A Variety of Functional Analysis Exercises: Part 1

As a helpful review, here are a variety of problem and solutions to exercises from an introductory functional analysis class.  This is the first installment in a series of functional analysis exercises.


Problem 1

Show that for any real inner product space

\langle u, v \rangle + \langle v, u \rangle = \frac{1}{2} \left ( \left \| u + v \right \|^2 - \left \| u - v \right \|^2 \right ),

and for a complex inner product space

\langle u, v \rangle - \langle v, u \rangle = \frac{i}{2} \left (\left \| u + iv \right \|^2 - \left \| u - iv \right \|^2 \right ).


Recall that if \langle \cdot, \cdot \rangle is an inner product on a vector space \mathscr{X}, the following hold:

  1. \langle \alpha x + \beta y, z \rangle = \alpha \langle x, z \rangle + \beta \langle y, z \rangle;
  2. \langle x, \alpha y + \beta z \rangle = \overline{\alpha} \langle x, y \rangle + \overline{\beta} \langle x, z \rangle;
  3. \langle x, x \rangle \geq 0 and \langle x, x \rangle =0 \Leftrightarrow x=0;
  4. \langle x, y \rangle = \overline{\langle y, x \rangle}.

Moreover, let us recall that the norm induced by the inner product is given by \left \| x \right \|= \langle x, x \rangle^{1/2} for all x \in \mathscr{X}. With this in mind, let us proceed in demonstrating the desired identities:


Let \mathscr{X} be a real inner product space. Recall that \overline{\alpha} = \alpha for all \alpha \in \mathbb{R}, and observe that the following holds for all u,v \in \mathscr{X}:

Now, let \mathscr{X} be a complex inner product space. Then, observe that the following holds for all u,v \in \mathscr{X}:

Problem 2

Let X = C[0,1] with supremum norm and define E \subset X to be the set of functions such that

\int_0^{\frac{1}{2}} f - \int_{\frac{1}{2}}^1 f = 1.

Prove that E is a closed, convex subset of X, but that E has no element of minimal norm. This gives an alternative proof of the fact that C[0,1] is not a Hilbert space.


Let X = C[0,1] with supremum norm and define E:= \left \{ f \in X : \int_0^{\frac{1}{2}} f - \int_{\frac{1}{2}}^1 f = 1 \right \}. Let us consider the linear continuous function G: C[0,1] \rightarrow \mathbb{R} defined by G(g) = \int_0^{\frac{1}{2}} g - \int_{\frac{1}{2}}^1 g. As the preimage of a closed set under a continuous function is closed, the preimage of the closed set \left \{1 \right \} must also be closed under the continuous function G. Following from the fact that G^{-1} \left ( \left \{ 1 \right \} \right ) = E by definition, we may see that E is closed.


Now, let g, h \in E be arbitrary and fix t \in [0,1]. Then, we may observe:

It follows that, for all g, h \in E, it must be the case that tg+(1-t)h \in X whenever t \in [0,1]. Thus, E is convex.


Further, for all f \in E, let us observe that the following holds:

That is, \left \|f \right \| \leq 1 for all f \in E.


Observe that if we desire to show that C[0,1] is not a Hilbert space, it suffices to show that there exists no unique element f_0 in E such that d(0,E) = \left \| 0-f_0 \right \|.

To do so, we construct a sequence in E which approaches the minimum of the norm, but which does not converge to an element of C[0,1]. Setting n= \frac{1}{t} for convenience, let us then consider the sequence \left \{ f_t \right \}_{t=1}^\infty defined by:

While rather undesirably messy when expressed as such, this is simply the following function:


Observe by inspection (for convenience, as this is already cumbersome enough) that f_t \in E for all $t > 0$, and \left \| f_l \right \| > \left \| f_p \right \| whenever l > p.


Moreover, note that \left \| f_t\right \| = 1 + \frac{2n}{1-2n} and \left \| f_t\right \| \rightarrow 1 as t \rightarrow \infty. But, observe as well that f_t(X) \rightarrow \chi_{[0,1/2)} (x) - \chi_{[1/2,1]} (x) point-wise as t \rightarrow \infty, which is not a continuous function. Thus, we have constructed a sequence of functions in E which grow arbitrarily close to the minimum of the norm, but which do not converge point-wise to an element of C[0,1]. Correspondingly, there cannot exist an element of minimal norm in E with regard to 0, and thus C[0,1] is not a Hilbert space.    \blacksquare

Problem 3


\underset{a,b,c}{min}\int_{-1}^1 \left | x^3 - a -bx -cx^2 \right |^2 dx.


We seek to compute \underset{a,b,c}{min}\int_{-1}^1 \left | x^3 - a -bx -cx^2 \right |^2 dx.

Let us begin by observing that the set \mathscr{X} := \left \{ \alpha_1x^2+\alpha_2x +\alpha_3 \mid \alpha_i \in \mathbb{R} \right \} is a closed, linear subspace of the Hilbert space L^2[-1,1] with inner product \langle f, g \rangle = \int_{-1}^1 (fg)(x) dx. Further, note that it may be easily checked via routine computation that \mathscr{E} = \left \{ \frac{\sqrt{2}}{2}, \frac{\sqrt{6}}{2}x, \frac{3\sqrt{10}}{4}(x^2-\frac{1}{3}) \right \} is an orthonormal basis for \mathscr{X}.


Observing that x^3 \in L^2[-1,1], let f_0 \in \mathscr{X} be the unique element of \mathscr{X} with minimal norm in regard to x^3. That is, let f_0 be such that

Then, we may compute f_0 by taking the projection of x^3 onto \mathscr{X}. As \mathscr{E} is an orthonormal basis for \mathscr{X}, it follows that

As a Hilbert space is necessarily complete, the following then holds

Thus, the given integral is minimized when a =0, b= \frac{3}{5}, and c=0.   \blacksquare

Problem 4

If the closed unit ball of \mathscr{H} is compact, show that \textrm{dim} \mathscr{H} < \infty.


Suppose \mathscr{H} is a Hilbert space, and suppose the closed unit ball of \mathscr{H}, denoted by \mathcal{B}_\mathscr{H}, is compact. For some index set I, let \mathscr{E} = \left \{e_i \right \}_{i \in I} be an orthonormal basis for \mathscr{H}. Then, it follows that e_i \in \mathcal{B} for all i \in I. Moreover, as \langle e_i, e_j \rangle =0 for any two e_i, e_j \in \mathscr{E} with i\neq j, observe that we may deduce:

Let us denote the open ball of radius \delta about x \in \mathscr{H} by \mathcal{B}(x; \delta) and construct the set:

Observe that each open ball in C contains precisely one element of \mathscr{E}. Further, by the polar identity, let us note that, for all x \in \mathcal{B}_\mathscr{H}, the following holds:

From this, we may see that each x \in \mathcal{B}_\mathscr{H} lies within at least one open ball in the set C. Thus, \mathcal{B}_\mathscr{H} \subseteq \bigcup_{i \in I}\mathcal{B}(e_i ; 1), and it follows that C is indeed an open cover of \mathcal{B}_\mathscr{H}.


But, as \mathcal{B}_\mathscr{H} is compact, every open cover admits a finite subcover. As no open ball in C may be omitted without also omitting an element of the basis \mathscr{E}, it must then be the case that C is finite. Correspondingly, as the cardinality of C is the same as the cardinality of \mathscr{E}, it must be the case that our orthonormal basis is finite as well. Thus, dim\mathscr{H} < \infty.   \blacksquare

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