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Let $ ‎X=C[-1,1]‎$‎‎ be inner product space with definition $$‎\langle f,g‎‎‎\rangle =‎\int_{-1}^1 f‎‎ \overline{g}‎ ‎dt ‎‎.$$ Let $M$ be the subspace defined by ‎$$ ‎M= ‎‎\left\{f‎ \in ‎X\mid ‎f(t)=0 ,‎ ‎‎-1 \leq‎ t ‎‎\leq ‎0 \right\}. ‎$$

What is $M^{\perp}$?

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the set of g orthogonal to M, that is, $\{g\in X:\forall f\in M, \langle g,f\rangle=0\}$. –  Brady Trainor Mar 27 '13 at 8:04
    
bardy: I have another question, Is true equality $ M \bigoplus M^{\perp}=X$? –  nim Mar 27 '13 at 8:29
    
Sorry, I'm not completely fluent in these matters. I would have to review, but roughly, that is somewhat true. But there may be counterexamples. If so, I'm sure there are conditions when it is guaranteed. For instance, we would have to say what type of space $M$ is possibly. –  Brady Trainor Mar 27 '13 at 8:33
    
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Also, the definition of $M^\perp$ may depend on the context. –  Brady Trainor Mar 27 '13 at 8:41
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Let $N := \{\, g \in X \mid \text{$g(t) = 0$ for all $t \in (0, 1)$} \,\}$. I'll show that $M^\perp = N$.

From the definition of inner product, $N \subset M^\perp$ is obvious. So assume $g \in M^\perp$ and prove $g \in N$ by contradiction.

Suppose $g(\tau) \neq 0$ for some $\tau \in (0, 1)$. Without loss of generality, we may assume that $g(\tau) > 0$. Since $g$ is a continuous function, there exist a (small) $\delta > 0$ such that $g(t) > 0$ for all $\lvert t - \tau \rvert < \delta$. Then take a function $f \in M$ defined by

$$ f(t) = \begin{cases} 0 & \text{if $\lvert t - \tau \rvert \geq \delta$} \\ 1 + (t - \tau)/\delta & \text{if $\tau - \delta \leq t \leq \tau$} \\ 1 - (t - \tau)/\delta & \text{if $\tau \leq t \leq \tau + \delta $}. \end{cases}$$ (Its graph is something like triangle wave which has peek at $t = \tau$ and its amplitude is $1$.) Since $f \in M$ and $g \in M^\perp$, it should be $\langle f, g \rangle = 0$. However, from the choice of $\delta$ and $f$, it's also $\langle f, g \rangle > 0$. Contradiction.

%% Though I show in case $X$ is a set of continuous real-valued functions, the same argument would work in complex-valued case by separation of integral to real part and imaginary part.

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Taro: Is true equality $ M \bigoplus M^{\perp}=X$? thanks –  nim Mar 27 '13 at 11:34
    
@nim If the inner product space is finite, the similar statement is well-known and it's true. I'm unsure about what happens in general (i.e., infinite inner product space). I recommend you ask that question independently. (I feel it's false in this case.) –  Taro Mar 27 '13 at 11:56
    
If the inner product space is finite or Subspace of $ M $ be finite dimension, then, there is equality? which? thanks –  nim Mar 27 '13 at 12:13
    
@nim Let $W$ be a subspace of a finite inner product space $V$. Then $V = W \oplus W^\perp$ holds. –  Taro Mar 27 '13 at 12:27
    
I found a proof. Assume constant function $1 \in X$ can be uniquely decomposed into $1 = f + g$ where $f \in M$ and $g \in M^\perp$. Then $1 = f(0) + g(0) = g(0)$ and $g(\varepsilon) = 0$ for all $\varepsilon > 0$. Thus, $g$ becomes discontinuous. Contradiction. Hence $M \oplus M^\perp \neq X$. –  Taro Mar 28 '13 at 4:56
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I would conjecture that $M^\perp=\{g\in X:\forall t\in [0,1], g(t)=0\}$. Try to show equality of sets, and you may discover if this is indeed the case.

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