# 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 second installment in a series of functional analysis exercises (Part 1, Part 2, Part 3, Part 4).

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

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

Solution

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.

Solution

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

Compute

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

Solution

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 $\textrm{dim} \mathscr{H} < \infty$.

Solution

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$\mathscr{H} < \infty$.   $\blacksquare$