Orthogonal complement of a subspace of $C[-1,1]$

Consider $$C[1,1]$$ equipped with the scalar product induced from $$L^{2}[-1,1]$$, i.e. $$$$ = $$\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.

• Isn't any function $g$ with $g(x) = 0$ for $x\geq 0$ in $M_1^\perp$? Jan 12, 2021 at 13:00
• Oh yes of course thats silly of me. The second one is the one im having trouble with Jan 12, 2021 at 13:05
• So the orthogonal complement of $M_{1}$ is just the functions $g$ with $g(x)=0$ for $x\gt0$ right? Jan 12, 2021 at 13:07
• @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 $0$. Jan 12, 2021 at 13:18
• For the second one, see here : math.stackexchange.com/questions/1046047/… Jan 12, 2021 at 13:39