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Let $f:\left[\frac{1}{2} ,1\right] \rightarrow \mathbb R$ be a continuous function, $\{g_n\}_{n=1}^{\infty}$ a sequence of functions where $g_n(x) = x^n f(x)$, with $x \in \left[\frac{1}{2} ,1\right]$ and $n \in \mathbb{N}$.

Show that

a) the sequence of functions $\{g_n\}$ is convergent on $\left[\frac{1}{2} ,1\right]$.

b) $\{g_n\}$ is uniformly convergent on $\left[\frac{1}{2} ,1\right]$ if and only if f is bounded on $\left[\frac{1}{2} ,1\right]$ and $f(1)=0$.

I'm struggling with this problem and looking for any help. Thanks!

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We have $$\lim_{n\to\infty}g_n(x)=\lim_{n\to\infty}x^nf(x)=\left\{\begin{array}{cl} 0&\mathrm{if}\, x\in[\frac{1}{2},1)\\ f(1)&\mathrm{if}\, x=1 \end{array}\right.$$ Since $f$ is continuous on the compact $[\frac{1}{2},1]$ then it's bounded so we'll prove that $(g_n)$ is uniformly convergent if and only if $f(1)=0$.

If $(g_n)$ is uniformly convergent and since it's continuous then it's pointwise limit is also continuous and hence $f(1)=0$.

Now suppose that $f(1)=0$ and since $|g_n|$ is continuous on the compact $[\frac{1}{2},1]$ then its supremum is atteined on a point $a_n\in[\frac{1}{2},1)$ then $$||g_n||_\infty=a_n^n|f(a_n)|$$ By Weierstrass theorem $(a_n)$ has a convergent subsequence $(a_{\varphi(n)}$) and if its limit is $1$ then $$0\leq||g_{\varphi(n)}||_\infty\leq|f(a_{\varphi(n)})|\to|f(1)|=0$$ and if its limit is $\ell\not=1$ then $$0\leq||g_{\varphi(n)}||_\infty\leq(\ell+\epsilon)^n||f||_\infty\to0$$ where $\epsilon $ is such that $\ell+\epsilon<1$.

Hence we proved that $\lim_{n\to\infty}||g_n||_\infty=0$ and this allows us to conclude.

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a) Deal with the cases $x<1$ and $x=1$.

b) $f$ is continuous on $[1/2,1]$ hence bounded on this interval, so this condition is empty. In the previous question, we can identify the limit. An uniform limit of continuous functions is continuous, hence the limit function obtained function has to be continuous. For the other direction, use continuity of $f$ at $1$.

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$f$ is a continuous function on a compact set, so it has a maximum. Then what is the limit for $x=1$ and $x < 1$?

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