B S u p x i n [ a , b ] | P n ( x ) − f ( x ) | < ϵ

[feerrr]

why sup x in [a,b] |Pn(x) - f(x) | < ϵ , Pn(x)=a0+a1x+...+anx^n
why f(x)-ϵ<Pn(x)<f(x)+ϵ

 

[PeroK]

Are you able to express that in a way that let's other people understand what you are asking?

 

[feerrr]

f in C[a,b] and n=0,1,2,...
we have : ∫ x^n*f(x)dx=0 (integral from a to b )
im trying to show that f(x)=0 in [a,b]
can i use : sup x in [a,b] |Pn(x) - f(x) | < ϵ ( Pn(x)=a0+a1x+...+anx^n ) ? and why ?

 

[PeroK]

I don't follow that you can go straight to that.

Do you know the Weierstrass approximation theorem?

I suspect this is hard to prove from first principles.

 

[feerrr]

no i don't know the Weierstrass approximation theorem .
can i show that f(x)=0 without using the Weierstrass approximation theorem ? i need some help

 

[PeroK]

no i don't know the Weierstrass approximation theorem .
can i show that f(x)=0 without using the Weierstrass approximation theorem ? i need some help

What I suggest is to look up the theorem on line and find a proof using it. It's still not easy.

I can't immediately see another way.

Also, you need to know or prove that $f$ is bounded on $[a,b]$.