wirtinger's inequality proof complex valued function
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The standard Wirtinger's inequality requires that ∫bafdx=0, which implies that c0=0, and hence the difficulty you are encounter does not exist.
If instead we assume that f(a)=f(b)=0, then we exploit this by expanding f in a sine series, i.e., for a=0 and b=π,
f(x)=∑n=1∞ansinnx,
and hence
∫π0f2=π2∑n=1∞a2n,
while
∫π0(f′)2=π2∑n=1∞n2a2n,
and hence
∫π0(f′)2≥∫π0f2,
with best constant c=1.
For arbitrary b>a,
f(x)=∑n=1∞ansin(nπ(x−a)b−a),
and hence
∫baf2=b−a2∑n=1∞a2n,
while
∫ba(f′)2=b−a2∑n=1∞(πb−a)2n2a2n,
and hence
∫ba(f′)2≥π2(b−a)2∫π0f2.
If instead we assume that f(a)=f(b)=0, then we exploit this by expanding f in a sine series, i.e., for a=0 and b=π,
f(x)=∑n=1∞ansinnx,
and hence
∫π0f2=π2∑n=1∞a2n,
while
∫π0(f′)2=π2∑n=1∞n2a2n,
and hence
∫π0(f′)2≥∫π0f2,
with best constant c=1.
For arbitrary b>a,
f(x)=∑n=1∞ansin(nπ(x−a)b−a),
and hence
∫baf2=b−a2∑n=1∞a2n,
while
∫ba(f′)2=b−a2∑n=1∞(πb−a)2n2a2n,
and hence
∫ba(f′)2≥π2(b−a)2∫π0f2.
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