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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.

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  • $\begingroup$ 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? $\endgroup$ – Greg Martin Mar 16 at 6:46
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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.

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