Let $\phi: [0,1] \to \mathbb{R}$ be continuous. Let $f_n :[0,1] \to \mathbb{R}$ be defined by $f_n(x) = x^n \phi(x)$. Prove that $f_n$ converges uniformly on $[0,1]$ iff $\phi(1) = 0$
The forward implication part was easy but I'm having trouble in proving the converse part.
Attempt: Conversely, suppose $\phi(1)=0$ then $f_n(x)= x^n\phi(x) \to 0$ for each $x\in [0,1]$. Let $g:[0,1] \to \mathbb{R}$ be defined by $g(x) = 0, \ \forall x \in [0,1]$, Then $f_n \to g$ pointwise on $[0,1]$
And $|f_n(x)- g(x)| = |f_n(x)|=|x^n \phi(x)|$
(I'm stuck here, I tried estimating the inequality by using the maximum of $\phi$ but didn't get anything. Give me hints so that I can proceed, Thanks)
Best Answer
Let $\varepsilon > 0$. Since $\phi$ is continuous and $\phi(1) = 0$, there exists $\delta > 0$ such that $$x \in [1-\delta,1] \implies |\phi(x)| \le \varepsilon.$$
Now, denoting $\|\phi\|_\infty = \sup_{x\in[0,1]}|f(x)|$, pick $n \in \Bbb{N}$ large enough such that $(1-\delta)^n\|\phi\|_\infty \le \varepsilon$.
Then for $x \in [0,1-\delta]$ we have $$|x^n\phi(x)| \le (1-\delta)^n\|\phi\|_\infty \le \varepsilon$$ while for $x \in [1-\delta,1]$ we have $$|x^n\phi(x)| \le \varepsilon.$$ Therefore for all $x \in [0,1]$ we have $|x^n\phi(x)| \le \varepsilon$ which proves uniform convergence.