# Is it true that $\lim\limits_{r \rightarrow 1}f(r x) = f(x)$ in $L^1$?

Suppose $f \in L^1(\mathbb{R})$ with Lebesgue measure and $r > 0$. Does $f(rx)$ converges to $f$ in $L^1$ as $r \rightarrow 1$ ? Put differently, does $$\| f(rx) - f(x)\|_1 \rightarrow 0$$ as $r \rightarrow 1$.

Using Lebesgue dominated convergence, we can pass the limit under the integral sign but then I'm stuck because it doesn't feel right to expect $$|\lim\limits_{r \rightarrow 1} f(rx) - f(x)\big| = 0$$ whenever $f$ is not continuous at $x$.

• This just says that $f$ is continuous a.e. $\lambda$ – Matematleta May 30 '16 at 23:41
• What is $\lambda$ ? – Joe G. May 30 '16 at 23:48
• I'm not entirely sure of the answer, but sometimes you can pass from a property held by the continuous, compactly supported functions in $L^1$ to a property for all functions using the fact that these functions are dense in $L^1$. Just a thought. – J.G May 30 '16 at 23:51
• $\lambda$ is Lebesgue measure sorry, – Matematleta May 31 '16 at 1:03

Elaborating on my comment, I believe the answer is yes. Recall that the continuous, compactly supported functions are dense in $L^1$: that is, for any $f \in L^1(\mathbb{R})$ and $\epsilon>0$, we can find $g$ continuous with compact support such that $\vert\vert g-f \vert\vert_1<\epsilon$. Suppose that you proved the result for continuous functions with compact support.
With that in mind, fix $\epsilon>0$ and choose $g$ such that this holds for our $f$. Now, we can compute, using the Triangle Inequality: \begin{align} \int \vert f(rx)-f(x) \vert&=\int \vert f(rx)-g(rx)+g(rx)-g(x)+g(x)-f(x)\vert\\ &\leq \int \vert f(rx)-g(rx)\vert +\int \vert g(rx)-g(x)\vert +\int \vert g(x)-f(x)\vert\\ &< \frac{1}{r}\int \vert f(u)-g(u)\vert+\int \vert g(rx)-g(x)\vert+\epsilon\\ &<\frac{(1+r)\epsilon}{r}+\int \vert g(rx)-g(x)\vert \end{align} The first inequality is just the Triangle Inequality, and the second follows from assumption and using the substitution $u=rx$ (I believe this holds for the Lebesgue Integral). The final integral can be made arbitrarily small if you have proved it for continuous functions, so the bound can be made arbitrarily small by choosing $r$ small first so that the last integral is as small as you like, and then letting $\epsilon$ go to 0.
• Slight correction: "proved it for compactly supported continuous functions." Otherwise, this is the proof I would do as well. $+1$! – Cameron Williams May 31 '16 at 0:16