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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 $\displaystyle\int_0^{1/2} f(x)g(x) \, d x$

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I have enclosed your TeX in dollar signs $\$$ which make the output more readable. –  JavaMan Dec 6 '12 at 20:39
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Do 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? –  snarski Dec 6 '12 at 20:46
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.... but on another hand, it is a semi-inner product. –  N. S. Dec 6 '12 at 20:52
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1 Answer

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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You want $x-1/2$, otherwise $f \notin C[0,1]$. –  JSchlather Dec 6 '12 at 20:49
    
ty, I've corrected it. –  user127.0.0.1 Dec 6 '12 at 20:51
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