Let f : R + R be continuous, with 00 s_, If(x)l dx < o. Show that there is a sequence (x,,) such that x, -+ 00, x, f (x,) 4 0, and x, f(-xc,) 4 0 as n -+ o.

Let f : R + R be continuous, with
00 s_, If(x)l dx < o.
Show that there is a sequence (x,,) such that x, -+ 00, x, f (x,) 4 0, and
x, f(-xc,) 4 0 as n -+ o.

To show that there is a sequence (xn)(x_n)(xn​) such that xn→∞x_n \to \inftyxn​→∞, xnf(xn)→0x_n f(x_n) \to 0xn​f(xn​)→0, and xnf(−xn)→0x_n f(-x_n) \to 0xn​f(−xn​)→0 as $n\to \infty$, we will leverage the fact that the integral of $|f(x)|$ over the entire real line is finite. Here’s a step-by-step proof:

Proof:

1. Given Conditions:
   • $f:\mathbb{R}\to \mathbb{R}$ is continuous.
   • ∫−∞∞∣f(x)∣ dx<∞\int_{-\infty}^{\infty} |f(x)| \, dx < \infty∫−∞∞​∣f(x)∣dx<∞.
2. Implications of the Integral Condition:
   • Since ∫−∞∞∣f(x)∣ dx<∞\int_{-\infty}^{\infty} |f(x)| \, dx < \infty∫−∞∞​∣f(x)∣dx<∞, the integrand $|f(x)|$ must approach zero as $x\to \pm \infty$. This follows from the fact that if $|f(x)|$ did not approach zero, the integral would diverge.
   • Formally, for any $ϵ>0$, there exists $M>0$ such that ∫∣x∣>M∣f(x)∣ dx<ϵ\int_{|x| > M} |f(x)| \, dx < \epsilon∫∣x∣>M​∣f(x)∣dx<ϵ.
3. Constructing the Sequence xnx_nxn​:
   • We need to find a sequence xn→∞x_n \to \inftyxn​→∞ such that both xnf(xn)→0x_n f(x_n) \to 0xn​f(xn​)→0 and xnf(−xn)→0x_n f(-x_n) \to 0xn​f(−xn​)→0.
4. Using the Integral Condition to Define xnx_nxn​:
   • Consider the function $g(x)=xf(x)$. We need to show that $g(x)\to 0$ as $x\to \pm \infty$.
   • Assume for contradiction that $xf(x)$ does not approach zero as $x\to \infty$. Then there exists $ϵ>0$ and a subsequence (xnk)(x_{n_k})(xnk​​) such that xnk∣f(xnk)∣≥ϵx_{n_k} |f(x_{n_k})| \geq \epsilonxnk​​∣f(xnk​​)∣≥ϵ.
5. Estimating the Integral on the Contradicting Subsequence:
   • Let xnkx_{n_k}xnk​​ be a subsequence such that xnk→∞x_{n_k} \to \inftyxnk​​→∞ and xnk∣f(xnk)∣≥ϵx_{n_k} |f(x_{n_k})| \geq \epsilonxnk​​∣f(xnk​​)∣≥ϵ.
   • Consider the intervals Ik=[xnk−δ,xnk+δ]I_k = [x_{n_k} – \delta, x_{n_k} + \delta]Ik​=[xnk​​−δ,xnk​​+δ] for some small $\delta >0$. The length of each interval is $2\delta$.
   • Since $|f(x)|$ is continuous, there exists $\delta >0$ such that ∣f(x)∣≥ϵ2xnk|f(x)| \geq \frac{\epsilon}{2x_{n_k}}∣f(x)∣≥2xnk​​ϵ​ for x∈Ikx \in I_kx∈Ik​.
6. Calculating the Contribution to the Integral:
   • The contribution of $|f(x)|$ over the interval IkI_kIk​ is at least ϵ2xnk⋅2δ=ϵδxnk\frac{\epsilon}{2x_{n_k}} \cdot 2\delta = \frac{\epsilon \delta}{x_{n_k}}2xnk​​ϵ​⋅2δ=xnk​​ϵδ​.
   • Summing over the subsequence IkI_kIk​, we have: ∑k∫Ik∣f(x)∣ dx≥∑kϵδxnk.\sum_k \int_{I_k} |f(x)| \, dx \geq \sum_k \frac{\epsilon \delta}{x_{n_k}}.k∑​∫Ik​​∣f(x)∣dx≥k∑​xnk​​ϵδ​.
   • Since xnk→∞x_{n_k} \to \inftyxnk​​→∞, the terms 1xnk\frac{1}{x_{n_k}}xnk​​1​ become arbitrarily small, making the sum finite. However, this contradicts the assumption that the integral of $|f(x)|$ is finite unless xnkf(xnk)→0x_{n_k} f(x_{n_k}) \to 0xnk​​f(xnk​​)→0.
7. Conclusion:
   • Therefore, $xf(x)\to 0$ as $x\to \infty$.
   • By a symmetric argument, $-xf(-x)\to 0$ as $x\to -\infty$.
   • Thus, we can construct the sequence xnx_nxn​ such that xn→∞x_n \to \inftyxn​→∞, xnf(xn)→0x_n f(x_n) \to 0xn​f(xn​)→0, and xnf(−xn)→0x_n f(-x_n) \to 0xn​f(−xn​)→0 as $n\to \infty$.