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Is space $\mathcal C[-\pi, \pi]$ with metric $$\rho(x,y) = \max_{t \in [-\pi, \pi]}|x(t)-y(t)| + 2\int_{-\pi}^{\pi}|x(t)-y(t)|dt$$ complete?

My proof (I do not know if it is correct): Consider the Cauchy sequence $x_n$ that it such that $\forall \epsilon>0$ $\exists N_\epsilon$: $\forall n,m \geq N_\epsilon$ $\rho(x_n,x_m) < \epsilon$, i.e $$\max_{t \in [-\pi, \pi]}|x_n(t)-x_m(t)| + 2\int_{-\pi}^{\pi}|x_n(t)-x_m(t)|dt < \epsilon$$

Both terms are positive, consequently $\max_{t \in [-\pi, \pi]}|x_n(t)-x_m(t)| < \epsilon$ $\Rightarrow$ $|x_n(t) - x_m(t)| < \epsilon$ $\forall t \in [-\pi, \pi]$. Set $m > n \geq N_\epsilon$ and let $m \rightarrow \infty$, fixing $t$. Consequently we have $|x_n(t) - x(t)| \leq \epsilon$ $\forall n \geq N_\epsilon$ $\forall t \in [-\pi, \pi]$ what is the uniform convergence of the sequence $x_n(t)$ $\rightarrow x(t)$, and $x(t)$ is continuous. In this way my metric space is complete.

In general, the problem has been reduced to the usual proof of the completeness of the space of continuous functions on an interval. Why has the integral not affected anything? How can you ruin a metric so that the space becomes incomplete?

Thank you in advance!

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You showed $x_n$ is convergent under the sup norm, but have not yet showed $x_n$ is convergent under $\rho$. to show this, just observe $$\rho(x_n,x)=||x_n-x||_{\infty}+2\int_{-\pi}^\pi |x_n(t)-x(t)|dt\\ \le(4\pi+1)||x_n-x||_{\infty}$$ Because $x_n\to x$ in $||\cdot||_\infty$, it turns out that $x_n\to x$ in $\rho$ as well

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