Suppose $f_n \rightarrow f$ pointwise on $[0,1]$ and $\int_0^1{|f_n(x)-g(x)|}^2dx \rightarrow0$ as $n \rightarrow \infty$. Prove that if $g,f$ are continuous on $[0,1]$ then $f=g$.

So I tried to prove this claim by contradiction. Suppose there exists a point $x_0 \in [0,1]$ such that $f(x_0) \neq g(x_0)$. Since $f,g$ are continuous, there is an interval $I_0 \subset [0,1]$ containing $x_0$ such that $\int\limits_{I_0}|f(x)-g(x)|^2dx \neq 0$, which violates the assumption that $\lim_{n \to \infty}\int_0^1{|f_n(x)-g(x)|}^2dx =0 $. I don't know how to write this rigorously. Am I doing it right?


You haven't shown how $\int_{I_0} |f(x) - g(x)|^2\, dx \neq 0$ contradicts $\lim\limits_{n\to \infty} \int_0^1 |f_n(x) - g(x)|^2\, dx = 0$.

Since $\{\lvert f_n - g\rvert^2\}_{n\in \Bbb N}$ is a sequence of non-negative measurable function that converges pointwise to $\lvert f - g\rvert^2$ on $[0,1]$, Fatou's lemma gives

$$\int_0^1 \lvert f(x) - g(x)\rvert^2\, dx \le \varliminf_{n\to \infty} \int_0^1 \lvert f_n(x) - g(x)\rvert^2\, dx = 0.$$

Hence $\int_0^1 \lvert f(x) - g(x)\rvert^2\, dx = 0$, and thus $0 \le \int_{I_0} \lvert f(x) - g(x)\rvert^2\, dx \le \int_0^1 \lvert f(x) - g(x)\rvert^2\, dx = 0$. This implies $\int_{I_0} \lvert f(x) - g(x)\rvert^2\, dx = 0$. There's your contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.