Prove that this sequence does NOT converge uniformly on [0,1] AND (0,1).

Question

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Answer 1

The function $$ \phi_n(x) = \frac{x}{2^n - (2^n-1)x} $$ has a pointwise limit $\phi(x)$ at every $x$. If it's going to converge to something uniformly, it's going to be to $\phi(x)$.

You should figure out what $\phi(x)$ is as a function of $x$. (Your work for (a) shows how to find $\phi(1)$, since you substitute $x=1$ in the first step.)

For either part (a) or part (b), your approach can be used to show that $\phi_n$ does not uniformly converge to $\phi$. Except instead of using $1$, you should use the correct value of $\phi(x)$. If you show that the sequence does not converge uniformly on $(0,1)$, then it does not converge uniformly on $[0,1]$ either. The only difference between the two solutions is that the value $$ \sup\{|\phi_n(x) - \phi(x)| : x \in S\} $$ allows $x$ to be $0$ or $1$ when $x\in[0,1]$, and does not allow it when $x\in(0,1)$. If you don't need $0$ or $1$ for computing the supremum, then the same argument works in both cases.

However, for (a) a shorter argument also exists, using the fact that $\phi$ is not continuous on $[0,1]$. And for (b), you can use (a) together with this result. So even though there is an argument that solves both cases simultaneously, there is a different argument that does not, which might be easier.