Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $M^{\perp}$?

share|cite|improve this question
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
Also, the definition of $M^\perp$ may depend on the context. – Brady Trainor Mar 27 '13 at 8:41
up vote 0 down vote accepted

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.

share|cite|improve this answer
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.) – Orat 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. – Orat 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 $M \oplus M^\perp \neq X$. – Orat Mar 28 '13 at 4:56

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.