Uniform convergence implies Convergence in the mean

real-analysis

If $f_n$ is a sequence of Riemann integrable functions on $[a,b]$ and f is Riemann integrable on $[a,b]$ we say that "$f_n$ converges in the mean to f" if $\lim_{n\to\infty} \int_{a}^{b} (f_n – f)^2 = 0$.

Show that if $f_n$ is uniformly convergent to f on $[a,b]$ then $f_n$ converges in the mean to f.

I know the definitions of uniformly convergent, namely that there exists N such that if $n \geq N$ then $|f_n(x) – f(x)| < \epsilon$. But where can I go from here?

Best Answer

Assume $(f_n)$ converge uniformly to $ f $ at $ [a b]$.

Let $\epsilon>0$ given.

$$\exists N\ge 0 : \forall n\ge N \; \forall x\in[a,b] \;$$ $$ |f_n(x)-f(x)|<\sqrt{\frac{\epsilon}{b-a}}$$

$$\implies$$ $$\exists N\ge 0 : \forall n\ge N \; \forall x\in[a,b] \;$$ $$ (f_n(x)-f(x))^2<\frac{\epsilon}{b-a}$$ $$\implies$$

$$\exists N\ge 0 \; \;:\; \forall n\ge N \; \int_a^b(f_n(x)-f(x))^2dx<\epsilon$$

$$\implies$$

$$\lim_{n\to+\infty}\int_a^b(f_n(x)-f(x))^2dx=0$$