int f(x) dx = \int f(x - \frac{1}{n}) dx.$ But on a second thought, that does not seem to give you convergence in $\mathscr{L}^1.$ You may need to use that you can approximate $f$ by some nice function (e.g. continuous that vanish at infinity) and then for the nice function you can use that the sequence of translations converge uniformly. – William M. Jan 31 '22 at 15:51
  • @WilliamM. What do you mean by a nice function? Does approximating by supremum of simple function works here? Does it help to show that any simple function that approximates f has the same integral of a simple function that approximate $f_n$? – Itai Cohen Jan 31 '22 at 15:54
  • If you can approximate $f$ in $\mathscr{L}^1$ with a sequence $\varphi_k$ in $\mathscr{C}0$ (continuous that vanish at infinity) then $\varphi_k(t - \cdot) \to f(t - \cdot)$ in $\mathscr{L}^1,$ so you'd write $$f(x) - f\left(x - \frac{1}{n}\right) = f(x) - \varphi_k(x) + \varphi_k(x) - \varphi_k\left(x - \frac{1}{n}\right) + \varphi_k\left(x - \frac{1}{n}\right) - f\left(x - \frac{1}{n}\right).$$ The first and last pairs of functions here converge to zero in $\mathscr{L}^1$ and the second pair _will converge to zero uniformly because functions un $\mathscr{C}0$ are _uniformly continuous. – William M. Jan 31 '22 at 16:07
  • @WilliamM. Thanks! Why can I approximate $f$ with a sequence of continuous function? $f$ isn't said to be continuous. Why is it needed for them to vanish at infinity? – Itai Cohen Jan 31 '22 at 16:13
  • I'm gonna leave it here, you need to also think a bit. Notice the emphasis I placed, try to follow the argument given "if this nice approximation exists, you are done" (why?, what is the crux of the proof?, etc.) – William M. Jan 31 '22 at 16:16
  • (By the way, the set $\mathscr{K}$ of continuous with compact support is a subset of $\mathscr{C}0,$ you may want to _actually approximate $f$ with a sequence in $\mathscr{K}$ rather than in $\mathscr{C}_0$. This does not matter in your exercise but $\mathscr{C}_0$ is the closure of $\mathscr{K}$ in the space $\mathscr{B}$ of bounded function with the sup-norm.) – William M. Jan 31 '22 at 16:19
  • @WilliamM. It looks like I missed a whole section in my book! So, any $L^1$ function can be approximated by a continuous function with a compact support, which makes it uniformly continuous.
    Now it all makes sense, since those functions are continuous, when $n\rightarrow\infty$ I can use uniformly continuity, and the equation above works. Thanks!     – Itai Cohen Jan 31 '22 at 16:32
  • Yes, and it is called "uniform continuity" by the way. – William M. Jan 31 '22 at 16:34
  • @WilliamM. Yeah, it looked weird to me and I changed it a moment later (not a native English speaker unfortunately) By missing this section of the book, no wonder I couldn't think of a solution for this question. Thanks again – Itai Cohen Jan 31 '22 at 16:36

0 Answers0