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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 $\|l\|$.

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 $H$.

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 $H$.