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Consider $C[1,1]$ equipped with the scalar product induced from $L^{2}[-1,1]$, i.e. $<f,g>$ = $\int_{-1}^{1} f(x)\overline{g(x)} dx$. Compute the orthogonal complement of the following subspaces of $C[-1,1]$:

$M_{1} = \{f \in C[-1,1] : f(x) = 0$ for $x\le 0$$\}$

$M_{2} = \{f \in C[-1,1] : f(0) = 0\}$

For the first one I see now (after Olof's comment) that the complement would be $\{g \in C[-1,1] : g(x) = 0$ for $x \gt 0\}$

For the second one I am really not sure how to proceed. Since the interval is symmetric my initial thought was to find a $g$ that turns an arbitrary $f$ $\in$ $M_{2}$ into an odd function but I don't think such a $g$ exists.

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    $\begingroup$ Isn't any function $g$ with $g(x) = 0$ for $x\geq 0$ in $M_1^\perp$? $\endgroup$
    – OgvRubin
    Jan 12, 2021 at 13:00
  • $\begingroup$ Oh yes of course thats silly of me. The second one is the one im having trouble with $\endgroup$
    – duelspace
    Jan 12, 2021 at 13:05
  • $\begingroup$ So the orthogonal complement of $M_{1}$ is just the functions $g$ with $g(x)=0$ for $x\gt0$ right? $\endgroup$
    – duelspace
    Jan 12, 2021 at 13:07
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    $\begingroup$ @duelspace It's not $x>0$, it's $x \ge 0$, as Olof R said in his comment. Yes, that's the orthogonal complement of $M_1$, but you'll have to prove it. Hint: If a continuous function $g$ is not $0$ at a point $x$, then there is a whole neighborhood of $x$ where it's not $0$. $\endgroup$
    – jjagmath
    Jan 12, 2021 at 13:18
  • $\begingroup$ For the second one, see here : math.stackexchange.com/questions/1046047/… $\endgroup$ Jan 12, 2021 at 13:39

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