# A working example wrt the supremum norm

## Main Question or Discussion Point

Folks,

Could anyone give me a working example of a sequence of functions that converges to a function wrt to the supremum norm?

Thank you.

Depends on what you mean with "working example". A nice example is probably

$$f_n(x)=\sum_{k=0}^n{\frac{x^k}{k!}}.$$

This converges to $f(x)=e^x$. This convergence is uniform on each bounded set.

Depends on what you mean with "working example". A nice example is probably

$$f_n(x)=\sum_{k=0}^n{\frac{x^k}{k!}}.$$

This converges to $f(x)=e^x$. This convergence is uniform on each bounded set.
Would this converge to the same function wrt to the integral norm?

Yes. It is actually a nice exercise to show that it does. In fact, it suffices in this case to show that

$$\int_a^b|f_n|\rightarrow \int_a^b |f|$$

LCKurtz
Homework Helper
Gold Member
Pick a number a with ##0 < a < 1##. Let ##f_n(x) = x^n\hbox{ on }[0,a]##. Then ##\|f_n - 0\|\rightarrow 0##. Note that this fails if ##a=1##.

LCKurtz' example is a really nice one!! It is nice to notice that it can be generalized into what is known as Dini's theorem:

If $(X,d)$ is a compact metric space and if $f_n:X\rightarrow \mathbb{R}$ are a sequence of functions such that
- They are monotonically decreasing/increasing
- They pointswize converge to a function f
- f is continuous
then the convergence is uniform.

It is one of the few cases where pointsiwize convergence implies uniform convergence.

LCKurtz' example follows with X=[0,a] (with 0<a<1), $f_n(x)=x^n$ and $f(x)=0$. It is a nice exercise to see what goes wrong in the theorem if a=1.

Depends on what you mean with "working example". A nice example is probably

$$f_n(x)=\sum_{k=0}^n{\frac{x^k}{k!}}.$$

This converges to $f(x)=e^x$. This convergence is uniform on each bounded set.
Would this converge to the same function wrt to the integral norm?
Would it be something like this...

$\displaystyle ||f_n(x)||_1=\int_{a}^{b}| \sum_{n=0}^{\infty} \frac{x^k}{k!}|dx$ for c[a,b]

Pick a number a with ##0 < a < 1##. Let ##f_n(x) = x^n\hbox{ on }[0,a]##. Then ##\|f_n - 0\|\rightarrow 0##. Note that this fails if ##a=1##.
Hmmm..but regarding post 3 ie a function f_n the converges to the same function f on both sup and integral norms...this function does not converge to a function but to zero.....?

LCKurtz' example is a really nice one!! It is nice to notice that it can be generalized into what is known as Dini's theorem:

If $(X,d)$ is a compact metric space and if $f_n:X\rightarrow \mathbb{R}$ are a sequence of functions such that
- They are monotonically decreasing/increasing
- They pointswize converge to a function f
- f is continuous
then the convergence is uniform.

It is one of the few cases where pointsiwize convergence implies uniform convergence.

LCKurtz' example follows with X=[0,a] (with 0<a<1), $f_n(x)=x^n$ and $f(x)=0$. It is a nice exercise to see what goes wrong in the theorem if a=1.