5.1 Uniform boundedness principle

Theorem 5.1 Uniform bounded principle

• $X$
  • banach space
• $Y$
  • norm space
• $\mathcal{A} \subset \mathcal{L}(X, Y)$

\[\forall u \in X, \ \sup_{A \in \mathcal{A}} \|A u \| < \infty \Rightarrow \sup_{A \in \mathcal{A}} \|A\| < \infty .\]

proof

\[F_{n} := \left\{ u \in X \mid \sup_{A \in \mathcal{A}} \|A u\| \le n \right\} \ (n \in \mathbb{N}) .\]

$F_{n}$ satisfies the condition of Baire’s category theorem. Indeed, we first show that $F_{n}$ is closed in $X$. Let $u_{n} \in F_{n}$, $u_{n} \rightarrow u \in X$.

\[\forall A \in \mathcal{A}, \ \|Au_{k}\| \le n\]

Then by taking the limit $k \rightarrow \infty$, we have

\[\forall A \in \mathcal{A}, \ \|Au\| \le n .\]

It follows that $u \in F_{n}$. Hence $F_{n}$ is closed. We next show that $X := \bigcup_{n \in \mathbb{N}}F_{n}$. Let $u \in X$. By our assumption, there exists $n \in \mathbb{N}$ such that

\[\sup_{A \in \mathcal{A}} \|Au\| \le n .\]

It follows that $u \in F_{n}. Hence \(X := \bigcup_{n \in \mathbb{N}}F_{n}\).

Since $X$ is Banach space, $F_{n}$ contains at least a interior point by Baire’s category theorem. Therefore,

\[\exists N \in \mathbb{N}, \ a \in X, \ r > 0, \ \text{ s.t. } \ \overline{B(a, r)} \subseteq F_{N} .\]

Let $u \in X$, $|u | \le r$. Since $a + u, a \in \overline{B(a, r)}$,

\[\forall A \in \mathcal{A}, \ \|A u \| \le \|A(u + a)\| + \|A a \| \le 2 N .\]

Hence

\[\forall A \in \mathcal{A}, \ \sup_{\|u \| \le r} \|A u\| \le 2N .\]

It follows that

\[\forall A \in \mathcal{A}, \ \sup_{\|u \| \le 1} \|A u\| \le \frac{2N}{r} .\]

Therefore,

\[\|A\| \le \frac{2N}{r} < \infty .\]

Since $A \in \mathcal{A}$ is arbitrary, the statement holds.

$\Box$

Example

• $X := L^{2}(\Omega)$,

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proposition 5.1

• $\phi: \Omega \rightarrow \mathbb{R}$,
  • measurable function

If $u \overline{\phi}$ is integrable for all $u \in L^{2}(\Omega)$,

\[\phi \in L^{2}(\Omega) .\]

proof

It is enough to show that $|\phi(x)| < \infty$ for all $x$. Indeed,

\[\begin{eqnarray} \int_{\Omega} \phi(x) \phi(x) \ \mu(dx) & \le & \int_{\Omega} |\phi(x)|^{2} \ \mu(dx) \end{eqnarray}\]

Let

\[\Omega_{n} := \{ x \in \Omega \mid |x| < m, \quad |\phi(x)| < n \} \quad (n \in \mathbb{N}) .\]

$\Omega_{n}$ is a measurable set in $\Omega$. Let \(\phi_{n} := 1_{\Omega_{n}} \phi\). $\phi_{n}$ is measurable, bounded and equal to 0 out of \(\{x \in \Omega \mid |x| < n\}\).

\[\begin{eqnarray} f_{n}(u) & := & (u, \phi_{n})_{L^{2}(\Omega)} \nonumber \\ & = & \int_{\Omega} u(x) \overline{\phi_{n}(x)} \ dx . \end{eqnarray}\]

Then $f_{n} \in X^{*} := \mathcal{L}(X, K)$. From \(\eqref{chap03_04_03_48}\), we have \(\|\phi_{n}\|_{X} = \|f_{n}\|_{X^{*}}\).

We next show that \(\{f_{n}\}\) is bounded. Let $u \in X$ be fixed. By our assumption,

\[\begin{eqnarray} |f_{n}(u)| & \le & \int_{\Omega} |u(x)\overline{\phi_{n}(x)}| \ dx \nonumber \\ & = & \int_{\Omega_{n}} |u(x)\overline{\phi(x)}| \ dx \nonumber \\ & = & \int_{\Omega} |u(x)\overline{\phi(x)}| \ dx \nonumber \\ & =: & M_{u} \nonumber . \end{eqnarray}\]

$M_{u}$ is independent on $n$ so that

\[\sup_{n \in \mathbb{N}} |f_{n}(u)| \le M_{u} < \infty \quad (u \in X) .\]

It follows that the assumptions in theorem 5.1 are satisfied with $Y = K$, \(\mathcal{A} = \{f_{n}\}\). Therefore, by theorem 5.1, there exists $M \ge 0$ such that

\[\|f_{n}\| = \|\phi_{n}\| \le M .\]

Since \(\{\Omega_{n}\}_{n \in \mathbb{N}}\) is monotonically ingcreasing and \(\bigcup_{n=1}^{n} \Omega_{n} = \Omega\). Hence \(\{|\phi_{n}(x)|^{2}\}_{n \in \mathbb{N}}\) is a monotonically nondecreasing sequence of non negative functions. Moreover, $\phi_{n}(x) \rightarrow \phi(x)$.

\[\begin{eqnarray} \int_{\Omega} |\phi(x)|^{2} \ dx & = & \lim_{n \rightarrow \infty} \int_{\Omega} |\phi(x)|^{2} \ dx \nonumber \\ & \le & M^{2} . \nonumber \end{eqnarray}\]

Therefore, $\phi \in L^{2}(\Omega)$.

$\Box$

TBD

Theorem 5.2

• $X$,
  • Hilbert space
• $\Omega \supseteq \mathbb{C}$,
  • region
• $A: \Omega \rightarrow \mathcal{L}(\Omega)$,

If $(A(z)u, v)$ is holomorphic in $\Omega$ for all $u, v \in X$, then $A(z)$ is $\mathcal{L}(X)$ valued holomorphic function in $\Omega$.

proof

TBD

$\Box$

Definition 5.1

• $X, Y$,
  • norm space
• $A_{n} \in \mathcal{L}(X, Y)$,
• $A \in \mathcal{L}(X, Y)$,

$(A_{n})$ is strongly converge to $A$ if

\[\forall u \in X, \ \lim A_{n} = Au .\]

We denote strong convergence by

\[\mathop{\text{s-lim}}_{n \rightarrow \infty} A_{n} = A .\]

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Remark

\[\|A_{n} - A\| \rightarrow 0 \Rightarrow \mathop{\text{s-lim}}_{x} A_{n} = A .\]

Indeed,

\[\begin{eqnarray} & & \|A_{n} - A\| \rightarrow 0 \nonumber \\ & \Rightarrow & \|A_{n}u - A_{n}u\| \le \|A_{n} - A\| \|u\| \rightarrow 0 . \nonumber \end{eqnarray}\]

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Example 5.1

• $X := L^{p}(0, \infty)$,
• $1 \le p < \infty$,

\[(A_{n} u)(x) := u(x + n)\] \[\begin{eqnarray} \|A_{n}u\|^{p} & = & \int_{0}^{\infty} |u(x + n)|^{p} \ dx \nonumber \\ & = & \int_{n}^{\infty} |u(y)|^{p} \ dy \rightarrow 0 \quad (n \rightarrow \infty) . \nonumber \end{eqnarray}\]

Hence $A_{n}$ strongly converge to 0. On the other hand, if $x \le n$ then

\[\forall u \in L^{p}, \ u(x) = 0, \Rightarrow \|A_{n}u\| = \|u\| .\]

Therefore, \(\|A_{n}\| = \|A_{n} - 0\| = 1\).

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Theorem 5.3

• $X$,
  • Banach space
• $Y$,
  • norm space
• $A_{n} \in \mathcal{L}(X, Y)$,
  • $\forall u \in X$, $\lim_{n \rightarrow \infty}A_{n}u \in Y$,

Then

• (i) \(\{A_{n}\}_{n \in \mathbb{N}}\) is bounded in $\mathcal{L}(X, Y)$,
• (ii) $\mathop{s-\lim} A_{n} =: A \in \mathcal{L}(X, Y)$

proof

(i)

For all $u \in X$, \(\{A_{n} u\}_{n \in \mathbb{N}}\) is bounded sequence. By applying theorem 5.1, we have

\[\sup_{n \in \mathbb{N}} \|A_{n}\| < \infty .\]

(ii)

Let

\[Au := \lim_{n \rightarrow \infty} A_{n} u \ (u \in X) .\]

Obviously, $A: X \rightarrow Y$ is linear. From (i), there exists $M$ such that \(\|A_{n}\| \le M\).

\[\|Au \| = \lim_{n \rightarrow \infty} \|A_{n} u\| \le M \|u\| .\]

Therefore, $A \in \mathcal{L}(X, Y)$.

$\Box$

Proposition 5.2

• $A_{n}, A \in \mathcal{L}(X, Y)$,
• \(\{A_{n}\}_{n \in \mathbb{N}}\),
  • bounded
• $X_{0}$,
  • dense in $X$

\[\forall v \in X_{0}, \ \lim_{n \rightarrow \infty} A_{n}v = Av .\]

Then

\[\mathop{\text{s-lim}}_{n \rightarrow \infty} A_{n} = A .\]

proof

Since $A$ and \(\{A_{n}\}\) are bounded, we can take $M > 0$ such that

\[\|A\|, \|A_{n}\| \le M .\]

We show that

\[\forall u \in X, \ \forall \gamma > 0, \ \exists n_{0} \in \mathbb{N}, \text{ s.t. } \|A_{n} u - Au\| < \gamma \quad (n > n_{0}) .\]

Let $u \in X$ and $\gamma > 0$ be fixed. Since $X_{0}$ is dense in $X$, there exists $v \in X_{0}$ such that

\[\|u - v\| < \frac{ \gamma }{ 3M } .\]

By assumption \(\|A_{n}\| < M\),

\[\begin{eqnarray} \|A_{n} u - A u\| & \le & \|A_{n} u - A_{n}v\| + \|A_{n} u - A_{n}v\| + \|Av - Au\| \nonumber \\ & \le & 2M \|u - v\| + \|A_{n}v - Av\| \nonumber \\ & \le & \frac{2 \gamma}{3} + \|A_{n}v - Av\| \nonumber \end{eqnarray}\]

Since there exists $n_{0}$ such that $\forall n \ge n_{0}$, \(\|A_{n}v - Av\|< \frac{\gamma}{3}\).

$\Box$

Example 5.2

TBD

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Definition strong continuity

• $X$,
  • norm space
• $T_{a}:X \rightarrow \mathbb{R} \ a \in \mathbb{R}$

$T_{a}$ is said to be strong continuous if for all $a \in \mathbb{R}$

\[\mathop{s-lim}_{a^{\prime} \rightarrow a}T_{a^{\prime}} = T_{a} .\]

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Example 5.3

• $X := L^{p}(\mathbb{R})$,
• $1 \le p < \infty$,
• $T_{a}:X \rightarrow \mathbb{R} \ a \in \mathbb{R}$,

\[(T_{a}u)(x) := u(x - a) .\]

Then

• (1) $T_{a} \in \mathcal{L}(X)$, \(\|T_{a}\| = 1\),
• (2)

\[\mathop{s-lim}_{a^{\prime} \rightarrow a} T_{a^{\prime}} = T_{a} .\]

proof of (1)

This is ovbious.

proof of (2)

Let $X_{0} := C_{0}(\mathbb{R})$. Then $X_{0}$ is dense in $X$. Now, let $R > 0$ be fixed. There exists $v \in X_{0}$ such that

\[|x| > R, \ v(x) = 0 .\]

Since $v$ is uniformly continuous, $T_{a^{\prime}}v$ uniformly converge to $T_{a}v$. Indeed,

\[\forall \epsilon > 0, \ \exists \delta > 0, \ |a^{\prime} - a| < \epsilon \ \Rightarrow \ v(x - a^{\prime}) - v(x - a) < \epsilon .\]

On the other hand,

\[|a^{\prime} - a| < 1, \ \{x \in \mathbb{R} \mid |x - a| \le R + 1\}\]

■

. $$

TBD

$\Box$