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!
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.