Suppose that $f \in C[0,1]$ . Show that $\lim_{n\to \infty}\int_{0}^{1}(n+1)x^{n}f(x)dx=f(1)$ Suppose that $f \in C[0,1]$ . Show that $\lim_{n\to \infty}\int_{0}^{1}(n+1)x^{n}f(x)dx=f(1)$.
This is how I proceeded:
Suppose $x^{n+1}=t$ . Then $(n+1)x^ndx=dt$. Now the integral becomes $$\int_{0}^{1}dtf(t^{\frac{1}{n+1}})$$. Now If i take $g_n(t)=f(t^{\frac{1}{n+1}})$, then $\lim_{n\to \infty}g_{n}(t)=f(1)$. Now I know that $g_{n}(t)$ $\to$ $f(1)$ pointwise . All i need to show is that the convergence is uniform and then I am through. 
How do i show that the convergence is uniform??
Thanks for the help!!
 A: The convergence is not uniform, as for example $(2^{-(n+1)})^\frac{1}{n+1} = \frac{1}{2}$ for all $n$. But actually you do not need uniform convergence if you agree to use dominated convergence theorem.
A: Convergence cannot be uniform on $[0,1]$ or even $(0,1)$ since $0^{1/(n+1)} \to 0$ means there will be a tail coming down from $1$ to $0$ in each $t^{1/(n+1)}$ plot on $[0,1]$.
But we almost have uniform convergence. Let's consider an $\epsilon$ interval near zero and then the rest of the interval. Let $M$ be the max of $|f(x)|$ on our interval. Then
$$\limsup_{n \to \infty} \left| \int_0^1 [f(t^{1/n}) - f(1)]dt \right| \leq \limsup_{n \to \infty} \int_0^\epsilon |f(t^{1/n}) - f(1)|dt + \limsup_{n \to \infty} \int_\epsilon^1 |f(t^{1/n})-f(1)|dt \\ \leq 2 \epsilon M + 0$$
provided convergence is uniform on $[\epsilon, 1]$. This is the case since $\sup_{x \in [\epsilon, 1]} |t^{1/n}-1| \leq |\epsilon^{1/n}-1|$ and $f$ is uniformly continuous so as to make $f(t^{1/n})$ uniformly convergent to 1.
Finally, take $\epsilon \to 0$ to get the limsup to be 0, and hence the limit exists and is zero.
If you ever take a graduate analysis class then you can use the dominated convergence theorem. Look up the wiki article for it. The only additional assumption you need is a function $g(x)$ such that $|f(t^{1/n})|\leq g(x)$ and $\int_0^1 g(x) dx < \infty$. In our case $g(x) = M$ works.
A: Put $ \vert $$ \vert $f$ \vert $$ \vert $$ _{\infty} $ = $ \max\limits_{x \in [0,1]}  $ $ \vert $f(x)$ \vert $
f continuous on 1 so : $ \forall $ $ \varepsilon $$ > $0, $ \exists $ $ \eta $$ \in $ ]0,1[, 1-$ \eta $$ \leq $ x $ \leq $ 1 $ \Rightarrow $ $ \vert $ f(x) - f(1) $ \vert $ $ \leq $ $ \varepsilon $ 
$ \vert $ (n+1)$\int_{0}^{1}x^{n} f(x) dx $ -f(1) $ \vert $ = $ \vert $ (n+1)$\int_{0}^{1}x^{n} (f(x)-f(1)) dx $ $ \vert $ 
$\vert $ (n+1)$\int_{0}^{1}x^{n} f(x) dx $ -f(1)$ \vert $ $ \leq $ (n+1)$\int_{0}^{1-\eta}x^{n}\vert f(x)-f(1)\vert dx $ + (n+1)$\int_{1-\eta}^{1}x^{n} \vert f(x)-f(1) \vert dx $ 
$\vert $ (n+1)$\int_{0}^{1}x^{n} f(x) dx $ -f(1)$ \vert $ $ \leq $ (n+1)$\int_{0}^{1-\eta}(1-\eta)^{n}2 \vert \vert f \vert\vert_{\infty}dx$ + (n+1)$\int_{1-\eta}^{1}x^{n} \epsilon dx $ 
$\vert $ (n+1)$\int_{0}^{1}x^{n} f(x) dx $ -f(1)$ \vert $ $ \leq $ 2(n+1)(1-$ \eta $)$ ^{n+1} $ $ \vert $$ \vert $f$ \vert $$ \vert $$ _{\infty} $ + $ \varepsilon $ $ \underset{n\rightarrow +\infty}{\rightarrow}$ 0 
so  (n+1)$\int_{0}^{1}x^{n} f(x) dx $  $ \underset{n\rightarrow +\infty}{\rightarrow}$ f(1)
