Math, asked by sumitarora4167, 1 year ago

Weierstrass's theorem states that a continuous function on [0, 1] can be uniformly approximated by polynomials. can every continuous function on the closed unit disc be approximated uniformly by polynomials in the variable z?

Answers

Answered by shubhamsinhaiitg
0
It depends on what is meant by "polynomial".

If only ∑cnzn∑cnzn, then every function that is uniformly approximable by polynomials must be holomorphic on the interior of JJ.

Although that condition is trivially satisfied if JJ has empty interior, that doesn't mean that for such JJ every continuous function is the uniform limit of polynomials. For example the unit circle has empty interior, but a sequence of polynomials converging uniformly on the unit circle converges uniformly on the closed unit disk by the maximum principle, and thus if ff is a uniform limit of polynomials on the unit circle, then there is a holomorphic function hh on the unit disk that extends continuously to the unit circle, with boundary values ff. In particular, we have

∫|z|=1f(z)⋅zndz=0(1)(1)∫|z|=1f(z)⋅zndz=0

for all n⩾0n⩾0. (And, in this case, that condition is sufficient.)

That phenomenon generalises, if JJdisconnects the plane, that is, if C∖JC∖Jhas at least two connected components, then the bounded components of the complement of JJ impose restrictive conditions on the continuous functions that are uniform limits of polynomials similar to (1)(1).

Mergelyan's theorem asserts the converse, if JJ is a compact subset of CCwith empty interior such that C∖JC∖J is connected, then every continuous function on JJ can be uniformly approximated by polynomials (in zz only).

If "polynomial" means polynomial in zz and z¯¯¯z¯, or equivalently polynomial in RezRe⁡z and ImzIm⁡z, then the Weierstraß approximation theorem holds for all compact JJ.

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