What would you like to understand?

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 $\|l\|$.

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 $\|l\|$.

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 $\|l\|$.

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 $\|l\|$.

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 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 Cauchy-Schwarz inequality is used to bound the absolute difference of the inner products $|\langle x_n, v \rangle - \langle l, v \rangle|$ by the product of the norm of the difference $\|x_n - l\|$ and the norm of the test vector $\|v\|$. This step converts the problem of showing inner product convergence into showing that a product involving the norm goes to zero.

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

The Cauchy-Schwarz inequality is used to bound the absolute difference of the inner products $|\langle x_n, v \rangle - \langle l, v \rangle|$ by the product of the norm of the difference $\|x_n - l\|$ and the norm of the test vector $\|v\|$. This step converts the problem of showing inner product convergence into showing that a product involving the norm goes to zero.

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