# Thread: how can u prove this argument

1. ## how can u prove this argument

$f_{n},g_{n}$ are sequences of functions : $X\rightarrow R$ which converge uniformly to f,g respectively.

If $\lambda\in R$, prove $\lambda f_{n}$ converges uniformly to $\lambda f$

and

prove : $f_{n}g_{n}$ converges to $fg$ given f,g are bounded.

2. Originally Posted by szpengchao
$f_{n},g_{n}$ are sequences of functions : $X\rightarrow R$ which converge uniformly to f,g respectively.

If $\lambda\in R$, prove $\lambda f_{n}$ converges uniformly to $\lambda f$

and

prove : $f_{n}g_{n}$ converges to $fg$ given f,g are bounded.
You sent me a PM.
I responded with a question: “What do you do for yourself?”
Now I ask you the same question publicly.
It seems to me that you simply expect to be given the answers.
Am I wrong?
If so, explain yourself!

3. ## No. I have done part of this question.

No. I done part of this question. I am not sure if my proof is right. and I am not sure my link between two arguments is right.

i post my proof for the first part,

when $\lambda=0$, question is trivial:
$\epsilon_{1}>0, \ \forall x\in X, \ |\lambda f_{n}(x)-\lambda f(x)|=0<\epsilon_{1}$
So assume $\lambda\neq0$
$f_{n}$ converges uniformly to f:
$\forall \epsilon_{1}>0, \ \exists N, \ \forall x\in X, \forall n\geq N, |f_{n}(x)-f(x)|<\epsilon_{1}$
by writing $\epsilon_{2}=\frac{\epsilon_{1}}{|\lambda|}$:
$\forall \epsilon_{2}>0, \ \exists N, \ \forall x\in X, \forall n\geq N, |\lambda f_{n}(x)-\lambda f(x)|<\epsilon_{2} \ \ \square.$

4. ## and then

and then i just said, by setting:
$\lambda=g_{n}(x) \forall x\in X$
and use lemma just proved,
I got:
$\lambda f_{n}(x)$ converges uniformly to $
\lambda f$

and then set $\lambda=f(x)$, use lemma again,
$g_{n}(x) f(x)$ converges pointwise to $fg$

5. ## ok

ok. i posted all the proof i can get. so, could you please give me any correction of my proof if there is any problem.