Let $L$ be a set of all bounded sequences of $\mathbb{R}$. Then it is easy to show that $L$ is a vector space with respect to normal addition and scalar multiplication. Define a function on $L \times L$ given by $$\langle(a_i) , (b_i)\rangle = \sum _{i = 1}^{\infty} {\frac{a_i b_i}{i^2}}.$$

Verifying that the above function is an inner product (scalar product) is a routine calculation.

The orthogonal complement of a subset $U$ of a vector space $L$ is $\,$ $U^* = \{\, A\in L: \, \langle A,B\rangle = 0 \, \text{and}\, \,\text{ for any} \,B\in U \,\}$.

If vector space is finite dimensional and $\,U$ is subspace then we have many nice properties like $U^{**} = U \,$ and any vector in vector space can be written uniquely as sum of vectors $U$ and $U^*$ and many more. Here we observe that if $U$ is a subspace then $U^*$ cannot be $\{ 0\}$ because the orthogonal complement of $\{0\}$ is vector space itself.

Here obviously $L$ is not a finite dimensional vector space. Is there any proper non zero subspace of $L$ whose orthogonal complement is $\{0\}$ ?


1 Answer 1


Let $M:=\{(a_n):\exists m,\ n>m\Rightarrow a_n=0\}\subset L$ be the subspace of finite sequences. Then $M^\perp=\{0\}$.

Proof: Suppose $(b_n)\in M^\perp$ and consider for $m\in\mathbb{N}$, $$(a^m_n):=(1,2^2,\ldots,n^2,\ldots,m^2,0,\ldots)\in M$$

Then $$\forall m,\quad0=\langle (a^m_n),(b_n)\rangle=b_1+b_2+\cdots+b_m$$ implying $(b_n)=0$.

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    $\begingroup$ Can you please clarify how $b_1 + b_2 + \cdots b_m = 0$ implies each $b_n = 0$. This will be true only if all $b_i$ s are positive real number. It is not necessary that they are positive. $\endgroup$ Oct 9, 2020 at 17:01
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    $\begingroup$ It is true for $m=1$ so $b_1=0$, and for $m=2$, etc. $\endgroup$ Oct 9, 2020 at 17:02

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