2
$\begingroup$

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

  • 0
    I have enclosed your TeX in dollar signs $\$$ which make the output more readable.2012-12-06
  • 4
    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?2012-12-06
  • 1
    .... but on another hand, it is a semi-inner product.2012-12-06

1 Answers 1

2

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 $f\neq 0$

  • 1
    You want $x-1/2$, otherwise $f \notin C[0,1]$.2012-12-06
  • 0
    ty, I've corrected it.2012-12-06