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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 $H$.

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 $H$.