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SHANNON SAMPLING SERIES WITH AVERAGED KERNELS1 1 ...

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6 A.KIVUNUKK, G.TAMBERG<br />

Proof: Let f ∈ T V (R), hence f is bounded on R. Since sm ∈ B 1 π, the derivative s ′ m ∈ B 1 π also, and by<br />

Nikolskii inequality <br />

k∈Z |s′ m(u − k)| < ∞. Therefore, there exists the bounded derivative<br />

(SW,mf) ′ (t) = W<br />

∞<br />

k=−∞<br />

f( k<br />

W )s′ m(W t − k),<br />

which implies SW,mf ∈ AC(R). Now we have for the total variation<br />

Since by (15) the derivative<br />

we get<br />

(SW,mf) ′ (t) = W<br />

m<br />

∞<br />

k=−∞<br />

k=−∞<br />

VR[SW,mf] =<br />

s ′ m(t) = 1<br />

<br />

m<br />

f( k<br />

W )<br />

<br />

∞<br />

−∞<br />

s(t + m<br />

2<br />

s(W t − k + m<br />

2<br />

|(SW,mf) ′ (t)|dt. (20)<br />

m<br />

) − s(t −<br />

2 )<br />

<br />

,<br />

m<br />

) − s(W t − k −<br />

2 )<br />

<br />

= W<br />

<br />

∞<br />

f(<br />

m<br />

k=−∞<br />

k<br />

∞ m<br />

)s(W t − k + ) −<br />

W 2<br />

k=−∞<br />

= W<br />

∞<br />

<br />

f(<br />

m<br />

k − m<br />

) − f(k<br />

W W )<br />

<br />

s(W t − k + m<br />

2 )<br />

= W<br />

m−1 <br />

m<br />

∞<br />

j=0 k=−∞<br />

<br />

f(<br />

k − j<br />

W<br />

Now by (20) and Theorem 4 we obtain for the total variation<br />

VR[SW,mf] 1<br />

m−1 <br />

m<br />

j=0 k=−∞<br />

∞<br />

<br />

<br />

<br />

− j − j − 1<br />

f(k ) − f(k<br />

W W<br />

)<br />

<br />

<br />

<br />

<br />

f( k<br />

m<br />

)s(W t − k −<br />

W 2 )<br />

<br />

− j − 1<br />

) − f(k<br />

W<br />

)<br />

<br />

s(W t − k + m<br />

2 ).<br />

∞<br />

−∞<br />

<br />

<br />

s(W t − k + m<br />

2 )<br />

<br />

<br />

d(W t)<br />

1<br />

m−1 <br />

VR[f]s1 = s1VR[f] SW VR[f]. (21)<br />

m<br />

j=0<br />

Under the assumptions of Theorem 5 we have s1 = SW,1, therefore<br />

VR[SW,1f] SW,1VR[f]<br />

4. Examples of operators with variation detracting property<br />

4.1 Hann’s sampling operators<br />

We can provide more examples of operators with variation detracting property (9).<br />

2 πu<br />

Take the Hann window λH(u) := cos . Then by (3)<br />

sH(t) = 1<br />

<br />

<br />

sinc(t − 1) + 2 sinc t + sinc(t + 1)<br />

4<br />

and its corresponding averaged kernel by (15) is<br />

sH,1(t) = 1<br />

<br />

4<br />

2<br />

Sci(t + 1<br />

2<br />

) − Sci(t − 1<br />

2<br />

= 1<br />

<br />

<br />

r0(t − 1/2) + r0(t + 1/2) =<br />

2<br />

sinc t<br />

3<br />

3<br />

) + Sci(t + ) − Sci(t −<br />

2 2 )<br />

<br />

,<br />

2(1 − t 2 )<br />

(22)

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