Uniformly convergence of functions defined on closed intervals to a function defined on an open interval

Question

Let $f(x)$ be a function defined on $(0,1]$, possibly unbounded at $0$. i.e., $f(x)=\frac{1}{\sqrt{x}}$ on $(0,1]$.

Define $f_n(x) $ be a sequence of continuous functions defined on $[0,1]$: \begin{eqnarray*} f_n(x) &=& \frac{1}{\sqrt{x}} \quad \mbox{ if } x \in [\frac{1}{n}, 1]\\ && \sqrt{n} \quad \mbox{ if } x \in [0, \frac{1}{n}] \end{eqnarray*} I'd like to the uniform convergence of $f_n(x)$ to $f(x)$. Although the uniform convergence on a closed interval implies the continuity of $f(x)$, $f(x)$ is only defined on $(0,1]$. So I think we can say the uniform convergence of $f(x)$.

Let's use the standard epsilon delta argument.

For $\epsilon >0$, there exists $N$ such that for $x\in (0,1]$ and $n\geq N$, $$ |f_n(x) -f(x)|<\epsilon. $$

However, this approach also fails because any given $\epsilon, n$, we can find $x$ such that $|f_n(x)-f(x)|>\epsilon$.

Then, how to say the uniform convergence of $f(x) \in C(0,1]$, possibly unbounded??

Answer 1

No, the convergence is not uniform. If a sequence of bounded functions converges uniformly to a function $f$, the $f$ is bounded too. But your function $f$ is unbounded.

Answer 2

hint

$ f_n $ is well defined for $ n>0$ and for $ x\in [0,1] $ But,

$$f_n(0)=n, $$

$$\lim_{n\to+\infty}f_n(0)=+\infty$$ and for $ x\in (0,1],$ $$\lim_{n\to+\infty}f_n(x)=f(x)$$ So, the convergence of $ (f_n) $ to $ f $ is pointwise only at $ (0,1]$ and we cannot speak about uniform convergence at $ [0,1]$. For $ a\in(0,1)$, the convergence is uniform at $ [a,1] $ because

$$\forall x\in[a,1]\; $$ $$|f(x)-f_n(x)|=\frac{1}{\sqrt{x}(n\sqrt{x}+1)}\le \frac{1}{\sqrt{a}(n\sqrt{a}+1)}$$

and $$\lim_{n\to+\infty}\frac{1}{\sqrt{a}(n\sqrt{a}+1)}=0$$