How is weak convergence defined using inner products in a Hilbert space?

The Riesz representation theorem allows identifying elements of the dual space with elements of the Hilbert space $H$ itself via the inner product. This adapts the general definition of weak convergence (which uses dual space functions) to the inner product form used specifically in Hilbert spaces.

Conditions: $H$ is a Hilbert space.; The general definition of weak convergence involves continuous linear functionals from the dual space.

No, weak convergence alone does not imply strong convergence. A sequence can converge weakly to a limit without the norm of the difference converging to zero.

Conditions: $H$ is a Hilbert space.; $(x_n)$ is a sequence in $H$ converging weakly to $l$.; No assumption is made about the convergence of $\|x_n\|$ to $\|l\|$.

Strong convergence implies weak convergence because the Cauchy-Schwarz inequality bounds the difference of inner products by the product of the norm of the difference and the norm of the test vector. If the sequence converges strongly, the norm of the difference goes to zero, forcing the inner product difference to go to zero for any fixed test vector.

Conditions: $H$ is a Hilbert space.; $(x_n)$ is a sequence in $H$.; $(x_n)$ converges strongly to $l$ (i.e., $\|x_n - l\| \to 0$).; $v$ is an arbitrary element of $H$.

The Kakutani theorem states that every bounded sequence in a Hilbert space $H$ admits a subsequence that converges weakly in $H$. This provides a compactness principle for weak convergence in infinite-dimensional spaces.

Conditions: $H$ is a Hilbert space.; $(x_n)$ is a bounded sequence in $H$, meaning $\sup_n \|x_n\|_H < \infty$.

The extra assumption required is that the norms of the sequence converge to the norm of the limit, i.e., $\|x_n\| \to \|l\|$. When combined with weak convergence, this condition implies strong convergence.

Conditions: $H$ is a Hilbert space.; $(x_n)$ is a sequence in $H$ converging weakly to $l$.; $\|x_n\|$ converges to $\|l\|$.