# Is $\int_0^{1/2} \! f(x)g(x) \, \mathrm{d} x$ an inner product on $C[0,1]$

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 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 .... but on another hand, it is a semi-inner product. – N. S. Dec 6 '12 at 20:52 ## 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$f\neq 0$- 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