If $\{\boldsymbol{\varphi}_1, \boldsymbol{\varphi}_2, ... \}$ is an orthonormal system in $\mathcal{H}$, then the Bessel's inequality is:
$$\sum_{j=1}^{\infty} |\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle|^2 \leq \lVert \mathbf{x} \rVert^2 \quad \text{for every } \mathbf{x}\in \mathcal{H}.$$
My question is:
Any comments or answers are appreciated.
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A basis for a vector space is a set of vectors which you can construct unique linear combinations of any vector in that space:
$$\text{if } {\bf u} \in V = \{{\bf v_1,v_2,\cdots,v_n}\}$$ $$\text{then there exist unique } c_k \text{ so that } {\bf u} = \sum_{k=1}^{n} c_k{\bf v}_k$$
A basis can span a subspace of a larger vector space, if there exist vectors in the larger vector space which can not be represented as such a combination.
The inequality then happens if the basis is for a subspace which the vector we sum scalar products over does not lie in the subspace, but has components outside of it. Those components will then be kind of "invisible" to the basis.
It's like the Pythagorean theorem, but we might not have enough vectors in our orthonormal system to represent $x$ exactly. If $x \in \mathbb R^3$ and we have an orthonormal set of vectors $\{\phi_1,\phi_2,\phi_3\} \subset \mathbb R^3$, then we can express $x$ as a linear combination $x = \sum_{j=1}^3 \langle x,\phi_j\rangle \phi_j$. The Pythagorean theorem then tells us that $\|x\|^2=\sum_{j=1}^3 | \langle x,\phi_j\rangle |^2$. If we had only two vectors in our orthonormal set, we might not be able to represent $x$ exactly, and we would get an inequality rather than an equality.