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For each $n$, define the bounded linear functionalThe hypothesis is precisely that $x_n$ converges weakly to $x$. In particular, for each fixed $y\in H$, the scalar sequence $T_n(y)$ converges and is therefore bounded. Thus the family $\{T_n\}$ is pointwise bounded.

The Uniform boundedness principle givesBy the Riesz representation theorem, $\Vert T_n\Vert =\Vert x_n{\Vert }_H$. Henceso the sequence is bounded.

If an associative unital algebra has a complete norm and separately continuous multiplication, the Uniform boundedness principle applied to $L_x(y)=xy$, for $\Vert x\Vert \le 1$, yields $\Vert xy\Vert \le C\Vert x\Vert \Vert y\Vert$. The operator norm of $L_x$ then defines an equivalent algebra norm: $\Vert x\Vert /\Vert e\Vert \le \Vert L_x\Vert \le C\Vert x\Vert$, and $L_{xy}=L_x{L}_y$ proves submultiplicativity. It normalizes the identity to norm one.

For a real or complex matrix, the three displayed conditions are equivalent to being a regular matrix summability method. For necessity, first apply the Uniform boundedness principle to each row's finite partial sums on the convergent sequence space, obtaining absolute summability of that row. Apply the principle again to the full row functionals, obtaining a uniform bound. Coordinate vectors and the constant-one sequence give the remaining conditions. For sufficiency, write $x_j=L+(x_j-L)$ and split the error into a finite head, controlled by columnwise convergence, and a uniformly small tail, controlled by the absolute row bound.

For a countable minimal total family in a normed space in a Banach space, let $(x_n,f_n)$ be its unique biorthogonal system and let $P_{n}x={\sum }_{j\le n}{f}_j(x)x_j$. Uniform boundedness of these projections implies $P_{n}x\to x$ for every $x$: approximate $x$ by a finite linear combination $y$, eventually fixed by $P_n$, and bound the error by $(1+{\sup }_n\Vert P_n\Vert )\Vert x-y\Vert$. Conversely, if every vector has a norm-convergent expansion in this family, applying each continuous $f_j$ shows that its coefficients must be $f_j(x)$. Thus $P_{n}x\to x$, so the projections are pointwise bounded. The Uniform boundedness principle gives their uniform operator-norm bound. This proves the criterion and uniqueness of all expansions.

A subset of a normed space is norm bounded if it is bounded under every continuous linear functional. Apply the Uniform boundedness principle on the complete dual space to the evaluation family $f\mapsto f(x)$. Hahn-Banach theorem identifies the evaluation norm with $\Vert x\Vert$, even when the original normed space is not complete. As a consequence, two norms with exactly the same continuous dual as sets must be equivalent: apply this criterion to either norm's unit ball in the other norm.

For a weakly holomorphic Banach-valued function, choose a closed disk contained in its domain. Each continuous functional is bounded on the image of that disk by scalar continuity. The Uniform boundedness principle, applied to its evaluation functionals in the bidual, gives a common bound $M$ on the norms of the image vectors. Cauchy's formula on the disk's boundary then bounds the scalar difference between difference quotients at increments $h,k$ by a constant times $M\Vert \phi \Vert |h-k|$, uniformly for sufficiently small increments. Explicitly the difference kernel isThe Hahn-Banach theorem dual norm formula converts this into the same norm bound for the vector difference quotients. They are Cauchy, so completeness gives a norm derivative. This also explains why weak holomorphy suffices without a separately assumed norm-continuity hypothesis.

A subset $D$ of a normed vector space is weakly bounded when ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. It is norm bounded: its canonical embedding into the bidual is a pointwise bounded family of functionals on the complete continuous dual space, so the Uniform boundedness principle applies. The converse is immediate. A weakly compact set is weakly bounded because each scalar evaluation has compact image.

Every weakly compact set in a normed vector space is norm bounded, even if the space is incomplete. Its canonical embedding into the bidual is pointwise bounded on the continuous dual space; the Uniform boundedness principle on that Banach dual gives the norm bound.

Every weakly convergent sequence in a normed vector space is a bounded sequence. Apply the Uniform boundedness principle on the Banach space $X^{\ast }$ to the evaluation maps $J{x}_n:f\mapsto f(x_n)$. They are pointwise bounded, while the canonical embedding into the bidual gives $\Vert J{x}_n\Vert =\Vert x_n\Vert$.

Convergence in the displayed sense combines strong function convergence with weak-star convergence of derivative vector measures against $C_0$ tests. Strong $L^1$ convergence and a uniform variation bound imply the derivative convergence by integration by parts and uniform test-function approximation. Conversely the Uniform boundedness principle bounds the derivative measures of a convergent sequence.