# Continuous sum function?

#### cp05

is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it #### Drexel28

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is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it What about $$\displaystyle f_n(x)=\sum_{m=1}^{n}\frac{f(x)}{2^m}$$ the sum is uniformly convergent since $$\displaystyle \|f\|_{\infty}$$ exists, etc.

• cp05

#### cp05

How do you know it's uniformly convergent, because you don't know what f is?

Also, how does that show every continuous function on the interval [0,1] is a limit of a sequence of continuous functions? Or is it showing the statement is false?
I don't understand.
Don't I need to start with a continuous function on [0,1] and show it's not the limit of a sequence of continuous functions to prove it false?

#### Jose27

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is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it Actually you can say even more :Stone?Weierstrass theorem - Wikipedia, the free encyclopedia (look for the original Wierstrass approximation theorem).

That said, Drexel's argument works fine since $$\displaystyle \sum_{k=1}^{\infty } \frac{1}{2^k} =1$$ and you know $$\displaystyle f$$ is continous.

• cp05

#### mabruka

Read these posts again having in mind that you were wrong, the statement is true.

The proof should come out easily, take f continuous on [0,1],

use what drexel and jose said and prove f is the uniform limit of f_n.

Dont forget that since f's domain is compact then f is bounded there, it will come in handy somewhere.

Good luck!

• cp05

#### Bruno J.

MHF Hall of Honor
is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it I might be missing something here, but what about $$\displaystyle f_n = f$$?

• cp05

#### cp05

I can't prove something saying fn=f can I? Because that's being specific.

And why am I picking ? Isn't that being specific also?

#### cp05

Oh wait i think I might understand it.

So I'm taking an f continuous on [0,1]
And I have to show that there is A (like...ONE at least) sequence of continuous functions that converges to that uniformly

So taking , sup|f/2^m| = 1/2^m (I'm a little confused about this...because how do I know what f is? Would the supremum be f/2^m instead? )
Either way
then SUM(1/2^m) or SUM(f/2^m)...whichever one it is, will be geometric with r = 1/2, so it converges
So by Wierstrass M-test, fn is uniformly convergent.

So how do I show it converges to f(x)?? because as m -> infinity, the sum goes to f/[1-(1/2)], not f right?

But for continuity...we know f is continuous and 2^m is continuous, so division of two continuous functions is continuous and then the sum of continuous functions is continuous so fn(x) is continuous.

Am I anywhere on the right track? I'm just not sure about the supremum part of the M-test and how to show fn(x) converges to f.

#### cp05

I think I get the fn=f better now

f is continuous on [0,1]
so fn is continuous on [0,1] also since fn=f

and the sup|fn-f| = 0 which ->0 so fn converges to f uniformly on M

Is that right?

#### mabruka

first,

what bruno suggests is "trivially" true (because he proposes a constant sequence) so i think what cp05 needs to prove is that there exists a "non constant" sequence of continuous f_n... bla bla bla

second,

good try cp05, still you need to be sure exactly what and why you are doing the things you are doing :

take f continuous on [0,1].

Define a sequence by $$\displaystyle f_n(x)=\sum_{k=1}^n \frac{f(x)}{2^k}$$

Since f is continuous and f_n is a finite sum of continuous functions then each f_n is continuous on [0,1].

By what i said earlier f is bounded on [0,1] by, lets say, M. So $$\displaystyle |f(x)| \leq M$$ $$\displaystyle \forall x\in [0,1]$$

Now lets see the uniform convergence

Given $$\displaystyle \epsilon$$ we need to find an adequate N which does not depend on x, only on $$\displaystyle \epsilon$$.

Take ANY x\in [0,1]

$$\displaystyle |f(x)-f_n(x)|=|f(x)||1-\sum_{k=1}^n \frac{1}{2^k}|$$

We know that $$\displaystyle \sum_{k=1}^n \frac{1}{2^k}\to 1$$ when $$\displaystyle n\to \infty$$
so for $$\displaystyle \frac{\epsilon}{M}$$ there is an N for which

$$\displaystyle |1-\sum_{k=1}^n \frac{1}{2^k}|<\frac{\epsilon}{M}$$ if $$\displaystyle n\geq N$$

Using the first equalities we have that for $$\displaystyle n\geq N$$

$$\displaystyle |f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon$$

where we used that f is bounded by M. Take a good look, N does not depend on x, only on M (and on the bound but that is fixed anyway).

If you want to get the supremum way of seing it then think like this:

We just showed that for any $$\displaystyle \epsilon$$ there is an$$\displaystyle N(\epsilon)$$ such that
$$\displaystyle |f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon$$ if $$\displaystyle n\geq N$$ for any $$\displaystyle x\in [0,1]$$

Taking supremum over [0,1] grants you that

$$\displaystyle ||f-f_n||_{sup}<\epsilon$$ if $$\displaystyle n\geq N$$

i hope i didnt make any mistake

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• cp05