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$\phi : \mathbb{R} \to \mathbb{R}$ is a continuous, odd function satisfying the conditions: $$\phi(0) = \phi(1) = 0, \ \ \ \ \ \ \phi(x + 2) = \phi(x) \ \forall x \in \mathbb{R}$$

For a function $f$ that is Lebesgue integrable on $\mathbb{R}$, we define $$T_nf = \int_{\mathbb{R}} \phi(nx)f(x) \ d \lambda_1$$

Show that for every $n \in N$, $T_nf$ is a real number and $\lim_{n \to \infty} T_nf = 0$.

For every $n \in N$, $T_nf$ is a real number:

$\phi$ is continuous on the interval $[0,2]$, which is compact.

Therefore, $\phi$ is a bounded function (bounded by its maximum).

Larger proof:

Since $\phi$ is continuous on a compact set, by the Extreme Value Theorem, it attains its maximum and minimum values on $[0,2]$. Let's denote these values as $M$ and $m$, respectively.

Now, because $\phi$ is odd, we have $\phi(0) = \phi(2) = 0$ Therefore, $0$ must lie between $m$ and $M$. Formally, we have $m \leq 0 \leq M$.

Since $m$ and $M$ are the minimum and maximum values of $\phi$ on $[0,2]$, respectively, we can conclude that $∣\phi(x)∣ \leq \max⁡ \{ ∣m∣,∣M∣ \}∣\phi(x)∣ \leq \max \{ ∣m∣,∣M∣ \}$ for all $x \in [0,2]$.

Since $\phi$ has period $2$, this bound holds for all $x\in\mathbb{R}$

Therefore, $\phi$ is a bounded function.

$\lim_{n \to \infty} T_nf = 0$:

To show that for every $n \in N$, $T_nf$ is a real number, we can utilize the Dominated Convergence Theorem along with the fact that $\phi$ is a bounded function.

Given that $\phi$ is bounded, there exists a constant $M$ such that $∣\phi(x)∣ \leq M$ for all $x \in \mathbb{R}$.

Now, let's consider the sequence of functions $f_n(x)=\phi(nx)f(x)$ for $n \in \mathbb{N}$. Since $\phi(nx)$ is bounded, we have $∣f_n(x)∣ = ∣\phi(nx)f(x)∣ \leq M∣f(x)∣$ for all $x \in \mathbb{R}$.

Since $f$ is Lebesgue integrable, $M∣f∣$ is also integrable. Therefore, $f_n$ is integrable for all $n \in \mathbb{N}$.

Now, let's consider the limit of $f_n$ as $n \to \infty$. Since $x\mapsto\phi(nx)$ oscillates between $−M$ and $M$, and $f$ is bounded and Lebesgue integrable, $\lim_{⁡n \to \infty} \phi(nx)f(x)=0$ almost everywhere by the Dominated Convergence Theorem. This implies that $\lim_{⁡n \to \infty} f_n(x)=0$ almost everywhere as well.

Therefore, by the Dominated Convergence Theorem, we have: $$\lim_{n \to \infty}​T_n​f_n​ = \lim_{n \to \infty}​ \int_{\mathbb{R}}\phi(nx)f(x)d \lambda_1 ​= \int_{\mathbb{R}} ​\lim_{n \to \infty}​\phi(nx)f(x)d \lambda_1​ = \int_{\mathbb{R}} ​0 d \lambda_1​ = 0$$

Is that correct?

stange
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thefool
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    Hi. For the first part about $T_n f$ being real, you should also show that $\int_{\mathbb{R}}|\phi(nx)f(x)|, d\lambda_1$ is finite. For example, $g(x)=1$ is a real-valued measurable function but $\int_{\mathbb{R}}g(x), d\lambda_1=\infty$ is not a real number. Hint: show that $\phi$ is a bounded function and use Lebesgue's dominated convergence theorem. For the second part, its a variation of the Riemann−Lebesgue lemma. The proof in the Wikipedia article (https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma) is perhaps modifiable to deal with $\phi$ in your question. – Kai May 02 '24 at 01:43
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    Hi again. The proof in the Wikipedia article may be not easily modifiable to your case (it works with an additional assumption that $\phi(x+1)=-\phi(x)$). There are plenty of other proofs of the R−L lemma and some of them applies to your $\phi$, so please have a look at books around you! – Kai May 02 '24 at 02:07
  • @Kai thank you for help! I edited my solution. Is that correct? – thefool May 02 '24 at 14:04
  • Great! I see lots of progresses, especially, the proof of the finiteness of $T_n f$ is well done. (A tiny remark: $\phi$ is continuous on $\mathbb{R}$ by assumption so why divide it into $[0,1]$ and $[1,2]$? Also, I don't see why we need to argue $m\leq 0\leq M$. Simply take the maximum of the continuous function $|\phi|$ to bound $|\phi|$.) However, the proof of $\lim_{n\to \infty}T_n f=0$ is not correct unfortunately. For example, is your argument that $\lim_{n\to \infty}f_n(x)=0$ correct for $\phi(x)=\sin \pi x$ and $f(x)=1; (|x|\leq 1);, f(x)=0; (|x|>1)$? – Kai May 03 '24 at 05:57
  • I edited the first part of my solution - as you said, it can be argued simpler. Were it comes to the second one, I see your point. I tried to argue that each product of $/phi(nx)$ and $f(x)$ can be bounded and as $n$ grows, effectively go to $0$. I thought that somehow I will be able to use Dominated Convergence Theorem. Now, I don't see how to solve that. Could you help me? @Kai – thefool May 05 '24 at 23:40

1 Answers1

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This can be solved by direct application of Fejer's lemma:

Theorem (Féjer): Suppose $\phi$ is a bounded measurable $T$ periodic function ($T>0$). For any $f\in\mathcal{L}_1(\mathbb{R})$ and numeric sequence $\alpha_n\in\mathbb{R}$, $$ \lim_n\int f(x)\phi(nx+\alpha_n)\,dx=\Big(\frac{1}{T}\int^T_0\phi\Big)\int f \tag{1}\label{one} $$

In the OP it is assumed that $\phi$ mean that $\phi$ is $2$-periodic and that $\phi$ is odd. These assumptions imply that $$\int_{[0,2]}\phi(t)\,dt=\int_{[-1,1]}\phi(t)\,dt=0$$

Thus, for any $f\in L_1(\mathbb{R},\lambda_1)$, $$\lim_{n\rightarrow\infty}\int_{\mathbb{R}}\phi(nt)\,f(t)\,dt=\Big(\frac12\int_{[0,2]}\phi(t)\,dt\Big)\Big(\int_\mathbb{R} f(t)\,dt\Big)=0$$

The boundary conditions are not crucial here.

Mittens
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