$C[0, 1]$ denotes the set of all real-valued continuous functions on $[0, 1]$. Let $\{f_n\}$ be a sequence
of functions in $C[0, 1]$ such that $\lim_{\rightarrow \infty}f_n(x) = f(x)$ for each $x\in [0, 1]$. Then
which of the following statement is True ?
I thinks option c will be true by uniform convergence theorem
im not sure about other option
Any/hints solution will be appreciated
pliz help me
thanks u
Best Answer
a and b are clearly false since $f$ is not necessarily continuous. c is however true. We have $$\int_0^{1-1/n}f_n(x)\,dx=\int_0^{1}f_n(x)\,dx+\int_{1-1/n}^1 f_n(x)\,dx.$$ Since $[0,1]$ is compact and the functions $f_n$ are continuous, they are therefore bounded on the interval and since they converge uniformly, it follows that the sequence of functions is uniformly bounded, i.e. that there exists some number $M$ such that $|f_n(x)| \leq M$ for all $n$ and $x$. Thus $$\Bigg|\int_{1-1/n}^1 f_n(x)\,dx\Bigg|\leq\int_{1-1/n}^1 M\,dx=\frac{M}{n}\to 0\text{ as } n\to \infty.$$ Furthermore, since $\lim_{n\to\infty}\int_0^1 f_n(x)\,dx=\int_0^1 f(x)\,dx$, the validity of c follows by the above equality. This also implies that d is false.