# Pointwise convergence of a sequence of functions $g_n$ on $(0,1]$

As stated in the title I am trying to prove $$g_{n}=\sum_{k=1}^{2^n}\frac{2^{n}}{k}\chi_{\left(\left(\frac{k-1}{2^{n}}\right)^{2},\left(\frac{k}{2^{n}}\right)^{2}\right]}$$ converges pointwise to $$\frac{1}{\sqrt{x}}$$ on $$(0,1]$$. Obviously $$\left\lbrace\left(\left(\frac{k-1}{2^{n}}\right)^{2},\left(\frac{k}{2^{n}}\right)^{2}\right]\right\rbrace_{k=1}^{2^n}$$ is a partition of the interval $$(0,1)$$ so for some $$k\in\left\lbrace 1,2,\dots,2^n\right\rbrace$$, $$g_n(x) =\frac{2^n}{k}$$.

I was thinking about taking the largest $$n$$ such that for some $$k\in\left\lbrace 1,2,\dots,2^n\right\rbrace$$, $$x\in \left(\left(\dfrac{k-1}{2^n} \right)^2,\left(\dfrac{k}{2^n}\right)^2 \right)$$ and then breaking that up to find a $$N$$ such that $$x=\left(\dfrac{k}{2^N}\right)^2$$, but I am not sure such a $$n$$ exists. Any help is welcome. I tried squeeze theorem as well, but I failed to find one of the bounds.

• - Your sentence "obviously .. so for some $k\in\dots, g_n(x) = \frac {2^n}k.$" doesn't make sense to me, I don't know what you're trying to say. Probably because $x$ appears out of nowhere. - is it $\sum_{k=1}^{2n}$ or $\sum_{k=1}^{2^n}$? - There is no largest $n$ such that $x$ belongs to that interval (I suppose you mean $(a,b]$ instead of $(a,b)$? This is because the endpoints $k/2^n$ cover the entire interval $[0,1]$, so for each $n$, there's always a $k$. – Calvin Khor May 8 at 6:07
• $\sum_{k=1}^{2^n}$ – Michael Cook May 8 at 6:09

For any $$n$$ and $$x$$ there is a unique $$k_n$$ such that $$(\frac {k_n-1} {2^{n}})^{2} . Note that $$\frac {k_n-1} {2^{n}} <\sqrt x \leq \frac {k_n} {2^{n}}$$. This implies that $$\frac {k_n} {2^{n}} \to \sqrt x$$. Hence $$g_n(x)=\frac {2^{n}} {k_n} \to \frac 1 {\sqrt x}$$.