If $L^1$ average of $f$ is smaller than $t^2$ then $f=0$ a.e.

Suppose $$f\in L^1(\mathbb R)$$ and there exits $$\delta>0$$ such that $$\|f(\cdot + t)-f\|_{L^1}\leq |t|^2$$ for all $$|t|\leq \delta$$, where $$f(\cdot + t)$$ is the function $$x\mapsto f(x+t)$$. Show that $$f=0$$ a.e.

It is well known that $$\lim_{t\to 0}\|f(\cdot+t)-f\|_{L^1}=0$$ but this fact can't help here. Intuitively, the condition in the problem would deduce that $$\int |f'|=0$$ if the derivative is good enough so the conclusion must be right and $$|t|^2$$ can be improved to $$|t|^{1+\epsilon}$$. I also tried to use the separability of $$L^1$$, like when we prove the famous fact mentioned above, but it helps nothing. Any hints and thoughts are welcomed.

• A constant function seems to be a counter example. Aug 20 '19 at 8:24
• @quarague $f$ needs to be in $L^1$
– user678090
Aug 20 '19 at 8:25
• It is for all $|t|\leq \delta$ and all $x$ ? Aug 20 '19 at 8:41
• @ujsgeyrr1f0d0d0r0h1h0j0j_juj the norm is an integration over $x$ Aug 20 '19 at 8:42
• @P.Quinton What an horrible notation then. Aug 20 '19 at 8:44

Consider the Fourier transformation, $$\hat f(\xi)=\int_\mathbb Rf(x)e^{-2\pi ix\xi}\,dx$$ for $$\xi\in\mathbb R$$. Denote $$f_t(x)=f(x+t)$$, then $$\hat f_t(\xi)=e^{2\pi it\xi}\hat f(\xi)$$.

By hypothesis, $$\|\hat f(\cdot)(e^{2\pi it\cdot}-1)\|_{L^\infty}=\|\hat f_t-\hat f\|_{L^\infty}\leq \|f_t-f\|_{L^1}\leq |t|^2$$. So for a.e. $$\xi\in\mathbb R-\{0\}$$, $$|\hat f(\xi)||e^{2\pi it\xi}-1|\leq |t|^2,\ \ \text{for all } 0<|t|\leq\delta.$$ This implies $$|\hat f(\xi)|\frac{|e^{2\pi it\xi}-1|}{|t|}\leq |t|,\ \ \text{for all } 0<|t|\leq\delta.$$ Letting $$t\to 0$$ gives that $$|\xi\hat f(\xi)|=0$$ for a.e. $$\xi\in\mathbb R$$, so $$\hat f\equiv0$$ since $$\hat f$$ is continuous, which completes the proof.

• Perfect answer. :) Aug 20 '19 at 9:03
• Never thought about Fourier transformations, impressive!
– user678090
Aug 20 '19 at 9:31

$$|\int_a^{b} f(x)dx-\int_{a+t}^{b+t} f(x)dx| =|\int_a^{a+t} f(x)dx-\int_b^{b+t} f(x)dx|\leq t^{2}$$ for $$t>0$$ sufficiently small. [ Because $$\int_a^{b} f(x)dx-\int_{a+t}^{b+t} f(x)dx=\int_a^{b} f(x)dx-\int_a^{b} f(x+t)dx$$]. Divide by $$t$$ and let $$t \to 0$$. Using Lebesgue's Theorem we get $$f(a)-f(b)=0$$ whenever $$a$$ and $$b$$ are Lebesgue points of $$f$$. Can you now show that $$f=0$$ a.e.?

• Lebesgue points! I can't believe I forgot them. Totally understand your thoughts, thank you sir.
– user678090
Aug 20 '19 at 9:28
• Welcome. The result is trivial for smooth $f$ so it is very natural to prove this using Lebesgue points. Aug 20 '19 at 9:38

This is just a special case of the fact that if $$\alpha>1$$ then $$Lip_\alpha$$ contains only constants:

Say $$f_t(x)=f(x+t)$$, and define $$F:\Bbb R\to L^1$$ by $$F(t)=f_t$$.

Then $$F(t)-F(0)=\sum_{j=1}^n(F(jt/n)-F((j-1)t/n)),$$so if $$n$$ is large enough that $$t/n<\delta$$ we have $$||F(t)-F(0)||_1\le\sum_{j=1}^n||F(jt/n)-F((j-1)t/n)||_1\le n(t/n)^2=t^2/n.$$ So $$||F(t)-F(0)||_1=0$$; hence $$f$$ is constant, and so $$f\in L^1$$ implies $$f=0$$.

Applying the triangle inequality we have for, each $$n\in \mathbb{N}$$: $$\lVert f(x+n\lceil{\frac{1}{t}}\rceil t)-f(x)\rVert_1\leq n \lceil{\frac{1}{t}}\rceil t^2\leq 2nt$$ for $$t>0$$ sufficiently small. In particular, if $$f\neq 0$$ a.e. we can, for sufficiently large $$n$$, set $$t=\frac{||f||_1}{2n}$$ to conclude that $$\lVert f(x+n\lceil{\frac{1}{t}}\rceil t)-f(x)\rVert_1\leq \lVert f\rVert_1$$ for sufficiently large $$n$$ (1). Using the MCT we can find $$m>0$$ such that $$\int_{[-m,m]}|f| d\mu>(7/8)\cdot\lVert f\rVert_1$$. If $$m$$ was sufficiently large we then have $$\int|f(x+2m\lceil{\frac{1}{t}}\rceil t)-f(x)| d\mu(x)\geq \int_{[m,3m+1]}|f(x+2m\lceil{\frac{1}{t}}\rceil t)| d\mu(x)+\int_{[-m,m]}|f(x)| d\mu(x)-\int_{[m,3m+1]^C}|f(x+2m\lceil{\frac{1}{t}}\rceil t)| d\mu(x)-\int_{[-m,m]^C}|f(x)| d\mu(x)>(12/8)\lVert f\rVert_1$$.

• Thanks for the answer. But I can't follow you in all steps. Could you please explain how to deduce the first inequality? It seems to me that the RHS of the first inequality should be $(n\lceil{\frac1t}\rceil t)^2$. But I think it may be wrong because in my problem there is a $\delta$ and if $n$ is large $n\lceil{\frac1t}\rceil t\approx n>\delta$.