If $x_n(t)\to x(t)$ uniformly on $[0,1]$ ($x_n(t), \ x(t)\in \mathbb{R}^n)$, then $\int_0^1 \|\dot{x}_m(t)-\dot{x}(t) \|dt \to 0$.

Question

Is this statement true? If no, could you please give me a counterexample?

If $x_n(t)\to x(t)$ uniformly on $[0,1]$ ($x_n(t), \ x(t)\in \mathbb{R}^n)$, then $\int_0^1 \|\dot{x}_m(t)-\dot{x}(t) \|dt \to 0$.

Thank you so much.

Answer 1

No, it is not true. Consider, for example, the sequence $x_n(t) := \frac{\sin(2\pi nt)}{n}$. Then $(x_n)$ converges uniformly in $\mathbb{R}$ to $x(t) \equiv 0$, but $$ \int_0^1 |\dot{x}_n(t)-\dot{x}(t)|\, dt = 2\pi \int_0^1 |\cos(2\pi nt)|\, dt = \int_0^{2\pi} |\cos y|\, dy $$ does not converge to $0$.