This question is related to another question I asked earlier.
For reference, this is the relevant part of that question:
Let the sequence of continuous functions $\mathbf{\{x_{n}(t) \}_{n=1}^{\infty}}$, $\mathbf{0 \leq t < \infty}$ be uniformly bounded on $\mathbf{t \in [0, \infty)}$ and on each bounded interval $\mathbf{[0,A]}$ also satisfy the Lipschitz condition: $\mathbf{\forall t_{1}, t_{2} \in [0,A]}$, $\mathbf{|x_{n}(t_{1})-x_{n}(t_{2})|\leq L_{A} |t_{1}-t_{2}|}$, where the Lipschitz constant $\mathbf{L_{A}}$ is independent of the index $n \geq 1$ but depends on $A$.
In that other problem, I showed that the sequence $x_{n}$ was pre-compact (meaning its closure is compact) in the space of bounded continuous functions $C_{B}[0, \infty)$ with the supremum norm, and thus has a limit point $x \in C_{B}[0, \infty)$.
Now, in addition to those conditions, suppose that, $\exists \alpha > 0$ such that $\forall n$, $\mathbf{\int_{0}^{\infty}t^{\alpha}|x_{n}(t)|dt \leq \text{constant}}$.
I need to prove that $\mathbf{x_{n}\to x}$ in $\mathbf{L^{1}(0,\infty)}$ as $\mathbf{n\to \infty}$.
I suppose that I should consider the norm of the difference of $x_{n}-x$ in $L^{1}(0, \infty)$ of the integral from $0$ to $\infty$, and split the interval into two parts. How to do that, however, is part of my problem: I was given the hint to choose my interval in terms of small $\epsilon$, but I'm not entirely sure how to do that.
I really have no idea how to put all of this together. I've been playing around with this for days and am still no closer to coming up with a proof.
Please help.
,
To assert that the sequence $(x_n)$ converges in $L^1(0,\infty)$, we should require that the sequence is pointwise convergent (and by Ascoli, it is then locally uniformly convergent), otherwise sequences like $x_n(t) = (-1)^n\cdot f(t)$ for a nonzero $f$ give counterexamples. So either we strengthen the hypothesis and require that the full sequence is pointwise/locally uniformly convergent, or we weaken the conclusion and only state that a subsequence of $(x_n)$ converges in $L^1(0,\infty)$. Both ways are equivalent, the only difference is whether we extract a convergent subsequence before or after we state the proposition. Let's extract the subsequence before, and hence assume that $x_n \to x$ locally uniformly on $[0,\infty)$.
From the premise that there is an $\alpha > 0$ and a $C \in (0,\infty)$ such that
$$\int_0^{\infty} t^{\alpha}\lvert x_n(t)\rvert\,dt \leqslant C\tag{1}$$
for all $n$, the uniform boundedness of the sequence $(x_n)$ and the pointwise convergence, we obtain (via the dominated convergence theorem for example) that also
$$\int_0^{\infty} t^{\alpha} \lvert x(t)\rvert\,dt \leqslant C,$$
with the same $\alpha$ and $C$.
The function $t \mapsto t^{\alpha}$ is monotonically increasing, hence, for every $n$ and $A\in (0,\infty)$ we have
$$C \geqslant \int_A^{\infty} t^{\alpha}\lvert x_n(t)\rvert\,dt \geqslant A^{\alpha} \int_A^{\infty} \lvert x_n(t)\rvert\,dt,\tag{2}$$
and the same inequality holds for the limit function $x$. Thus
\begin{align} \lVert x_n - x\rVert_{L^1(0,\infty)} &= \int_0^{\infty} \lvert x_n(t) - x(t)\rvert\,dt \\ &= \int_0^A \lvert x_n(t) - x(t)\rvert\,dt + \int_A^{\infty} \lvert x_n(t) - x(t)\rvert\,dt \\ &\leqslant \int_0^A \lvert x_n(t) - x(t)\rvert\,dt + \int_A^{\infty} \lvert x_n(t)\rvert\,dt + \int_A^{\infty} \lvert x(t)\rvert\,dt\\ &\leqslant \int_0^A \lvert x_n(t) - x(t)\rvert\,dt + 2\cdot C \cdot A^{-\alpha}. \tag{by (2)} \end{align}
Now, given $\varepsilon > 0$, choose $A \in (0,\infty)$ so that $2\cdot C\cdot A^{-\alpha} < \varepsilon/2$, and then use the locally uniform convergence of $(x_n)$ to $x$ to find an $N \in \mathbb{N}$ such that
$$\sup \{ \lvert x_n(t) - x(t)\rvert : t \in [0,A]\} \leqslant \frac{\varepsilon}{2A}$$
for all $n \geqslant N$. Then $\lVert x_n - x\rVert_{L^1(0,\infty)} < \varepsilon$ for all $n \geqslant N$. Since $\varepsilon > 0$ was arbitrary, we have shown that $x_n \to x$ in $L^1(0,\infty)$.