TITLE: How to prove that this sequence of functions is bounded?
QUESTION [5 upvotes]: Let $(x_n)_n$ be a sequence of Lipschitzian functions (with same Lipschitz constant) over $(0, T)$ for which the inequality
$$ \left|\int_0^T x_n(t)^2 dt\right|\le \epsilon_n \|x_n\|_{\infty} +\int_0^T |x_n(t)| dt, dt $$
with $\varepsilon_n\to 0$.
It is possible to prove formally from the above inequality that $(x_n)_n$ is a bounded sequence? If it is true, how to do that?
About me it should be true since the LHS term has a quadratic growth and the RHS term has a linear growth, but it is just a naive idea.
Could someone please help with that?
Thank you in advance!

REPLY [2 votes]: I think your idea about quadratic growth is right if used in conjunction with the assumption that that the same Lipschitz contstant can be used for all $x_n$. Given that $x_n(t)$ all have the same Lipschitz constant $K$, we know that
$$  | x_n(t) - x_n(s)| < K|t-s| $$
for $s,t \in [0;T]$.  Since Lipschitz funtions are clearly continous we see that if $x_n(t_{n,\max})=|x_{\infty}|$ then for any $s \in [t_{n, \max}-1/K;t_{n,\max}+1/K]$ we have
$$   |x_n(t_{n,\max}) - x_n(s)| < 1$$
i.e.
$$ x_n(s) > ||x_n||_{\infty}-1.$$
But this obviously gives us that
$$   1_{[t_{n,\max}-1/K;t_{n,\max}+1/K]}(||x_n||_{\infty}-1)  < |x_n(t)| < 1_{[0;T]}||x_n||_{\infty},$$
hence we find
$$ 2/K (||x_n||_{\infty}-1)^2 <  \int_0^Tx_n(t)^2dt    $$
and
$$ \epsilon_n||x_n||_{\infty} + \int_0^T||x_n(t)||dt < (\epsilon_n + T)||x_n||_{\infty}$$
But if
$$2/K (||x_n||_{\infty}-1)^2 <  (\epsilon_n + T)||x_n||_{\infty}$$
we easily derive an upper bound for $||x_n||_{\infty}$, hence it cannot be unbounded.