Let \(f:\mathbb R\to\mathbb R\) be continuous and suppose \[ \int_{-\infty}^{\infty}|f(x)|\,dx<\infty. \] Show that there is a sequence \(x_n\to\infty\) such that \[ x_nf(x_n)\to0 \qquad\text{and}\qquad x_nf(-x_n)\to0. \]
Solution
For \(x\geq0\), define \[ h(x)=\abs{f(x)}+\abs{f(-x)}. \]
1. The function \(h\) is integrable on \([0,\infty)\):\[ \int_0^\infty h(x)\,dx<\infty. \]
One has \[ \begin{aligned} \int_0^\infty h(x)\,dx &= \int_0^\infty\abs{f(x)}\,dx + \int_0^\infty\abs{f(-x)}\,dx\\ &= \int_0^\infty\abs{f(x)}\,dx + \int_{-\infty}^0\abs{f(u)}\,du\\ &= \int_{-\infty}^{\infty}\abs{f(u)}\,du, \end{aligned} \] where the second integral uses the substitution \(u=-x\). The final quantity is finite by hypothesis.
2. For every integer \(n\geq1\), there exists \(x_n\geq n\) such that\[ x_nh(x_n)<\frac1n. \]
1. Therefore such an \(x_n\) exists.
3. The sequence from step 2 satisfies\[ x_n\longrightarrow\infty. \]
By construction, \[ x_n\geq n. \] Since \(n\to\infty\), the same is true of \(x_n\).
4. One has\[ x_n\abs{f(x_n)} \longrightarrow0. \]
By the definition of \(h\), \[ \abs{f(x_n)} \leq h(x_n). \] Therefore step 2 gives \[ 0 \leq x_n\abs{f(x_n)} \leq x_nh(x_n) < \frac1n. \] The squeeze theorem gives the limit.
5. One also has\[ x_n\abs{f(-x_n)} \longrightarrow0. \]
Again, \[ \abs{f(-x_n)} \leq h(x_n). \] Thus \[ 0 \leq x_n\abs{f(-x_n)} \leq x_nh(x_n) < \frac1n, \] and the squeeze theorem applies.
6. Consequently,\[ \boxed{ x_nf(x_n)\to0 \qquad\text{and}\qquad x_nf(-x_n)\to0. } \]
Steps <1>4–<1>5 show convergence of the absolute values of the two displayed expressions to zero, which implies convergence of the expressions themselves to zero.
7. Q.E.D.
Steps 3 and 6 give all required properties of the sequence.