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Suppose $f$ is a square-integrable function. There exists $\alpha>0$ satisfying $$ \|f(x+h)-f(x)\|_{L^2}=O(h^{1+\alpha}), h\rightarrow 0. $$ Prove that $f(x)$ equals a constant almost everywhere.

Let $F(x)=\int_0^x f(t)\mathrm{d}t$, I can prove that $$ F(x+h)-F(x)=o(h^{1/2}), h\rightarrow 0, $$ but it seems have no relation with the problem. I have no idea how to start.

I got the $o(h^{1/2})$ result by using the Holder Inequality directly and the absolute continuity of Lebesgue Integral (maybe no use for the problem). And I'm currently learning the $L^p$ spaces. For the original problem, the domian probably does not matter for the conclution only asks almost everywhere, but I think assuning the domain to be $\mathbb{R}$ is enough.

Any help! Thanks.

user823011
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    You will need to provide me more context. First, I'd like to know how you got $F(x+h)-F(x) = O(h^{\frac 12})$, because I feel I'm doing better. Second : do you have a source for this question? I feel like the domain matters here : it's probably of the form $\mathbb R$ or $[a,\infty)$ for some $a$, so that the translations can make sense for all $h$.Third : If necessary, feel free to add any background , in particular if you are aware of the fundamental theorem of calculus for the Lebesgue measure. Edit your question to reflect this, and I can try the question. – Sarvesh Ravichandran Iyer Dec 11 '21 at 12:23
  • I got the $o(h^{1/2})$ result by using the Holder Inequality directly and the absolute continuity of Lebesgue Integral (maybe no use for the problem). And I'm currently learning the $L^p$ spaces. For the original problem, the domian probably does not matter for the conclution only asks almost everywhere, but I think assuning the domain to be $\mathbb{R}$ is enough. – user823011 Dec 11 '21 at 12:44
  • Can you edit your question with this information? It's my necessity, you may consider it that way. I'm planning to write an answer because I feel like with those conditions, I can get the answer. – Sarvesh Ravichandran Iyer Dec 11 '21 at 12:46
  • Thanks. I will try to write an answer now. – Sarvesh Ravichandran Iyer Dec 11 '21 at 13:02

1 Answers1

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So, let the domain be $\mathbb R$. First, note that the definition of $$\|f(x+h)-f(x)\|_{2} = O(h^{1+\alpha}) \\\iff \exists h',C>0 , h<h' \implies \|f(x+h)-f(x)\|_{2} \leq Ch^{1+\alpha}$$

Now, it's quite clear from here that $\left\|\frac{f(x+h)-f(x)}{h}\right\|_{2} \leq Ch^{\alpha}$ for $h<h'$. Therefore, let $g_h(x) = \frac{f(x+h)-f(x)}{h}$, we know that $\|g_h(x)\|_{L^2} \to 0$ as $h \to 0$, and therefore, we know that $g_h(x) \to 0$ in $L^2(\mathbb R)$.

Now, let $\psi$ be any smooth function with compact support. We can use a change of variables to write :$$ \int_{\mathbb R} \psi(x)g_h(x)dx = \frac 1h\int_{\mathbb R} \psi(x)f(x+h)dx - \frac 1h \int_{\mathbb R} \psi(x)f(x)dx \\ = \frac 1h \int_{\mathbb R} \psi(x-h)f(x)dx - \frac 1h\int_{\mathbb R} \psi(x)f(x)dx \\ =\int_{\mathbb R} \frac{\psi(x-h)-\psi(x)}{h}f(x)dx $$

Now, as $h\to 0$ , the LHS goes to zero as $g_h \to 0$ in $L^2(\mathbb R)$, so a Cauchy-Schwarz will do the trick. Therefore, the RHS goes to zero as well. However, on the RHS, the point is that we can interchange the $h$-limit and the integral over $\mathbb R$.

To prove this, we use the bounded convergence theorem. Let $D = \overline{\{x : \psi(x) \neq 0\}}$ be the support of $\psi$. It's a compact set, and by Holder's inequality$$ \|f\|_{L^1(D)} \leq \|f\|_{L^2(D)}\|1\|_{L^2(D)} < \|f\|_{L^2(\mathbb R)}\mu(D)^{\frac 12} < \infty $$ We also have $$ \frac{\psi(x-h)-\psi(x)}{h} \leq \sup_{x \in D} |\psi'(x)|<\infty $$ therefore, the RHS is uniformly bounded in $x$ and $h$ by $\|f\|_{L^2(\mathbb R)}\mu(D)^{\frac 12}\sup_{x \in D} |\psi'(x)|$, a finite quantity. Thus, the interchange of the $h$-limit and the integral is justified and $$ \lim_{h \to 0} \int_{\mathbb R} \frac{\psi(x-h)-\psi(x)}{h}f(x)dx = \int_{\mathbb R}\lim_{h \to 0} \frac{\psi(x-h)-\psi(x)}{h}f(x)dx \\ = \int_{\mathbb R} \psi'(x)f(x)dx $$

Combining this with the previous observation that $\int_{\mathbb R} \psi(x)g_h(x) \to 0$, it follows that for all $\psi \in L^2(\mathbb R)$ smooth with compact support, we have $\int_{\mathbb R}\psi'(x)f(x)dx = 0$. It follows that $f$ is weakly differentiable with weak derivative zero on $\mathbb R$ : that implies that $f$ is constant a.e, using a mollifiers-based argument as presented here.

Sarvesh Ravichandran Iyer
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  • Note : I couldn't make deductions on the quantity $|F(x+h)-F(x)|$ because of the fact that attempts to do it resulted in the $o(h^{0.5})$ bound as the OP realizes. So that's clearly not the right approach (or both of us are not doing it tightly enough, but that's for another thread). – Sarvesh Ravichandran Iyer Dec 12 '21 at 11:58