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 $C[0,1]$
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$\begingroup$
linear-algebra
inner-product-space
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1.... but o$n$ another hand, it is a semi-inner product. – 2012-12-06
1 Answers
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No, take $f(x)=\begin{cases}0 & \text{ if } 0\le x\le\frac{1}{2}\\x-\frac{1}{2}&\text{ if }\frac{1}{2}\lt x\le1\end{cases} $
Then $f\cdot f = 0$ but $f\neq 0$
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0ty, I've corrected it. – 2012-12-06