# Let $f_n , f \in C({\Bbb R}/{\Bbb Z}, \Bbb C)$, show that if $f_n \to f$ uniformly, then $f_n \to f$ in the $L^2$ metric.

Let $$f_n , f \in C({\Bbb R}/{\Bbb Z}, \Bbb C)$$, show that if $$f_n \to f$$ uniformly, then $$f_n \to f$$ in the $$L^2$$ metric. where $$C({\Bbb R}/{\Bbb Z},\Bbb C)$$ is the space of complex-valued continuous Z-periodic functions

My work so far: $$f_n \to f$$ uniformly implies that $$\forall \epsilon >0$$, $$\exists N > 0$$ such that $$f_n(x) - f(x) < \epsilon \ \forall n>N, x\in \Bbb C$$

In order to show that $$f_n \to f$$ in $$L^2$$, I need to show $$\forall \epsilon > 0, \exists N>0$$ such that $$d_{L^2}(f_n,f)= (\int_{[0,1]} |f_n(x)-f(x)|dx)^{\frac{1}{2}} <\epsilon$$

I am a bit stucked. I feel there should be something to do with the fact that these two functions are periodic but I am not sure.

• Your second paragraph should say "$|f_n(x) - f(x)| < \epsilon$ for all $n>N, x\in \Bbb R/\Bbb Z$". In that situation, how large can $\int_{\Bbb R/\Bbb Z} |f_n(x)-f(x)| \,dx$ be? Mar 16, 2019 at 6:46

The crucial thing is that $$\Bbb R/\Bbb Z$$ has finite measure. Basically we are integrating over the interval $$[0,1]$$ which has measure $$1$$.
If $$|f_n(x)-f(x)|\le\varepsilon$$ for all $$x$$, then $$d_n(f_n,f)^2= \int_0^1|f_n(x)-f(x)|^2\,dx\le\int_0^1\varepsilon^2\,dx=\varepsilon^2\,dx$$ which means that $$d_2(f_n,f)\le\varepsilon$$, etc.