Removable Singularity of L_a^2(G): Proof of Limit at z=0

[Nusc]

Homework Statement

If G = { z in C: 0<|Z|<1} show that every f in L_{a}^{2}(G) has a removable singularity at z = 0

Proof:

We must show that lim z->0 z*f(z) = 0 for all f in L_{a}^{2}(G)

By a corollary 1.12, if f in L_{a}^{2}(G), a in G and 0<r<dist(a,bdr G), thne
|f(a)| <= 1/(r\sqrt(\pi))||f||_2,

for |z| < 1/2 we have that

|f(z)| <= 1/(|z|\sqrt(pi)/sqrt(2)) ||f||_2 = 2/(|Z|sqrt(pi) ||f||_2

so that |zf(z)| = |z||f(z)| <= 2/sqrt(pi) ||f||_2.

how do I proceed from there

Homework Equations

The Attempt at a Solution

 

[lippyka]

We must show that lim z->0 z*f(z) = 0 for all f in L_{a}^{2}(G)By a corollary 1.12, if f in L_{a}^{2}(G), a in G and 0<r<dist(a,bdr G), thne|f(a)| <= 1/(r\sqrt(\pi))||f||_2,for |z| < 1/2 we have that|f(z)| <= 1/(|z|\sqrt(pi)/sqrt(2)) ||f||_2 = 2/(|Z|sqrt(pi) ||f||_2so that |zf(z)| = |z||f(z)| <= 2/sqrt(pi) ||f||_2.Now as z approaches 0, zf(z) tends to 0 since the limit of the product of two functions is the product of their limits. Therefore, lim z->0 zf(z) = 0 and hence, f(z) has a removable singularity at z = 0.