# Math Help - Need examples for functions that fit the following

1. ## Need examples for functions that fit the following

Give an example, if possible, of a sequence { $f_{n}$} $_{n \in N}$ of discontinuous functions on R which converges uniformly to:

a.) a discontinuous function on R

b.) a continuous function on R

Give an example of a sequence of functions { $f_{n}$} $_n \in N$ such that:

c.) $f_{n}$ is continuous and not differentiable but the sequence converges pointwise to a differentiable function

d.) {| $f_{n}$|} $_{n \in N}$ converges pointwise but { $f_{n}$} $_{n \in N}$ does not

2. Originally Posted by thaopanda
Give an example, if possible, of a sequence { $f_{n}$} $_{n \in N}$ of discontinuous functions on R which converges uniformly to:

a.) a discontinuous function on R

b.) a continuous function on R

Give an example of a sequence of functions { $f_{n}$} $_n \in N$ such that:

c.) $f_{n}$ is continuous and not differentiable but the sequence converges pointwise to a differentiable function

d.) {| $f_{n}$|} $_{n \in N}$ converges pointwise but { $f_{n}$} $_{n \in N}$ does not
a) Take the constant sequence of a discontinuous function
b) Take the sequence f(x)=0 if x<0, f(x)=1/n otherwise
c) Do you mean nowhere differentiable or only at one (or more) point?
d) Take $(-1)^n$

3. ## Nowhere differentiable

That was all the problem said, so I take it that it should be nowhere differentiable. The only thing I can think of that is nowhere differentiable is:

g(x) = |x| on [-1,1]
h(x) = g(x) if x $\in$ [-1,1];
otherwise, h(x-2) if x > 0 or h(x+2) if x < 0

f(x) = $\sum (\frac{3}{4})^n$ h( $4^nx$) with n going from 0 to $\infty$

4. Originally Posted by thaopanda
That was all the problem said, so I take it that it should be nowhere differentiable. The only thing I can think of that is nowhere differentiable is:

g(x) = |x| on [-1,1]
h(x) = g(x) if x $\in$ [-1,1];
otherwise, h(x-2) if x > 0 or h(x+2) if x < 0

f(x) = $\sum (\frac{3}{4})^n$ h( $4^nx$) with n going from 0 to $\infty$
So maybe take $F_k(x)=\frac{f(x)}{k}$. It's a pretty stupid example, but it does converge to $0$. I believe the convergence is uniform, but I'm not sure.