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Let V={f∈C1([−1,1])|f(1) =f(−1)}, then〈f, g〉=∫(f g+f'g′)dx (from -1 to 1) is an inner product on V...

Let V={f∈C1([−1,1])|f(1) =f(−1)}, then〈f, g〉=∫(f g+f'g′)dx (from -1 to 1) is an inner product on V (you do not have to verify this).

Under this inner product, prove that:

{x2}⊥={f∈V | ∫(x^2−2)f(x)dx= 0} (from -1 to 1)

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