The Stiefel gradient procedure returns a feasible iterate within its finite budget

For a current $X\in \mathrm{St}(n,p)$, put $M=X^{\mathsf{T}}G$. From [F1] and $X^{\mathsf{T}}X=I_p$, the projected gradient satisfies $X^{\mathsf{T}}g+g^{\mathsf{T}}X=M+M^{\mathsf{T}}-2\mathrm{sym}(M)=0$. Thus $g\in T_X\mathrm{St}(n,p)$ by [F2], and multiplication by $-{\alpha }_k$ keeps $\eta$ tangent.

Put $S=\mathrm{sym}(X^{\mathsf{T}}G)$, so $S^{\mathsf{T}}=S$. If $Z\in T_X\mathrm{St}(n,p)$, then $A=X^{\mathsf{T}}Z$ is skew-symmetric by [F2]. Hence $⟨XS,Z{⟩}_F=\mathrm{tr}(SA)=0$: transposing and cyclically reordering gives $\mathrm{tr}(SA)=\mathrm{tr}(A^{\mathsf{T}}{S}^{\mathsf{T}})=-\mathrm{tr}(AS)=-\mathrm{tr}(SA)$. The derivative of the restriction on a tangent $Z$ is therefore $D\mathrm{\Phi }(X)[Z]=⟨G,Z{⟩}_F=⟨g,Z{⟩}_F$. Together with step1.1 this makes $g$ its tangent metric-dual vector. It is unique, since two such tangent vectors differ by a tangent vector whose pairing with itself is zero. This uses only the supplied first derivative, so $C^1$ suffices.

The initial frame is in $\mathrm{St}(n,p)$ by the input condition. Whenever a current frame is in $\mathrm{St}(n,p)$, step1.1 makes the trial tangent. Then $(X+\eta {)}^{\mathsf{T}}(X+\eta )=I_p+{\eta }^{\mathsf{T}}\eta$, so for $v\mathrm{\ne }0$ its quadratic form is $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\eta v{\mathrm{\parallel }}_2^2>0$. Thus $X+\eta$ has full column rank, the polar call is in the declared domain of [F3], and [F3] puts its candidate in $\mathrm{St}(n,p)\subset U$. In particular every subsequent objective, residual and oracle evaluation has an input in its supplied domain. An accepted candidate is the next current frame; a rejected candidate returns the previous frame. Hence every current frame and every possible returned frame remains in $\mathrm{St}(n,p)$ by finite induction.

The index begins at zero, increases by one after each accepted nonterminal trial, and the loop ends when $k=K$; every other branch returns immediately. Therefore at most $K$ trial evaluations occur. The procedure accepts only candidates with objective no greater than the current value, and its residual test labels an output target-met exactly when its residual is at most $\tau$; every other return branch follows a failed target test and has residual greater than $\tau$.