What role does the Riesz theorem play in defining weak convergence in Hilbert spaces?

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 $\|l\|$.

A sequence $(x_n)$ in a Hilbert space $H$ is weakly convergent to $l$ if for every element $v$ in $H$, the inner product of $x_n$ with $v$ converges to the inner product of $l$ with $v$ as $n$ approaches infinity. This is denoted by $x_n \rightharpoonup l$.

Conditions: $H$ is a Hilbert space.; $(x_n)$ is a sequence in $H$.; $l$ is an element of $H$.; $v$ is an arbitrary element of $H$.

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 $\|l\|$.