I know $\displaystyle\int_0^1 f(x)g(x) \, dx $ is an inner product for $ C[0,1]$ but I can't find a conclusive answer either way for $\displaystyle\int_0^{1/2} f(x)g(x) \, d x$
Is $\int_0^{1/2} \! f(x)g(x) \, \mathrm{d} x$ an inner product on $C[0,1]$
2
$\begingroup$
linear-algebra
inner-product-space
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0I have enclosed your TeX in dollar signs $\$$ which make the output more readable. – 2012-12-06
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4Do you mean inner product on $C[0,1/2]$? If so, yes, it'll be the same argument as for $[0,1]$. If you mean on $C[0,1]$, think about this: can you find a function on [0,1] that is not identically 0 whose 'inner product' norm is 0? – 2012-12-06
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1.... but on another hand, it is a semi-inner product. – 2012-12-06