Here's the complete question:
Define a squence of functions $\{f_{n}(x)\}$ on $[0,1]$ as:
$$f_{n}(x)=\left\{ \begin{array}[lll] 11 & \mbox{if} & x=0\\ 1 & \mbox{if}& x\in (\frac{2k}{2^{n}},\frac{2k+1}{2^{n}}],k=0,1,...,2^{n-1}-1\\
-1 & \mbox{if}& x\in (\frac{2k+1}{2^{n}},\frac{2k+2}{2^{n}}],k=0,1,...,2^{n-1}-1 \end{array}\right.$$
Prove or disprove that if g is continuous, then $\lim_{n\to \infty}\int_{0}^{1}f_{n}(x)g(x)dx=0$.
I have no idea since $f_{n}(x)$ are even not pointwise converge to some function. I tried to let $g$ be the continuous function approximation of $f_{n}$ so that $f_{n}(x)g(x)\sim f_{n}(x)^{2}=1$, but it's worng since $g$ is fixed.
Could someone give some advise? Thanks!
You can write $$\int_0^1 f_n(x)g(x) dx = \sum_{k=0}^{2^n-1}(-1)^k\int_{k2^{-n}}^{(k+1)2^{-n}}g(x)dx$$
We use the continuity of $g(x)$ to replace the integral with a single rectangle Riemann sum.
$$\sum_{k=0}^{2^n-1}(-1)^k\int_{k2^{-n}}^{(k+1)2^{-n}}g(x)dx \approx 2^{-n}\sum_{k=0}^{2^n-1}(-1)^kg(k2^{-n})$$
We expand the series, $$(g(0) - g(2^{-n}))+(g(2*2^{-n})-g(3*2^{-n}))+\cdots$$ if $n$ is large then each difference can be replaced by $\epsilon$ thanks to uniform continuity, then we have that $$2^{-n}\sum_{k=0}^{2^n-1}(-1)^kg(k2^{-n})\leq 2^{-n}2^n\epsilon = \epsilon$$
You should be able to fill in the details.