Uniform convergence of $x^n$ on interval $[0,b]$ for $0 \leq b < 1$.

Question

Suppose we have $f_n (x) = x^n$ for $x \in [0,b]$ for $0 \leq b < 1$. If $\epsilon > 1$, any value of $K$ will do because every $x^n$ for any $x$ or any $n$ is less than $1$.

If $0 < \epsilon < 1$, things are a bit trickier. We need: \begin{align*} |f_n (x) - f(x)| = |x^k - 0 | = x^k. \end{align*} By setting up $x^k < \epsilon$, I get $k > \frac{\log \epsilon}{\ln x}$. But, this depends on $x$, which clearly cannot be the case for uniform convergence. The textbook solution I'm reading sets $K = \frac{\log \epsilon}{\log b}$. This doesn't make any sense to me. Both $\epsilon$ and $b$ are between $0$ and $1$, so both logarithms will be negative and the ratio positive. But the log function is monotonically increasing. So $\log b > \log x$ for any other $x \in [0,b]$. Thus, the denominator of this expression will be at its maximum and the expression, $\frac{\log \epsilon}{\log b}$ at its minimum. It seems I would want the left endpoint, but $\log 0$ is clearly undefined.

Any help with this would be greatly appreciated.

Answer 1

Let $0<\varepsilon<1$. Then we have \begin{align*} |f_n(x)-f(x)|=|f_n(x)|=|x^n|=x^n<b^n \end{align*} and $b^n<\varepsilon$ if and only if $$n\ln(b)<\ln(\varepsilon)$$ if and only if (since $\ln(b)<0$) $$n>\frac{\ln(\varepsilon)}{\ln(b)}.$$

Answer 2

If $n\ge\frac{\ln(\varepsilon}{\ln(b)}$, then $n\ln(b)\le\ln(\varepsilon)$, hence $b^n\le\varepsilon$.

Now, for such an $n$ and for every $x\in[0,b]$, we have :

$$\vert x^n-0\vert=x^n\le b^n\le\varepsilon$$

This prove that the sequence $(x\mapsto x^n)_{n\ge0}$ converges uniformly to $x\mapsto0$ on $[0,b]$, for every $b\in(0,1)$.