Let $f_{n}(x) = \int_{0}^{x}\min\{n, \frac{1}{\sqrt{t}}\} dt$ and $f(x) = \int_{0}^{x}\frac{1}{\sqrt{t}}dt$. Show $f_n(x) \overset{n\to\infty}{\longrightarrow} f(x)$ uniformly.
Current attempt: \begin{align*} ||f_{n}(x) - f(x)||_{\infty} &= \underset{x\in[0,1]}{\sup}|f_{n}(x) - f(x)| \\ &= \underset{x\in[0,1]}{\sup}\Big| \int_{0}^{x}\min\{n, \frac{1}{\sqrt{t}}\} dt - \int_{0}^{x}\frac{1}{\sqrt{t}} dt\Big| \\ &\overset{n\to\infty}{\longrightarrow} \underset{x\in[0,1]}{\sup}\Big| \int_{0}^{x}\frac{1}{\sqrt{t}}dt - \int_{0}^{x}\frac{1}{\sqrt{t}} dt\Big| (\ast)\\ &= \underset{x\in[0,1]}{\sup}\Big| \int_{0}^{x}\frac{1}{\sqrt{t}} - \frac{1}{\sqrt{t}} dt\Big| \\ &= \underset{x\in[0,1]}{\sup}\Big| \int_{0}^{x}0dt\Big| \\ &= 0 \end{align*} Where I get concerned is at $(\ast)$, the line where I argue as $n\to\infty$, $\frac{1}{\sqrt{t}}$ $=$ $\min\{n,\frac{1}{\sqrt{t}}\}$. Should I be concerned for when $t \to 0$ as this would have $n = \min\{n,\frac{1}{\sqrt{t}}\}$? Thanks for any help!
Your argument is not valid because you took pointwise limit and just put sup on both sides. For a correct proof note that $|f_n(2x)-f(x)| \leq \int_0^{1}|\min {\{n,\frac 1 {\sqrt t}\}} - \frac 1 {\sqrt t}| dt$; to show that this integal tends to $0$ split the interval into two parts $(0,\frac 1 {n^{2}})$ and $(\frac 1 {n^{2}},1)$. The second part drops out. Now $\int_0^{1/n^{2}} |f_n(t)-f(t)|dt =\int_0^{1/n^{2}} |\frac 1 {\sqrt t} -n| dt$. Can you compute this explicitly to complete the proof?.