1
$\begingroup$

Question: Suppose $C[−1, 1]$ is the vector space of continuous real-valued functions on the interval $[−1, 1]$ with inner product given by $\langle f, g\rangle = \int_{a}^b f(x)g(x)dx$

Let $U = {f ∈ C[−1, 1] : f(0) = 0}$ be the subspace of $C[−1, 1]$. Which of the following statement(s) is(are) correct?Justify your answer.

(a) $C[−1, 1] = U ⊕ U^\bot$

(b) $U^\bot = \{0\}$

(c) $U^\bot$ is a proper and non-trivial subspace of $C[−1, 1]$

Difficulty: I am sure that option (b) is correct but not able to write a proof of it. Also about option (a) is valid for finite-dimensional subspace but it is not am I correct about it.

$\endgroup$
1
  • $\begingroup$ Welcome to MSE. In order to get $\{x\}$, you should type \{x\}. $\endgroup$ Sep 29 '20 at 7:38
0
$\begingroup$

Hint: If $g \in U^{\perp}$ then $\int fg=0$ for all $f \in U$. Let $f_n(x)=g(x)$ for $|x| >\frac1 n$, $f_n(0)=0$ and $f_n$ have a straight line graph in $[-\frac 1 n ,0]$ as well as $[0, \frac 1 n]$. Then $\int g f_n=0$ and letting $n \to \infty$ yields $\int g^{2}=0$. Hence $g=0$. This proved b).

It is obvious from (b) that (a) is false. The constant function $1$ is in LHS but not in RHS. (c) is also answered by (b).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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