# A subspace whose orthogonal complement is {0}

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 $$\{0\}$$ ?

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 $$(b_n)=0$$.
• 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. Oct 9, 2020 at 17:01
• It is true for $m=1$ so $b_1=0$, and for $m=2$, etc. Oct 9, 2020 at 17:02