Show that the space $$C_p[-\pi,\pi] = \{f \in C[-\pi,\pi] \; | \; f(-\pi)=f(\pi) \}$$ is complete knowing that the space $C[-\pi,\pi]$, i.e. continuous functions on $[-\pi,\pi]$, is complete.
Here is my proof.
$C[-\pi,\pi]$ is complete wrt the supremum norm $$\|f\| = \sup_{t \in [-\pi,\pi]}|f(t)|,$$ and $C_p[-\pi,\pi] \subset C[-\pi,\pi]$. Therefore, taking a Cauchy sequence $(f_n)$ of functions in $C_p[-\pi,\pi]$, it follows that $$\lim_{n\to\infty} f_n = f_0 \in C[-\pi,\pi].$$ Now it's just left to prove that $f_0(-\pi)=f_0(\pi)$. In order to do that, we know that the convergence with respect to the supremum norm implies uniform convergence $f_n \to f_0$, which, in turn, implies pointwise convergence $f_n(x) \to f_0(x), \; \forall x \in [-\pi,\pi]$. Therefore, we can write $$\begin{cases} \lim_{n\to\infty} f_n(-\pi) = f_0(-\pi), \\ \lim_{n\to\infty} f_n(\pi) = f_0(\pi). \end{cases}$$ Substracting $$\lim_{n\to\infty} f_n(-\pi) - \lim_{n\to\infty} f_n(\pi) = f_0(-\pi)-f_0(\pi) \\ \Rightarrow \lim_{n\to\infty}(f_n(-\pi)-f_n(\pi)) = f_0(-\pi)-f_0(\pi).$$ Now we use that $(f_n)$ is in $C_p[-\pi,\pi]$, therefore $f_j(-\pi)=f_j(\pi) \; \forall j \in \mathbb{N}.$ Hence by calling $$(a_n) = (f_n(-\pi)-f_n(\pi)) \; \forall \; n \in \mathbb{N} \quad \text{and} \quad a_0 = f_0(-\pi)-f_0(\pi)\\ \Rightarrow \lim_{n\to\infty} a_n = a_0,$$ but $(a_n)$ is a sequence of zeros and therefore: $$\lim_{n\to\infty} a_n = a_0 = 0$$