Show that $f_n$ converges uniformly to $f$ on $[0, 1]$

Question

Show that $f_n$ converges uniformly to $f$ on $[0, 1]$.

Please help me. I have no idea.

Answer 1

If i'm correct, note that $h_k$ depends on n, then denote it as $h_{k,n}$. Note also that $$h_{k,n}(x)=\begin{cases}\sqrt{n},& t\in[\frac{k}{n},\frac{k+1}{n}),\\ 0, &\text{otherwise}.\end{cases}$$

Therefore, $$\langle f,h_{k,n}\rangle=\int_{\frac{k}{n}}^{\frac{k+1}{n}} \sqrt{n}f(x)\,dx=\frac{\sqrt{n}}{n}f(\zeta_{k,n})$$ for some $\zeta_{k,n}\in (\frac{k}{n},\frac{k+1}{n})$, this latter given by the Mean Value Theorem for definite integrals. That shows $$f_n(x)=f(\zeta_{k,n})\ \text{if}\ x\in[\frac{k}{n},\frac{k+1}{n}).$$

Use uniform continuity of $f$ in $[0,1]$ to show uniform convergence.

PS. Excuse my bad english.