Proof of Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound

Throughout, subscripts on inner products and norms are omitted when the space is clear from the arguments, $0$ also denotes the zero vector of each vector space, and $x-y=x+(-y)$, where $-y=(-1)y$ by claim 5 of Elementary Identities in a Vector Space. Elementary real order and arithmetic is used as licensed by The Real Numbers: Standing Notation and Background.

Preliminaries. (P1) Complex arithmetic. Let $z,w\in\mathbb{C}$ and $a\in\mathbb{R}$, and write $z=x+yi$ and $w=p+qi$ with real $x,y,p,q$, as allowed by claim 3 of Canonical Form and Arithmetic of Complex Numbers. The rules of claim 4 of that lemma give $z+w=(x+p)+(y+q)i$, $az=(ax)+(ay)i$ (writing $a=a+0i$), $iz=(-y)+xi$ and $(-i)z=y+(-x)i$, and $\overline{z}=x+(-y)i$ by Complex Conjugate. By the uniqueness in claim 3 of Canonical Form and Arithmetic of Complex Numbers and by Real and Imaginary Parts of a Complex Number therefore

moreover $\overline{i}=-i$ (as $i=0+1i$), so $\overline{i}\,i=-(ii)=1$ by condition 2 of The Complex Numbers. From Properties of Complex Conjugation and Modulus we use: conjugation respects sums and products, is involutive, and fixes exactly the real numbers, so $\overline{a}=a$ and $\overline{-1}=-1$ (claim 1); $z+\overline{z}=2\operatorname{Re}z$ (claim 2); $|z|=0$ only if $z=0$ (claim 3); $|zw|=|z|\,|w|$ (claim 4); $\operatorname{Re}z\le|z|$ and $-\operatorname{Re}z\le|z|$, so the absolute value of the real number $\operatorname{Re}z$ is at most $|z|$ (claim 6); and $|a|=a$ when $0\le a$, while $|a|$ is $a$ or $-a$ for every real $a$ (claim 8).

(P2) Norms. Let $X$ be one of $V,W,U$ and $x,y\in X$. The induced norm is a norm by claim 2 of The Induced Norm is a Norm, and Induces a Metric, so by conditions 1-3 of Norm on a Complex Vector Space we have $0\le\lVert x\rVert$ with equality only if $x=0$, $\lVert\lambda x\rVert=|\lambda|\,\lVert x\rVert$ for $\lambda\in\mathbb{C}$, and $\lVert x+y\rVert\le\lVert x\rVert+\lVert y\rVert$; in particular $\lVert0\rVert=\lVert0\cdot0\rVert=0$. By Norm Induced by a Complex Inner Product, $\langle x,x\rangle=\lVert x\rVert^{2}$, and by claim 1 of The Induced Norm is a Norm, and Induces a Metric, $|\langle x,y\rangle|\le\lVert x\rVert\,\lVert y\rVert$. The metric of $X$ is $d(x,y)=\lVert x-y\rVert$. Every linear map $T$ satisfies $T0=T(0\cdot0)=0\cdot T0=0$ by claim 3 of Elementary Identities in a Vector Space and the homogeneity of $T$ (Linear Map).

(P3) Separation. If $x,y\in X$ satisfy $\langle x,v\rangle=\langle y,v\rangle$ for every $v\in X$, then $x=y$. Indeed, by claims 1 and 2 of Elementary Properties of a Complex Inner Product and $\overline{-1}=-1$ we get $\langle x-y,v\rangle=\langle x,v\rangle-\langle y,v\rangle=0$ for every $v$; taking $v=x-y$, claim 4 of that lemma gives $x-y=0$, hence $x=y$.

Claim 1 (Least bound). Let $T\in\mathcal{L}(V,W)$ and let $B_{T}$ be its set of bounds, so that $\lVert T\rVert_{\mathrm{op}}=\inf B_{T}$ by Bounded Linear Maps between Complex Normed Spaces and the Operator Norm §operator-norm, a greatest lower bound in the sense of Lower Bound and Greatest Lower Bound in a Totally Ordered Set. Every bound is nonnegative by Bounded Linear Maps between Complex Normed Spaces and the Operator Norm §bounded, so $0$ is a lower bound of $B_{T}$ and hence $0\le\lVert T\rVert_{\mathrm{op}}$. Let $v\in V$. If $v=0$, then $\lVert Tv\rVert=0=\lVert T\rVert_{\mathrm{op}}\lVert v\rVert$ by (P2). If $v\neq0$, then $0<\lVert v\rVert$ by (P2), and each $C\in B_{T}$ satisfies $\lVert Tv\rVert\le C\lVert v\rVert$, that is $\lVert Tv\rVert/\lVert v\rVert\le C$; so $\lVert Tv\rVert/\lVert v\rVert$ is a lower bound of $B_{T}$, hence at most the greatest lower bound $\lVert T\rVert_{\mathrm{op}}$, and multiplying by $\lVert v\rVert>0$ gives $\lVert Tv\rVert\le\lVert T\rVert_{\mathrm{op}}\lVert v\rVert$. Thus $\lVert T\rVert_{\mathrm{op}}$ is a bound for $T$. Every bound $C$ lies in $B_{T}$, so $\lVert T\rVert_{\mathrm{op}}\le C$ because $\lVert T\rVert_{\mathrm{op}}$ is a lower bound of $B_{T}$.

Now let $T:V\to V$ be linear, that is, a linear operator on $V$, and give $V$ the norm $\lVert\cdot\rVert_{V}$ of (P2). A real $C\ge0$ is a bound for $T$ in the sense of Bounded Linear Maps between Complex Normed Spaces and the Operator Norm §bounded exactly when $\lVert Tu\rVert\le C\lVert u\rVert$ for every $u\in V$, which is exactly the condition for $C$ to be a bound for $T$ in the sense of Bound for a Linear Operator and Bounded Linear Operator. Hence $T$ has a bound in one sense if and only if it has one in the other, i.e. $T\in\mathcal{L}(V)$ if and only if $T$ is a bounded linear operator. In that case the first paragraph shows that $\lVert T\rVert_{\mathrm{op}}$ is a bound for $T$ that is at most every bound for $T$, so it is an operator norm of $T$ in the sense of Operator Norm; since $T$ has exactly one operator norm by Existence and Uniqueness of the Operator Norm, it is $\lVert T\rVert_{\mathrm{op}}$.

Claim 2 (Operations). First, for any linear maps $S,T:V\to W$, $R:W\to U$ and $c\in\mathbb{C}$, the maps $S+T$, $cT$ and $RT$ are linear: for $u,u'\in V$ and $\lambda\in\mathbb{C}$, linearity of $S,T,R$ (Linear Map) and conditions 1, 2, 5 and 7 of Vector Space over a Field in $W$ and $U$ give $(S+T)(u+u')=(S+T)u+(S+T)u'$, $(S+T)(\lambda u)=\lambda Su+\lambda Tu=\lambda(S+T)u$, $(cT)(u+u')=cTu+cTu'$, $(cT)(\lambda u)=(c\lambda)Tu=(\lambda c)Tu=\lambda(cTu)$, $R(T(u+u'))=R(Tu)+R(Tu')$ and $R(T(\lambda u))=\lambda R(Tu)$.

Now let $S,T\in\mathcal{L}(V,W)$, $R\in\mathcal{L}(W,U)$, $c\in\mathbb{C}$ and $v\in V$. By (P2) and claim 1,

where the middle step of the third chain multiplies the bound for $Tv$ by $\lVert R\rVert_{\mathrm{op}}\ge0$. The multipliers on the right are nonnegative real numbers, so they are bounds; hence $S+T,cT\in\mathcal{L}(V,W)$, $RT\in\mathcal{L}(V,U)$ and $I_{V}\in\mathcal{L}(V)$ ($I_{V}$ is linear), and the second part of claim 1 gives $\lVert S+T\rVert_{\mathrm{op}}\le\lVert S\rVert_{\mathrm{op}}+\lVert T\rVert_{\mathrm{op}}$, $\lVert cT\rVert_{\mathrm{op}}\le|c|\,\lVert T\rVert_{\mathrm{op}}$, $\lVert RT\rVert_{\mathrm{op}}\le\lVert R\rVert_{\mathrm{op}}\lVert T\rVert_{\mathrm{op}}$ and $\lVert I_{V}\rVert_{\mathrm{op}}\le1$. For the reverse inequality $|c|\,\lVert T\rVert_{\mathrm{op}}\le\lVert cT\rVert_{\mathrm{op}}$: if $c=0$, then $|c|=0$ by claim 8 of Properties of Complex Conjugation and Modulus and $0\le\lVert cT\rVert_{\mathrm{op}}$ by claim 1. If $c\neq0$, then $0<|c|$ by (P1), $c^{-1}(cT)=T$ by conditions 5 and 6 of Vector Space over a Field in $W$, and $|c^{-1}|\,|c|=|1|=1$ by (P1); the inequality just proved, applied to $c^{-1}$ and $cT$, gives $\lVert T\rVert_{\mathrm{op}}\le|c^{-1}|\,\lVert cT\rVert_{\mathrm{op}}$, and multiplying by $|c|$ gives the claim.

The zero map $0:V\to W$ is linear and, since $\lVert0\rVert=0$ by (P2), has the bound $0$; so $0\in\mathcal{L}(V,W)$. Two maps $V\to W$ are equal when they agree at every $v\in V$. Evaluating at each $v$, conditions 1, 2, 5, 6, 7 and 8 of Vector Space over a Field for $\mathcal{L}(V,W)$ with the operations above follow from the same conditions in $W$; condition 3 holds with the zero map, as $Tv+0=Tv$; and condition 4 holds with $(-1)T\in\mathcal{L}(V,W)$, as $Tv+(-1)Tv=Tv+(-Tv)=0$ by claim 5 of Elementary Identities in a Vector Space. So $\mathcal{L}(V,W)$ is a complex vector space with zero vector the zero map. Finally, for $R,R'\in\mathcal{L}(W,U)$, $S,T\in\mathcal{L}(V,W)$, $c\in\mathbb{C}$ and $v\in V$, linearity of $R$ gives $R(Sv+Tv)=R(Sv)+R(Tv)$ and $R(cTv)=c\,R(Tv)$, while $(R+R')(Tv)=R(Tv)+R'(Tv)$ and $(cR)(Tv)=c\,R(Tv)$ by definition; hence $R(S+T)=RS+RT$, $R(cT)=c(RT)$, $(R+R')T=RT+R'T$ and $(cR)T=c(RT)$.

Claim 3 (Underlying real structure). Since $\mathbb{R}\subseteq\mathbb{C}$ with the same sums and products (condition 1 of The Complex Numbers) and the same identity $1$ (claim 1 of Canonical Form and Arithmetic of Complex Numbers), conditions 1-8 of Vector Space over a Field for $X$ with real scalars are instances of those with complex scalars, so $X$ is a vector space over $\mathbb{R}$. Let $u,u',v,w\in X$, $c\in\mathbb{C}$ and $a\in\mathbb{R}$. By the third hypothesis on $h$, the first two, and (P1),

By (P1), $\beta(v,u)=\operatorname{Re}\overline{h(u,v)}=\operatorname{Re}h(u,v)=\beta(u,v)$, so $\beta$ is symmetric; $\beta(u,v+w)=\operatorname{Re}\bigl(h(u,v)+h(u,w)\bigr)=\beta(u,v)+\beta(u,w)$ and $\beta(u,av)=\operatorname{Re}\bigl(a\,h(u,v)\bigr)=a\,\beta(u,v)$, so $v\mapsto\beta(u,v)$ is linear over $\mathbb{R}$ in the sense of Linear Map, and by symmetry so is $u\mapsto\beta(u,v)$; and since $h(v,v)$ is real, $\beta(v,v)=\operatorname{Re}h(v,v)=h(v,v)\ge0$. Thus $\beta$ is symmetric, bilinear and positive semidefinite as in Cauchy Sequences for a Positive Semidefinite Symmetric Bilinear Form: Cauchy-Schwarz, the Space of Cauchy Sequences, Convergence of Pairings, and Null Sequences. By $(\ast)$, the second hypothesis and (P1),

so $\beta(u,iv)=-\beta(iu,v)$, and $h(u,v)=\operatorname{Re}h(u,v)+(\operatorname{Im}h(u,v))\,i=\beta(u,v)-i\,\beta(u,iv)$.

Now let $X=V$ and $h=\langle\cdot,\cdot\rangle_{V}$; the hypotheses on $h$ are conditions 2, 3, 1 and 4 of Complex Inner Product Space. Conditions (a), (b), (c) of Real Inner Product Space §inner-product are the symmetry of $\beta$ and the linearity of $u\mapsto\beta(u,v)$ proved above, and (d) holds because $\beta(x,x)=\langle x,x\rangle_{V}\ge0$, which vanishes only for $x=0$ by condition 4 of Complex Inner Product Space. So $\beta=\langle\cdot,\cdot\rangle_{V,\mathbb{R}}$ is an inner product. Its norm $|x|$ is the nonnegative real number with $|x|^{2}=\langle x,x\rangle_{V,\mathbb{R}}=\langle x,x\rangle_{V}$, and $\lVert x\rVert_{V}$ is a nonnegative real number with the same square by Norm Induced by a Complex Inner Product, so $|x|=\lVert x\rVert_{V}$ by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root. The zero vector and additive inverses are determined by the addition alone (claims 1 and 2 of Elementary Identities in a Vector Space), so $x-y$ is the same vector in both structures, and the distance $|x-y|$ equals $\lVert x-y\rVert_{V}$, the metric of Complex Hilbert Space. Being Cauchy (Cauchy Sequence in a Metric Space) or convergent (Convergent Sequence in a Metric Space) refers only to the metric, so the two metric spaces are complete together (Complete Metric Space); hence if $V$ is a complex Hilbert space, $V$ with $\langle\cdot,\cdot\rangle_{V,\mathbb{R}}$ is a real Hilbert space.

Claim 4 (Uniqueness of adjoints). Let $S,S':W\to V$ be adjoints of $T$. For $w\in W$, Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint gives $\langle Sw,v\rangle=\langle w,Tv\rangle=\langle S'w,v\rangle$ for every $v\in V$, so $Sw=S'w$ by (P3). Hence $S=S'$.

Claim 5 (Existence of adjoints). Let $V$ be a complex Hilbert space and $T\in\mathcal{L}(V,W)$. By claim 3, $V$ with $\langle\cdot,\cdot\rangle_{V,\mathbb{R}}=\operatorname{Re}\langle\cdot,\cdot\rangle_{V}$ is a real Hilbert space with norm $\lVert\cdot\rVert_{V}$. Fix $w\in W$ and put $\ell_{w}(v)=\operatorname{Re}\langle w,Tv\rangle_{W}$ for $v\in V$. For $v,v'\in V$ and $a\in\mathbb{R}$, linearity of $T$, conditions 2 and 3 of Complex Inner Product Space and (P1) give $\ell_{w}(v+v')=\ell_{w}(v)+\ell_{w}(v')$ and $\ell_{w}(av)=\operatorname{Re}\bigl(a\langle w,Tv\rangle\bigr)=a\,\ell_{w}(v)$; and by (P1), (P2) and claim 1,

So $\ell_{w}$ is a bounded linear functional on this real Hilbert space, and by The Riesz Representation Theorem for a Real Hilbert Space §existence we may choose $z_{w}\in V$ with $\ell_{w}(v)=\langle v,z_{w}\rangle_{V,\mathbb{R}}$ for every $v\in V$; by the symmetry in claim 3, $\ell_{w}(v)=\operatorname{Re}\langle z_{w},v\rangle_{V}$. Applying the identity $h(u,v)=\beta(u,v)-i\,\beta(u,iv)$ of claim 3 first to $\langle\cdot,\cdot\rangle_{V}$ and then to $\langle\cdot,\cdot\rangle_{W}$, and using $T(iv)=i\,Tv$,

Define $T^{*}:W\to V$ by $T^{*}w=z_{w}$ (one choice for each $w$). For $w,w'\in W$, $a\in\mathbb{C}$ and $v\in V$, claims 1 and 2 of Elementary Properties of a Complex Inner Product give

so $T^{*}(w+w')=T^{*}w+T^{*}w'$ and $T^{*}(aw)=a\,T^{*}w$ by (P3). Thus $T^{*}$ is linear and, by the displayed identity, an adjoint of $T$ in the sense of Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint; by claim 4 it is the adjoint.

Bounds. Let $w\in W$. The real number $\lVert T^{*}w\rVert^{2}=\langle T^{*}w,T^{*}w\rangle_{V}=\langle w,T(T^{*}w)\rangle_{W}$ is nonnegative, so it equals its modulus by (P1), and by (P2) and claim 1

If $T^{*}w=0$, then $\lVert T^{*}w\rVert=0\le\lVert T\rVert_{\mathrm{op}}\lVert w\rVert$; otherwise $0<\lVert T^{*}w\rVert$ by (P2), and dividing gives $\lVert T^{*}w\rVert\le\lVert T\rVert_{\mathrm{op}}\lVert w\rVert$. Hence $\lVert T\rVert_{\mathrm{op}}\ge0$ is a bound for $T^{*}$, so $T^{*}\in\mathcal{L}(W,V)$ and $\lVert T^{*}\rVert_{\mathrm{op}}\le\lVert T\rVert_{\mathrm{op}}$ by claim 1. Conversely, for $v\in V$ the adjoint identity with $w=Tv$ gives $\lVert Tv\rVert^{2}=\langle Tv,Tv\rangle_{W}=\langle T^{*}(Tv),v\rangle_{V}$, and in the same way $\lVert Tv\rVert^{2}\le\lVert T^{*}(Tv)\rVert\,\lVert v\rVert\le\lVert T^{*}\rVert_{\mathrm{op}}\lVert Tv\rVert\,\lVert v\rVert$, whence $\lVert Tv\rVert\le\lVert T^{*}\rVert_{\mathrm{op}}\lVert v\rVert$ (trivially if $Tv=0$). By claim 1, $\lVert T\rVert_{\mathrm{op}}\le\lVert T^{*}\rVert_{\mathrm{op}}$, so the two norms are equal.

Claim 6 (Calculus of adjoints). All maps below are linear by the first paragraph of claim 2, and each identity is checked against Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint; by claim 4 the adjoint found is the adjoint. Let $v\in V$, $w\in W$ and $x\in U$. By condition 1 of Complex Inner Product Space, $\langle Tv,w\rangle=\overline{\langle w,Tv\rangle}=\overline{\langle T^{*}w,v\rangle}=\langle v,T^{*}w\rangle$, so $T$ is the adjoint of $T^{*}:W\to V$. By claims 1 and 2 of Elementary Properties of a Complex Inner Product, conditions 2 and 3 of Complex Inner Product Space and $\overline{\overline{c}}=c$,

so $S+T$, $cT$, $RT$ and $I_{V}$ have the adjoints $S^{*}+T^{*}$, $\overline{c}\,T^{*}$, $T^{*}R^{*}$ and $I_{V}$. For a linear map $A:V\to V$, which is a linear operator, the condition of Self-Adjoint Operator ($\langle Au,v\rangle=\langle u,Av\rangle$ for all $u,v\in V$) is literally the condition of Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint for $A$ to be an adjoint of $A$ (with $W=V$), so the two are equivalent. Next, by condition 1 of Complex Inner Product Space, the adjoint identity and (P1),

the last number being real. Finally, the computation for $RT$ uses only that the three spaces are complex inner product spaces; applied to $T:V\to W$ and to $T^{*}:W\to V$ (in the roles of $T$ and $R$, with $V$ in the role of $U$; the adjoint of $T^{*}$ is $T$), it shows that $T^{*}T$ has the adjoint $T^{*}T$, so $T^{*}T$ is self-adjoint by the equivalence just proved; and $\langle v,T^{*}Tv\rangle=\lVert Tv\rVert^{2}$ is a nonnegative real number for every $v$, so $T^{*}T$ is positive semi-definite.

Claim 7 (Completeness). Let $W$ be a complex Hilbert space and $(T_{k})$ as stated. By claim 2 each $T_{k}-T_{l}$ lies in $\mathcal{L}(V,W)$, and $(T_{k}-T_{l})v=T_{k}v-T_{l}v$ by claim 5 of Elementary Identities in a Vector Space, so by claim 1

Pointwise limit. Fix $v\in V$. Given $\varepsilon>0$, choose $N$ from the hypothesis for $\varepsilon/(\lVert v\rVert+1)$; then for $k,l\ge N$, $(\dagger)$ gives $\lVert T_{k}v-T_{l}v\rVert\le\varepsilon\lVert v\rVert/(\lVert v\rVert+1)<\varepsilon$. So $(T_{k}v)$ is a Cauchy sequence in $W$, which converges because $W$ is complete (Complex Hilbert Space, Complete Metric Space); let $Tv$ be its limit, unique by Uniqueness of Limits in a Metric Space.

$T$ is linear. Let $u,u'\in V$ and $\lambda\in\mathbb{C}$. Given $\varepsilon>0$, choose $N$ so large that $\lVert T_{k}u-Tu\rVert<\varepsilon/2$, $\lVert T_{k}u'-Tu'\rVert<\varepsilon/2$ and $\lVert T_{k}u-Tu\rVert<\varepsilon/(|\lambda|+1)$ for $k\ge N$. Then for $k\ge N$, by (P2),

So $(T_{k}(u+u'))$ converges to $Tu+Tu'$ and $(T_{k}(\lambda u))$ to $\lambda\,Tu$; since they converge to $T(u+u')$ and $T(\lambda u)$ by definition, Uniqueness of Limits in a Metric Space gives $T(u+u')=Tu+Tu'$ and $T(\lambda u)=\lambda\,Tu$.

Uniform estimate. Let $\varepsilon>0$ and choose $N$ from the hypothesis for $\varepsilon$. Fix $k\ge N$ and $v\in V$, and let $\delta>0$. Choose $M$ with $\lVert T_{l}v-Tv\rVert<\delta$ for $l\ge M$, and put $l=\max(N,M)$. By (P2) and $(\dagger)$,

As $\delta>0$ was arbitrary, Comparison of Real Numbers with Arbitrary Positive Slack §slack-above gives $\lVert(T_{k}-T)v\rVert=\lVert T_{k}v-Tv\rVert\le\varepsilon\lVert v\rVert$. So $\varepsilon$ is a bound for the linear map $T_{k}-T$ (linear by claim 2), whence $T_{k}-T\in\mathcal{L}(V,W)$ and $\lVert T_{k}-T\rVert_{\mathrm{op}}\le\varepsilon$ for all $k\ge N$ by claim 1. In particular $T_{N}-T\in\mathcal{L}(V,W)$, and $T=T_{N}+(-1)(T_{N}-T)$ (evaluate at each $v$), so $T\in\mathcal{L}(V,W)$ by claim 2. Given $\varepsilon>0$, the above with $\varepsilon/2$ yields $N$ with $\bigl|\lVert T_{k}-T\rVert_{\mathrm{op}}-0\bigr|=\lVert T_{k}-T\rVert_{\mathrm{op}}\le\varepsilon/2<\varepsilon$ for $k\ge N$, so $\bigl(\lVert T_{k}-T\rVert_{\mathrm{op}}\bigr)$ converges to $0$ in the sense of Limit of a Sequence of Real Numbers.

Pointwise convergence for every such $T$. Let $T\in\mathcal{L}(V,W)$ with $\bigl(\lVert T_{k}-T\rVert_{\mathrm{op}}\bigr)$ converging to $0$, and let $v\in V$ and $\varepsilon>0$. Choose $N$ with $\lVert T_{k}-T\rVert_{\mathrm{op}}<\varepsilon/(\lVert v\rVert+1)$ for $k\ge N$. Then claim 1 gives, for $k\ge N$, $d(T_{k}v,Tv)=\lVert(T_{k}-T)v\rVert\le\lVert T_{k}-T\rVert_{\mathrm{op}}\lVert v\rVert<\varepsilon$, so $(T_{k}v)$ converges to $Tv$ in $W$ (Convergent Sequence in a Metric Space).

Claim 8 (Quadratic-form bound). Let $A$ and $c$ be as stated; $A$ is a linear operator and self-adjoint as in Self-Adjoint Operator. For $x\in V$, $\langle x,Ax\rangle=\langle Ax,x\rangle=\overline{\langle x,Ax\rangle}$ by self-adjointness and condition 1 of Complex Inner Product Space, so $\langle x,Ax\rangle$ is real by (P1), and $\langle x,Ax\rangle\le c\lVert x\rVert^{2}$ and $-\langle x,Ax\rangle\le c\lVert x\rVert^{2}$ because its modulus is $\langle x,Ax\rangle$ or $-\langle x,Ax\rangle$ (P1).

Expansion. For a linear $B:V\to V$ and $u,w\in V$, condition 2 of Complex Inner Product Space and claim 1 of Elementary Properties of a Complex Inner Product give the first identity below; writing $u-w=u+(-1)w$ and $B(u-w)=Bu+(-1)Bw$, condition 3 of Complex Inner Product Space, claim 2 of Elementary Properties of a Complex Inner Product and $\overline{-1}=-1$ give the second:

With $B=A$, subtracting, and using $\langle w,Au\rangle=\langle Aw,u\rangle=\overline{\langle u,Aw\rangle}$ (self-adjointness, condition 1 of Complex Inner Product Space) and claim 2 of Properties of Complex Conjugation and Modulus,

With $B=I_{V}$, adding, and (P2): $\lVert u+w\rVert^{2}+\lVert u-w\rVert^{2}=2\lVert u\rVert^{2}+2\lVert w\rVert^{2}$. By the first paragraph applied to $x=u+w$ and to $x=u-w$,

Now fix $w\in V$. If $Aw=0$, then $\lVert Aw\rVert=0\le c\lVert w\rVert$. Otherwise $w\neq0$ (as $A0=0$ by (P2)), so $0<\lVert Aw\rVert$ and $0<\lVert w\rVert$; put $s=\lVert w\rVert/\lVert Aw\rVert>0$ and $u=s\,Aw$. By claim 2 of Elementary Properties of a Complex Inner Product, (P1) and (P2), $\langle u,Aw\rangle=s\lVert Aw\rVert^{2}=\lVert w\rVert\,\lVert Aw\rVert$, a real number, so $\operatorname{Re}\langle u,Aw\rangle=\lVert w\rVert\,\lVert Aw\rVert$, and $\lVert u\rVert=|s|\,\lVert Aw\rVert=\lVert w\rVert$. The last inequality becomes $4\lVert w\rVert\,\lVert Aw\rVert\le4c\lVert w\rVert^{2}$, and dividing by $4\lVert w\rVert>0$ gives $\lVert Aw\rVert\le c\lVert w\rVert$. Hence the real number $c\ge0$ is a bound for $A$, so $A\in\mathcal{L}(V)$, and $\lVert A\rVert_{\mathrm{op}}\le c$ by claim 1.