Consider the sequence of functions $$h_n(x)=\begin{cases}2nx&\text{if}\;x \in [0,\frac{1}{2n}]\\\\2-2nx&\text{if}\;x \in [\frac{1}{2n},\frac{1}{n}]\\\\0&\text{if}\;x\geq \frac{1}{n} \end{cases}$$
- Prove that $h_n(x)$ does not converge uniformly on $[0,1]$
- Prove that $h_n(x)$ converges uniformly to $0$ on $[\delta,1]$, for $0<\delta \leq 1$
- Prove, if $f:[0,1] \to \Bbb R$ is continuous, then $\lim_{n \to \infty} \int_0^1 f(h_n(t))dt=f(0)$
My try
Here, $$\lim_{n \to \infty }h_n(x)=0=h(x)$$ Also $\frac{1}{2n} \in [0,1]$ such that $$\lim_{n \to \infty}\left[h_n\left(\frac{1}{2n}\right)-h\left(\frac{1}{2n}\right)\right]=1 \neq 0$$ so $h_n$ does not converge uniformly on $[0,1]$
For the third one, (assuming the second) , I know $f(h_n(x)) $ converges uniformly to $f(h(x))$ on $[\delta,1]$. But here the problem is $0$ to $1$ limit. Also I'm stuck on proving the second one.
Can I have a hint for these two?
For any $\delta>0$, pick some $N\ge 1/\delta$. Then, for any $n\ge N$, we have $1/n\le \delta\Rightarrow h_n(x)=0$ for $x>\delta$, whence $h_n\to 0$ uniformly on $[\delta,1]$. From this we have $$\int_0^1 f(h_n(x))=\int_0^\delta f(h_n(x))+\int_\delta^1 f(0)=\int_0^\delta f(h_n(x))+(1-\delta)f(0)$$ For all large enough $n$. Now use the fact that $f$ is bounded to discard the first term by letting $\delta\to 0$.