Suppose that $f_n:[a,b] \rightarrow \Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?
- No discontinuities
- At most ten discontinuities
- At least ten discontinuities
- Uncountably many discontinuities
- Countably many discontinuities
- No jump discontinuities
- No oscillating discontinuities
My try :
For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $\varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.
For fourth bullet: To disprove this , consider $$f_n(x)=\begin{cases} \frac{1}{n} & \text{if}\; x \in \Bbb Q \cap [0,1]\\\\0 &\text{otherwise}\end{cases}$$
Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform
For fifth bullet: To disprove this, consider $$f_n(x)=\begin{cases} \frac{1}{n} & \text{if}\; 0<x<\frac{1}{n} \\\\0 &\text{if}\;x=0 \wedge \frac{1}{n} \leq x \leq 1\end{cases}$$
Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=\frac{1}{n}$ but $f$ continuous on $[0,1]$.
This link answers the sixth bullet
Is my arguments correct ? Can I have a hint for others ?
Best Answer
First bullet point is correct.
Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.
Hint for third: Very similar construction to the 4th, which is correct.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.
To do that, consider such a point $x_0$ and a given $\epsilon > 0$. We know that because of the uniform convergence of $f_n$ to $f$, there is a $N \in \mathbb N$ such that
$$ \forall x \in [a,b], \forall n > N: |f_n(x) - f(x)| < \frac {\epsilon}3$$
We also know that $f_{N+1}$ is continuous at $x_0$, which means
$$ \exists \delta > 0: |x-x_0|<\delta \Rightarrow |f_{N+1}(x) - f_{N+1} (x_0)| < \frac {\epsilon}3$$
Together this means
$$ |x-x_0|<\delta \Rightarrow |f(x) - f(x_0)| = |f(x) - f_{N+1}(x) + f_{N+1}(x) - f_{N+1}(x_0)+ f_{N+1}(x_0) -f(x_0)| \le |f(x) - f_{N+1}(x)| + |f_{N+1}(x) - f_{N+1}(x_0)| + |f_{N+1}(x_0) -f(x_0)|< \frac {\epsilon}3 + \frac {\epsilon}3 + \frac {\epsilon}3 < \epsilon$$
In a nutshell, this is the proof that uniform convergence of a function sequence of continuous functions implies continuity of the limit.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.