Many equations of analysis are most naturally stated in the form “find \(u\) such that \(a(u,v)=\ell(v)\) for every test vector \(v\) ”. Both sides are measurements of \(v\) : the right side is a given functional, and the left side is a functional depending linearly on the unknown \(u\) . The equation asks for the \(u\) that makes two functionals equal. Boundary value problems for differential equations take this form once derivatives have been moved onto the test function by integration by parts, and so do the normal equations of least squares.

In a general Banach space this is an equation in the dual space, a different space from the one where \(u\) lives. In a Hilbert space the distinction disappears: every bounded functional is the inner product with a vector. The equation between functionals then becomes an equation \(Au=f\) between vectors, and the question of solvability becomes the question of whether one operator is invertible. This chapter proves the identification of functionals with vectors, which is the Riesz representation theorem, derives from it the adjoint of an operator, and then proves the theorem of Lax and Milgram, which says that the equation above has exactly one solution whenever the form \(a\) is bounded and coercive.

8.1 Every bounded functional is an inner product

For a fixed vector \(y\) in a Hilbert space \(H\) , the map \(x\mapsto\langle x,y\rangle\) is a bounded linear functional of norm \(\lVert y\rVert\) , by the Cauchy–Schwarz inequality and its case of equality. There are no others.

The proof is the projection theorem applied to the kernel. A nonzero functional is determined, up to a scalar, by its kernel, and in a Hilbert space the kernel has a normal direction \(z\) ; the representing vector points along it. Completeness enters only through Theorem 7.4, and it cannot be omitted: on the incomplete inner product space \(c_{00}\) the bounded functional \(x\mapsto\sum_kx_k/2^k\) is not the inner product with any element of \(c_{00}\) (Exercise 14 of Chapter 7Chapter 7).

The map \(y\mapsto\langle\cdot,y\rangle\) is thus a norm-preserving bijection of \(H\) onto \(H^*\) . It is conjugate-linear, since \(\langle x,\alpha y\rangle=\overline\alpha\langle x,y\rangle\) . Two consequences follow at once.

In particular every orthonormal sequence converges weakly to \(0\) , by Bessel’s inequality, which gives a supply of weakly convergent sequences that do not converge in norm.

8.2 The adjoint of an operator

The adjoint reverses products, \((ST)^*=T^*S^*\) , and is conjugate-linear, \((\alpha T)^*=\overline\alpha\,T^*\) . For a matrix acting on \(\mathbb K^n\) it is the conjugate transpose. An operator \(T\in\mathcal B(H)\) is self-adjoint if \(T^*=T\) ; it is unitary if \(T^*T=TT^*=I\) . Orthogonal projections are self-adjoint, by Theorem 7.4. The adjoint also governs solvability.

In words, the equation \(Tx=y\) can be solved, at least approximately, exactly for those \(y\) that are orthogonal to every solution of the homogeneous adjoint equation \(T^*w=0\) . In finite dimensions this is the familiar statement about the four fundamental subspaces of a matrix.

8.3 Bounded forms and the operators behind them

The inner product is the model: bounded with \(M=1\) and coercive with \(\alpha=1\) . Every bounded operator \(A\) gives a bounded form \(a(u,v)=\langle Au,v\rangle\) , and the Riesz theorem shows that every bounded form arises this way.

8.4 The Lax–Milgram theorem

Each hypothesis has a visible role. Boundedness of \(a\) produces the operator \(A\) . Coercivity is used twice, once to bound \(A\) below and once to make its range dense; the same inequality kills both the kernel of \(A\) and the kernel of \(A^*\) . Completeness of \(H\) enters through the Riesz theorem and through the closed range. Symmetry of \(a\) is not assumed, and that is the point of the theorem.

When \(a\) is symmetric it is itself an inner product on \(H\) , with a norm equivalent to the original one because \(\alpha\lVert u\rVert^2\le a(u,u)\le M\lVert u\rVert^2\) , and Theorem 8.7 reduces to the Riesz theorem for that inner product (Exercise 15). The theorem of Lax and Milgram extends the Riesz theorem to forms that are not symmetric, for which there is no energy to minimize.

In applications \(H\) is infinite-dimensional and the solution cannot be computed exactly. One replaces \(H\) by a finite-dimensional subspace and solves there. The next result says that nothing is lost beyond a constant factor.

When \(H_n\) has a basis \(\phi_1,\dots,\phi_n\) , the Galerkin equation is the linear system \(\sum_ja(\phi_j,\phi_i)c_j=\ell(\phi_i)\) for the coefficients of \(u_n=\sum_jc_j\phi_j\) . Coercivity guarantees that its matrix is invertible, for every choice of subspace.

Example 8.6: A boundary value problem

Work in the real Hilbert space \(\mathcal H\) of Example 8.2 and consider

\[ a(u,v)=\int_0^1u'v'\,dx+\int_0^1u'v\,dx,\qquad\ell(v)=\int_0^1v\,dx . \]

Boundedness. For \(v\in\mathcal H\) we have \(\lvert v(x)\rvert\le\sqrt x\,\lVert v\rVert_{\mathcal H}\) by Example 8.2, so \(\lVert v\rVert_2^2\le\int_0^1x\,dx\,\lVert v\rVert_{\mathcal H}^2=\tfrac12\lVert v\rVert_{\mathcal H}^2\) . Therefore \(\lvert\int u'v\rvert\le\lVert u'\rVert_2\lVert v\rVert_2\le\tfrac1{\sqrt2}\lVert u\rVert_{\mathcal H}\lVert v\rVert_{\mathcal H}\) , and \(a\) is bounded with \(M=1+\tfrac1{\sqrt2}\approx 1.707\) . The same estimate gives \(\lvert\ell(v)\rvert\le\lVert v\rVert_2\le\tfrac1{\sqrt2}\lVert v\rVert_{\mathcal H}\) .

Coercivity. The second term of \(a(u,u)\) is \(\int_0^1u'u=\tfrac12u(1)^2\ge 0\) ; this is the fundamental theorem of calculus for \(\tfrac12u^2\) when \(u'\) is continuous, and it extends to all of \(\mathcal H\) because both sides are continuous in the norm of \(\mathcal H\) and such \(u\) are dense. So \(a(u,u)\ge\lVert u\rVert_{\mathcal H}^2\) and \(\alpha=1\) . The form is not symmetric.

By Theorem 8.7 there is exactly one \(u\in\mathcal H\) with \(a(u,v)=\ell(v)\) for all \(v\in\mathcal H\) , and \(\lVert u\rVert_{\mathcal H}\le 1/\sqrt2\approx 0.707\) . To identify it, suppose \(u\) is twice differentiable and integrate the first term by parts, using \(v(0)=0\) : \(\int u'v'=u'(1)v(1)-\int u''v\) . The equation becomes \(\int(-u''+u'-1)v+u'(1)v(1)=0\) for all \(v\) , which holds if

\[ -u''+u'=1\ \text{on }(0,1),\qquad u(0)=0,\qquad u'(1)=0 . \]

Solving, \(u'=1-e^{x-1}\) and \(u(x)=x-e^{x-1}+e^{-1}\) . This function lies in \(\mathcal H\) and satisfies the weak equation, so by uniqueness it is the solution; its norm is \(\lVert u\rVert_{\mathcal H}=\lVert 1-e^{x-1}\rVert_2\approx 0.410\) . The condition \(u(0)=0\) was built into the space, whereas \(u'(1)=0\) was not imposed anywhere: it emerged from the weak equation.

Now the Galerkin method. With the single basis function \(\phi_1=x\) , the equation \(a(\phi_1,\phi_1)c=\ell(\phi_1)\) is \(\tfrac32c=\tfrac12\) , giving \(u_1=\tfrac13x\) . With \(\phi_1=x\) and \(\phi_2=x^2\) the system is

\[ \begin{pmatrix}\tfrac32&\tfrac53\\[2pt]\tfrac43&\tfrac{11}6\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}=\begin{pmatrix}\tfrac12\\[2pt]\tfrac13\end{pmatrix},\qquad c_1=\tfrac{13}{19},\quad c_2=-\tfrac6{19}, \]

so \(u_2=\tfrac1{19}(13x-6x^2)\) . The errors in the norm of \(\mathcal H\) are \(\lVert u-u_1\rVert_{\mathcal H}\approx 0.184\) and \(\lVert u-u_2\rVert_{\mathcal H}\approx 0.0233\) . The best approximation to \(u\) from the span of \(x\) and \(x^2\) has error \(0.0231\) , so the Galerkin solution is within one percent of optimal, comfortably inside the factor \(M/\alpha\approx 1.707\) allowed by Céa’s lemma. At the endpoint, \(u(1)=e^{-1}\approx 0.3679\) and \(u_2(1)=\tfrac7{19}\approx 0.3684\) .

This example is a one-dimensional model of the weak formulation of elliptic boundary value problems (math.partial-differential-equations.weak-dirichlet-formulation). Partial Differential Equations constructs the appropriate Hilbert spaces in several variables and verifies boundedness and coercivity for the forms that arise there; the existence and uniqueness of weak solutions (math.partial-differential-equations.weak-elliptic-existence) is then precisely Theorem 8.7.

8.5 Positivity is not coercivity

Exercises

1. Find the Riesz representers in \(\ell^2\) of the functionals \(f(x)=x_1+x_2+x_3\) and \(g(x)=\sum_kx_k/k\) , and their norms.

   Answer

   \(f\) is represented by \(y=(1,1,1,0,0,\dots)\) , with \(\lVert f\rVert=\sqrt3\) . \(g\) is represented by \(y=(1/k)\) , with \(\lVert g\rVert=\pi/\sqrt6\approx 1.2825\) .

2. Find the polynomial \(r\in\mathcal P_2\) with \(\int_0^1p(x)r(x)\,dx=p(0)\) for every \(p\in\mathcal P_2\) , and the norm of evaluation at \(0\) as a functional on \(\mathcal P_2\subseteq L^2[0,1]\) .

   Solution

   Write \(r(x)=\alpha+\beta x+\gamma x^2\) . The condition \(\int_0^1p\,r=p(0)\) for all \(p\in\mathcal P_2\) is linear in \(p\) , so it suffices to impose it on the basis \(1,x,x^2\) :

   \[ \alpha+\tfrac12\beta+\tfrac13\gamma=1,\qquad\tfrac12\alpha+\tfrac13\beta+\tfrac14\gamma=0,\qquad\tfrac13\alpha+\tfrac14\beta+\tfrac15\gamma=0 . \]

   The solution is \(\alpha=9\) , \(\beta=-36\) , \(\gamma=30\) , so

   \[ r(x)=30x^2-36x+9 . \]

   Check: \(\int_0^1r=10-18+9=1\) , \(\int_0^1xr=\tfrac{30}4-12+\tfrac92=0\) , \(\int_0^1x^2r=6-9+3=0\) .

   By Theorem 8.1 applied to the Hilbert space \(\mathcal P_2\) with the \(L^2\) inner product, the norm of the evaluation functional equals \(\lVert r\rVert_2\) . There is a shortcut for computing it: since \(r\in\mathcal P_2\) , the defining property with \(p=r\) gives \(\lVert r\rVert_2^2=\int_0^1r\cdot r=r(0)=9\) . Hence the norm of evaluation at \(0\) on \(\mathcal P_2\) is \(3\) , the same value \(n+1\) that Example 8.1 found for evaluation at the other endpoint. The representer is large near \(x=0\) , where \(r(0)=9\) , and oscillates elsewhere so as to annihilate \(x\) and \(x^2\) .

3. Let \(T\in\mathcal B(H)\) be bounded below, with \(T^*\) injective. Prove that \(T\) is invertible, and identify where the proof of Theorem 8.7 uses exactly this argument.

   Hint

   Bounded below gives injectivity and closed range (Proposition 5.4). \(T^*\) injective gives dense range (Proposition 8.4). In the proof of Theorem 8.7, coercivity supplies both: \(\operatorname{Re}\langle Aw,w\rangle=\operatorname{Re}\langle w,A^*w\rangle\) , so the same inequality bounds \(A^*\) below.

4. On \(\ell^2\) let \(Tx=(x_1,\,x_1+x_2,\,x_2+x_3,\,\dots)\) . Express \(T\) through the right shift and find \(T^*\) explicitly.

   Answer

   \(T=I+S\) , so \(T^*=I+S^*=I+L\) , that is, \((T^*y)_k=y_k+y_{k+1}\) .

5. Let \((Kf)(x)=\int_0^1xy^2f(y)\,dy\) on \(L^2[0,1]\) . Find \(K^*\) , and verify \(\langle Kf,g\rangle=\langle f,K^*g\rangle\) for \(f=1\) and \(g(x)=x\) .

   Solution

   The kernel is \(k(x,y)=xy^2\) , real-valued, so by Example 8.3 the adjoint has kernel \(k(y,x)=yx^2\) :

   \[ (K^*g)(x)=\int_0^1x^2y\,g(y)\,dy=x^2\int_0^1y\,g(y)\,dy . \]

   For \(f=1\) : \((Kf)(x)=x\int_0^1y^2\,dy=\tfrac13x\) , so with \(g(x)=x\) ,

   \[ \langle Kf,g\rangle=\int_0^1\tfrac13x\cdot x\,dx=\tfrac19 . \]

   On the other side, \((K^*g)(x)=x^2\int_0^1y\cdot y\,dy=\tfrac13x^2\) , and

   \[ \langle f,K^*g\rangle=\int_0^11\cdot\tfrac13x^2\,dx=\tfrac19 . \]

   The two agree. Both \(K\) and \(K^*\) have rank one: \(Kf=\langle f,y^2\rangle\,x\) and \(K^*g=\langle g,x\rangle\,y^2\) , where the functions \(x\) and \(y^2\) have exchanged roles. In general the adjoint of the rank-one operator \(f\mapsto\langle f,\phi\rangle\psi\) is \(g\mapsto\langle g,\psi\rangle\phi\) , which one verifies in one line from the definition.

6. For the Volterra operator \(V\) on \(L^2[0,1]\) compute \(V^*V\) and \(VV^*\) as integral operators, giving their kernels, and show that \(V^*V\ne VV^*\) by applying both to the constant function \(1\) .

   Solution

   By Example 8.3, \((Vf)(t)=\int_0^tf\) and \((V^*g)(x)=\int_x^1g\) . Then

   \[ (V^*Vf)(x)=\int_x^1\int_0^tf(s)\,ds\,dt=\int_0^1f(s)\bigl(1-\max\{x,s\}\bigr)\,ds , \]

   because for fixed \(s\) the variable \(t\) runs over \(t\ge x\) and \(t\ge s\) , an interval of length \(1-\max\{x,s\}\) . So \(V^*V\) is the integral operator with kernel \(1-\max\{x,y\}\) . Similarly

   \[ (VV^*f)(x)=\int_0^x\int_t^1f(s)\,ds\,dt=\int_0^1f(s)\min\{x,s\}\,ds , \]

   since \(t\) runs over \(t\le x\) and \(t\le s\) . So \(VV^*\) has kernel \(\min\{x,y\}\) , the operator of Example 2.5. The exchange of the order of integration is justified by the Fubini theorem.

   Applied to the constant function \(1\) : \((V^*V1)(x)=\int_x^1t\,dt=\tfrac12(1-x^2)\) and \((VV^*1)(x)=\int_0^x(1-t)\,dt=x-\tfrac12x^2\) . At \(x=0\) these are \(\tfrac12\) and \(0\) , so \(V^*V\ne VV^*\) : the Volterra operator is not normal. Both products are self-adjoint with symmetric kernels, and the substitution \(x\mapsto 1-x\) carries one kernel into the other, which is why Chapter 10 will find that they have the same eigenvalues.

7. Let \(P\in\mathcal B(H)\) satisfy \(P^2=P=P^*\) . Prove that \(P\) is the orthogonal projection onto its range.

   Hint

   \(M=\operatorname{ran}P=\ker(I-P)\) is closed. For any \(x\) , \(Px\in M\) , and \(x-Px\perp M\) because \(\langle x-Px,Pz\rangle=\langle P(x-Px),z\rangle=0\) using \(P^*=P\) and \(P^2=P\) . Now invoke uniqueness in Theorem 7.4.

8. Use \(\lVert T\rVert^2=\lVert T^*T\rVert\) to compute the Euclidean operator norm of \(A=\begin{pmatrix}1&2\\0&1\end{pmatrix}\) exactly.

   Answer

   \(A^{\top}A=\begin{pmatrix}1&2\\2&5\end{pmatrix}\) has largest eigenvalue \(3+2\sqrt2=(1+\sqrt2)^2\) , so \(\lVert A\rVert=1+\sqrt2\approx 2.414\) .

9. On \(\mathbb R^2\) let \(a(u,v)=3u_1v_1-2u_2v_1+2u_1v_2+u_2v_2\) . Find the matrix \(A\) with \(a(u,v)=v^{\top}Au\) , constants \(\alpha\) and \(M\) as in Theorem 8.7, and the solution \(u\) of \(a(u,v)=v_1+4v_2\) for all \(v\) ; check the bound \(\lVert u\rVert\le\lVert\ell\rVert/\alpha\) .

   Solution

   Collecting the coefficients of \(v_1\) and \(v_2\) : \(a(u,v)=v_1(3u_1-2u_2)+v_2(2u_1+u_2)\) , so

   \[ A=\begin{pmatrix}3&-2\\2&1\end{pmatrix}. \]

   Coercivity. \(a(u,u)=3u_1^2-2u_2u_1+2u_1u_2+u_2^2=3u_1^2+u_2^2\ge\lVert u\rVert^2\) , with equality at \(u=e_2\) . So \(\alpha=1\) and no larger constant works.

   Boundedness. The best constant is \(M=\lVert A\rVert\) . Here \(A^{\top}A=\begin{pmatrix}13&-4\\-4&5\end{pmatrix}\) with eigenvalues \(9\pm4\sqrt2\) , so \(M=\sqrt{9+4\sqrt2}\approx 3.828\) .

   Solution. With \(\ell(v)=v_1+4v_2\) the system \(Au=(1,4)\) is \(3u_1-2u_2=1\) , \(2u_1+u_2=4\) . Adding twice the second equation to the first gives \(7u_1=9\) , so

   \[ u=\bigl(\tfrac97,\tfrac{10}7\bigr),\qquad\lVert u\rVert=\tfrac17\sqrt{181}\approx 1.922 . \]

   The bound. \(\lVert\ell\rVert=\sqrt{1+16}=\sqrt{17}\approx 4.123\) , and \(\lVert u\rVert\le\lVert\ell\rVert/\alpha=4.123\) holds. The bound is far from sharp. Equality in the coercivity inequality \(a(u,u)\ge\alpha\lVert u\rVert^2\) occurs in the direction \(e_2\) , but the estimate \(\lVert Au\rVert\ge\alpha\lVert u\rVert\) that the proof derives from it is attained nowhere: \(\lVert Ae_2\rVert=\sqrt5\) , and the smallest value of \(\lVert Au\rVert\) over unit vectors is the smallest singular value \(\sqrt{9-4\sqrt2}\approx 1.828\) , well above \(\alpha=1\) . The skew part of \(A\) , which coercivity ignores, makes \(A\) larger in every direction, so the sharp bound here is \(\lVert u\rVert\le\lVert\ell\rVert/1.828\approx 2.255\) .

10. For which real \(\beta\) is the form \(a(u,v)=\langle u,v\rangle+\beta\langle Su,v\rangle\) on real \(\ell^2\) coercive? Show that for \(\beta=-1\) it is positive, in the sense that \(a(u,u)>0\) for \(u\ne 0\) , but not coercive.

    Solution

    By the Cauchy–Schwarz inequality and \(\lVert Su\rVert=\lVert u\rVert\) , \(\lvert\langle Su,u\rangle\rvert\le\lVert u\rVert^2\) . Hence \(a(u,u)\ge(1-\lvert\beta\rvert)\lVert u\rVert^2\) , and the form is coercive with \(\alpha=1-\lvert\beta\rvert\) whenever \(\lvert\beta\rvert<1\) .

    For \(\lvert\beta\rvert\ge 1\) it is not coercive. Let \(u^{(n)}=n^{-1/2}(1,\dots,1,0,\dots)\) with \(n\) ones. Then \(\langle Su^{(n)},u^{(n)}\rangle=(n-1)/n\) , so \(a(u^{(n)},u^{(n)})=1+\beta(n-1)/n\to 1+\beta\) , which is \(\le 0\) for \(\beta\le-1\) . For \(\beta\ge 1\) use the same vectors with alternating signs, for which \(\langle Su,u\rangle=-(n-1)/n\) and \(a\to 1-\beta\le 0\) . In neither case can a positive \(\alpha\) exist. So the form is coercive exactly when \(\lvert\beta\rvert<1\) .

    For \(\beta=-1\) : since \(\lVert Su\rVert=\lVert u\rVert\) ,

    \[ \lVert u-Su\rVert^2=2\lVert u\rVert^2-2\langle Su,u\rangle=2a(u,u), \]

    so \(a(u,u)=\tfrac12\lVert(I-S)u\rVert^2\) . This is positive for \(u\ne 0\) because \(I-S\) is injective (Exercise 7 of Chapter 5Chapter 5), and \(a(u^{(n)},u^{(n)})=\tfrac12\cdot\tfrac2n=\tfrac1n\to 0\) shows that it is not coercive. The operator behind the form is \(I-S\) , which is injective with non-closed range: positivity gives injectivity, and only coercivity gives the closed range that solvability needs.

11. A student writes: “For \(a(u,v)=\sum_ku_kv_k/k\) on \(\ell^2\) we have \(a(u,u)>0\) whenever \(u\ne 0\) , so \(a\) is coercive and every bounded \(\ell\) gives a unique solution.” Identify the error and give an \(\ell\) for which no solution exists.

    Answer

    Positivity is not coercivity: \(a(e_k,e_k)=1/k\to 0\) with \(\lVert e_k\rVert=1\) , so no \(\alpha>0\) exists. For \(\ell(v)=\sum_kv_k/k\) a solution would have to be \((1,1,1,\dots)\) , which is not in \(\ell^2\) .

12. In the space \(\mathcal H\) of Example 8.2 find the Riesz representer of \(\ell(v)=\int_0^1v(x)\,dx\) and the norm of \(\ell\) on \(\mathcal H\) .

    Solution

    For \(v\in\mathcal H\) write \(v(x)=\int_0^xv'(s)\,ds\) and exchange the order of integration over the triangle \(0\le s\le x\le 1\) :

    \[ \ell(v)=\int_0^1\int_0^xv'(s)\,ds\,dx=\int_0^1v'(s)\,(1-s)\,ds . \]

    This is \(\langle v,w\rangle_{\mathcal H}\) for the element \(w\in\mathcal H\) with \(w'(s)=1-s\) , that is,

    \[ w(x)=\int_0^x(1-s)\,ds=x-\tfrac12x^2 . \]

    By Theorem 8.1, \(w\) is the representer and

    \[ \lVert\ell\rVert=\lVert w\rVert_{\mathcal H}=\Bigl(\int_0^1(1-s)^2\,ds\Bigr)^{1/2}=\frac1{\sqrt3}\approx 0.577 . \]

    This is smaller than the bound \(1/\sqrt2\approx 0.707\) obtained in Example 8.6 from the inequality \(\lVert v\rVert_2\le\lVert v\rVert_{\mathcal H}/\sqrt2\) ; the Riesz representer gives the exact norm. Note that \(w\) solves the boundary value problem \(-w''=1\) , \(w(0)=0\) , \(w'(1)=0\) : representing \(\ell\) in \(\mathcal H\) is the same as solving the equation \(\int_0^1w'v'=\int_0^1v\) for all \(v\) , which is the symmetric case of Theorem 8.7.

13. In \(\mathcal H\) solve \(\int_0^1u'v'\,dx=\int_0^1xv\,dx\) for all \(v\in\mathcal H\) : state the boundary value problem that a smooth solution satisfies, including the condition at \(x=1\) , and solve it.

    Answer

    Integrating by parts as in Example 8.6, a smooth solution satisfies \(-u''=x\) on \((0,1)\) with \(u(0)=0\) and the natural condition \(u'(1)=0\) . Then \(u'=\tfrac12(1-x^2)\) and \(u(x)=\tfrac12x-\tfrac16x^3\) .

14. In the symmetric case of Theorem 8.7, show that the Galerkin solution \(u_n\) minimizes \(a(u-v,u-v)\) over \(v\in H_n\) , and deduce the improved estimate \(\lVert u-u_n\rVert\le\sqrt{M/\alpha}\,\operatorname{dist}(u,H_n)\) .

    Hint

    For symmetric \(a\) the relation \(a(u-u_n,v)=0\) on \(H_n\) says that \(u_n\) is the orthogonal projection of \(u\) onto \(H_n\) in the inner product \(a\) . So \(a(u-u_n,u-u_n)\le a(u-v,u-v)\) for \(v\in H_n\) ; bound the left side below by \(\alpha\lVert u-u_n\rVert^2\) and the right side above by \(M\lVert u-v\rVert^2\) .

15. Let \(a\) be a bounded, coercive, symmetric bilinear form on a real Hilbert space \(H\) . Show that \(a\) is an inner product whose norm is equivalent to that of \(H\) , and derive Theorem 8.7 for this \(a\) from Theorem 8.1.

    Hint

    Symmetry, bilinearity, and \(a(u,u)\ge\alpha\lVert u\rVert^2>0\) make \(a\) an inner product, and \(\alpha\lVert u\rVert^2\le a(u,u)\le M\lVert u\rVert^2\) makes its norm equivalent, so \(H\) is complete for it and \(\ell\) is bounded for it. Theorem 8.1 in the Hilbert space \((H,a)\) gives \(u\) with \(\ell(v)=a(v,u)=a(u,v)\) .

16. Let \((e_k)\) be an orthonormal basis of a Hilbert space \(H\) . Prove that \(x_n\rightharpoonup x\) if and only if \(\sup_n\lVert x_n\rVert<\infty\) and \(\langle x_n,e_k\rangle\to\langle x,e_k\rangle\) for each \(k\) .

    Hint

    Apply Proposition 6.3 with \(\Gamma\) the set of functionals \(\langle\cdot,e_k\rangle\) ; their span corresponds, under the Riesz identification, to the span of the \(e_k\) , which is dense.

17. Let \(T\in\mathcal B(H)\) on a complex Hilbert space satisfy \(\lvert\langle Tx,x\rangle\rvert\ge c\lVert x\rVert^2\) for some \(c>0\) and all \(x\) . Prove that \(T\) is invertible with \(\lVert T^{-1}\rVert\le 1/c\) .

    Solution

    By the Cauchy–Schwarz inequality, \(c\lVert x\rVert^2\le\lvert\langle Tx,x\rangle\rvert\le\lVert Tx\rVert\lVert x\rVert\) , so

    \[ \lVert Tx\rVert\ge c\lVert x\rVert\qquad\text{for all }x . \]

    Thus \(T\) is bounded below, and by Proposition 5.4 it is injective with closed range.

    The same hypothesis holds for the adjoint: \(\lvert\langle T^*x,x\rangle\rvert=\lvert\langle x,Tx\rangle\rvert=\lvert\langle Tx,x\rangle\rvert\ge c\lVert x\rVert^2\) . So \(T^*\) is bounded below as well, in particular injective. By Proposition 8.4 the range of \(T\) is dense. A subspace that is both closed and dense is all of \(H\) , so \(T\) is surjective.

    Hence \(T\) is a bijection, and the inequality \(\lVert Tx\rVert\ge c\lVert x\rVert\) applied to \(x=T^{-1}y\) gives \(\lVert T^{-1}y\rVert\le\lVert y\rVert/c\) , that is, \(\lVert T^{-1}\rVert\le 1/c\) . This generalises the Lax–Milgram theorem: the condition \(\operatorname{Re}\langle Tx,x\rangle\ge\alpha\lVert x\rVert^2\) is replaced by the weaker \(\lvert\langle Tx,x\rangle\rvert\ge c\lVert x\rVert^2\) , which also covers, for example, \(T=iI\) , for which the real part vanishes identically.

18. For the form and functional of Example 8.7, write down the Galerkin system in the span of \(e_1,\dots,e_n\) , solve it, and compute \(\lVert u_n\rVert\) ; explain why Theorem 8.9 does not apply.

    Answer

    The system is diagonal: \(u_k/k=1/k\) for \(k\le n\) , so \(u_n=(1,\dots,1,0,\dots)\) and \(\lVert u_n\rVert=\sqrt n\) . Theorem 8.9 presupposes coercivity, which fails here, and it compares \(u_n\) with a solution \(u\) that does not exist.