$\phi$ is continuous and odd, $f$ is Lebesgue integrable, we define $T_nf = \int_{R} \phi(nx)f(x)$. Show that: $lim_{n \to \infty} T_nf = 0$

$\phi : \mathbb{R} \to \mathbb{R}$ 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?