A question about uniformly bounded set of operators

Let $\ (E, \lVert \cdot \rVert_E) \$ and $\ (F , \lVert \cdot \rVert_F ) \$ be real normed vector spaces, $\ \big( \mathcal{L}(E,F) , \lVert \cdot \rVert_L \big) \$ be the real normed vector space of continuous linear transformations from $E$ to $F$ with the usual norm induced by $\ \lVert \cdot \rVert_E \$ and $\ \lVert \cdot \rVert_F \$ and $\ \mathcal{C} \subset \mathcal{L}(E,F) \$. Suppose that $\ \mathcal{C} \neq \varnothing \$ and $\mathcal{C}$ is (uniformly) bounded, ie, there exists $\ R > 0 \$ such that $\ \mathcal{C} \subset B_L (0_L , R) \subset \mathcal{L}(E,F)$. Where $\ 0_L : E \to F \$ is the zero function ⏤ the origin of $\ L = \mathcal{L}(E,F)$ ⏤ and $\ B_L (0_L , R) \$ is the open ball of $L$ centered in $0_L \,$.

Let $\ \alpha > 0 \,$. I want to know if (with possibly some restrictions on $\alpha$) the set $$G = \bigcap_{T \in \mathcal{C}} T^{-1} \big[ B_F(0_F, \alpha) \big]$$ is open in $E$.

Here are my efforts.

First it is obvious that $\ T^{-1} \big[ B_F(0_F, \alpha) \big] \$ is an open set of $E$, $\forall T \in \mathcal{C} \,$. Hence the answer is clearly positive in the case where $\mathcal{C}$ is finite.

From now on let us suppose that $\, \mathcal{C} \,$ is infinite.

So we have that $L$ is infinite and $\ \mathcal{C} \setminus \{ 0_L \} \neq \varnothing \,$. It follows that $\ \mathcal{C} \neq \{ 0_L \}$, $\ E \neq \{ 0_E \} \$ and $\ F \neq \{ 0_F \}$.

Obs.: when $\mathcal{C}$ is countable, the set $G$ is a $G_{\delta}$ of $E$.

I was able to prove that:

• $\ B_E \left( 0_E , \alpha/R \right) \subset G \$;
• For all $\ x \in G \$ and all $\ T \in \mathcal{C} \,$, one has $\ \lVert T(x) \rVert_F < \alpha \$;
• For all $\ x \in E \$ and all $\ T \in \mathcal{C} \,$, one has $\ \lVert T(x) \rVert_F \leq \lVert T \rVert_L \cdot \lVert x \rVert_E < R \cdot \lVert x \rVert_E \$.

Let $\ D = \Big\{ \lVert T(x) \rVert_F \in \mathbb{R}_{^+} \ : \ T \in \mathcal{C} \ \ \text{ and } \ \ x \in G \Big\} \,$. Since $\ 0_E \in G \$ and $\ \mathcal{C} \neq \varnothing \,$, we have that $\ 0 \in D \$. Thus $\ D \neq \varnothing \,$. Moreover, by the second item above, $D$ is bounded. So there exists the supremum (least upper bound) of $D$, say $\ s = \sup(D) = \text{lub}(D) \,$, with $\ 0 \leq s \leq \alpha \,$.

Here is a digression: actually $\ s>0 \,$.

Suppose that $\ s = 0 \$. Then one has $\ D = {0} \,$. Let $\ x \in G \,$. We have $\ \lVert T(x) \rVert_F = 0 \ \Rightarrow \ T(x) = 0_F \ \Rightarrow \ x \in ker(T)$, $\forall T \in \mathcal{C} \,$. It follows that $\ x \in \bigcap_{T \in \mathcal{C}} ker(T) \,$. Consequently, $\ B_E \left( 0_E , \alpha/R \right) \subset G \subset \bigcap_{T \in \mathcal{C}} ker(T) \,$. Now $\ \bigcap_{T \in \mathcal{C}} ker(T) \$ is a vector subspace of $E$ that contains an open ball. Whence $\ \bigcap_{T \in \mathcal{C}} ker(T) = E \,$. Thereafter, for all $\ T \in \mathcal{C} \,$, we have that $\ E = \bigcap_{A \in \mathcal{C}} ker(A) \subset ker(T) \ \Rightarrow \ ker(T) = E \ \Rightarrow \ T = 0_L \,$. So $\ \mathcal{C} = \{ 0_L \} \,$, a contradiction with the fact that $\mathcal{C}$ is infinite.

Similarly, fix $\ p \in G \$ and let $\ D_p = \Big\{ \lVert T(p) \rVert_F \in \mathbb{R}_{^+} \ : \ T \in \mathcal{C} \Big\} \,$. Since $\ \mathcal{C} \neq \varnothing \,$, we have that $\ D_p \neq \varnothing \,$. Moreover, $\forall T \in \mathcal{C}$, again by the second item above, we have that $\ \lVert T(p) \rVert_F < \alpha \,$. Hence $D_p \,$ is bounded. So there exists the supremum $\ s_p = \sup(D_p) \,$, with $\ 0 \leq s_p \leq \alpha \,$. Since $\ D_p \subset D \,$, we have that $\ 0 \leq s_p \leq s \leq \alpha \,$. Indeed it is easy to show that $\ \displaystyle \sup_{p \in G} s_p = s \,$.

If $\ p = 0_E \,$, then $\ D_p = \{ 0 \} \ \Rightarrow \ s_p = 0 \,$. But we can choose $\ p \neq 0_E \,$, since $\ B_E \left( 0_E , \alpha/R \right) \subset G \$ and $\ B_E \left( 0_E , \alpha/R \right) \neq \{ 0_E \} \,$, as $\ E \neq \{ 0_E \} \,$. In that case, $\ 0 < s_p \leq s \leq \alpha \,$. Like above, again we proved that $\ s>0 \,$.

My first attempt was restricting $\ \alpha \neq s \,$. Then $\ 0 < s_p \leq s < \alpha \,$ and we have that $$0< \frac{\alpha - s}{R} \leq \frac{\alpha - s_p}{R} \ \ .$$

But then I am afraid that I arrived at a contradiction. I will explain it below.

Let $\ p \in G \$ and $\ x \in B_E \left( p , \frac{\alpha - s_p}{R} \right) \,$. Hence, for all $\ T \in \mathcal{C} \,$, one has $\ \lVert T(p) \rVert_F \leq s_p \$ and then

\begin{eqnarray*} \lVert T(x) \rVert_F & = & \lVert T(x - p) + T(p) \rVert_F \\ & \leq & \lVert T(x - p) \rVert_F + \lVert T(p) \rVert_F \\ & \leq & \lVert T \rVert_L \cdot \lVert x - p \rVert_E + \lVert T(p) \rVert_F \\ & \leq & \lVert T \rVert_L \cdot \lVert x - p \rVert_E + s_p \\ & < & R \cdot \lVert x - p \rVert_E + s_p \\ & < & R \cdot \frac{(\alpha - s_p)}{R} + s_p \\ &=& \alpha \ . \end{eqnarray*}

Thus $\ x \in T^{-1} \big[ B_F (0_F , \alpha) \big] \,$, $\forall T \in \mathcal{C}$, that is, $x \in G$. Therefore $\ B_E \left( p , \frac{\alpha - s_p}{R} \right) \subset G \ \Rightarrow \ p \in int(G) \,$. Since $p$ is arbitrary, it follows that $G$ is open.

But a similar argument shows a stronger fact:

Let $\ x \in G \$ and $\ u \in B_E \left( x , \frac{\alpha - s}{R} \right) \,$. Hence, for all $\ T \in \mathcal{C} \,$, one has $\ \lVert T(x) \rVert_F \leq s \$ and then

\begin{eqnarray*} \lVert T(u) \rVert_F & = & \lVert T(u - x) + T(x) \rVert_F \\ & \leq & \lVert T(u - x) \rVert_F + \lVert T(x) \rVert_F \\ & \leq & \lVert T \rVert_L \cdot \lVert u - x \rVert_E + \lVert T(x) \rVert_F \\ & \leq & \lVert T \rVert_L \cdot \lVert u - x \rVert_E + s \\ & < & R \cdot \lVert u - x \rVert_E + s \\ & < & R \cdot \frac{(\alpha - s)}{R} + s \\ &=& \alpha \ . \end{eqnarray*}

Thus $\ u \in T^{-1} \big[ B_F (0_F , \alpha) \big] \,$, $\forall T \in \mathcal{C}$, that is, $u \in G$. Therefore $\ B_E \left( x , \frac{\alpha - s}{R} \right) \subset G \,$.

This reasoning gives a broader result because the radius $\ \frac{\alpha - s}{R} \$ do not depend on $x \,$:

For all $\ x \in G \$ we have that $\ B_E \left( x , \frac{\alpha - s}{R} \right) \subset G \,$.

Let $\ x \in E$. If $\ x \in B_E (0_E , \alpha/R)$, then $\ x \in G$. Suppose that $\ x \notin B_E (0_E , \alpha/R) \,$. Then $\ x \neq 0_E \$ and $\ 0 < \frac{\alpha}{2R} < \frac{\alpha}{R} \leq \lVert x \rVert_E \ \Rightarrow \ 2R \lVert x \rVert_E - \alpha > 0 \,$. Let $\ (x_n) \in E^{\mathbb{N}} \$ be such that, $\forall n \in \mathbb{N}$, $$x_n = \frac{[\alpha + n(\alpha - s)]}{2R \lVert x \rVert_E} \cdot x$$ So we have $\ \lVert x_0 \rVert_E = \frac{\alpha}{2R} < \frac{\alpha}{R} \ \Rightarrow \ x_0 \in B_E \left( 0_E , \frac{\alpha}{R} \right) \subset G \ \Rightarrow \ x_0 \in G \ \Rightarrow \ B_E \left( x_0 , \frac{\alpha - s}{R} \right) \subset G \,$. Let $\ N = \Big\{ n \in \mathbb{N} \ : \ B_E \left( x_n , \frac{\alpha - s}{R} \right) \subset G \Big\}$. Hence $\ 0 \in N$. Let $\ n \in N$, ie, $B_E \left( x_n , \frac{\alpha - s}{R} \right) \subset G$. We have that

\begin{eqnarray*} \lVert x_{n+1} - x_n \rVert_E & = & \left\lVert \frac{[\alpha + (n+1)(\alpha - s)]}{2R \lVert x \rVert_E} \cdot x - \frac{[\alpha + n(\alpha - s)]}{2R \lVert x \rVert_E} \cdot x \right\rVert_E \\ & = & \frac{[\alpha + (n+1)(\alpha - s)] - [\alpha + n(\alpha - s)]}{2R} \\ & = & \frac{\alpha - s}{2R} \\ & < & \frac{\alpha - s}{R} \\ & \Rightarrow & x_{n+1} \in B_E \left( x_n , \frac{\alpha - s}{R} \right) \subset G \\ & \Rightarrow & x_{n+1} \in G \\ & \Rightarrow & B_E \left( x_{n+1} , \frac{\alpha - s}{R} \right) \subset G \\ & \Rightarrow & n+1 \in N \ . \end{eqnarray*}

By induction, we conclude that $\ N = \mathbb{N}$, that is, $\ (x_n) \in G^{\mathbb{N}} \,$, with $\ B_E \left( x_n , \frac{\alpha - s}{R} \right) \subset G$, $\forall n \in \mathbb{N}$.

Note that $\ \frac{2R \lVert x \rVert_E - \alpha}{\alpha - s} > 0 \,$. Since $\mathbb{N}$ is unbounded, the set $\ J = \Big\{ n \in \mathbb{N} \ : \ n > \frac{2R \lVert x \rVert_E - \alpha}{\alpha - s} \Big\} \$ is nonempty. Off course we have that $\ 0 \notin J$. Since $\mathbb{N}$ is well-ordered, there exists the least element (minimum) of $J$, say $\ j = \min(J)$, with $\ j \in J$, that is, $j > \frac{2R \lVert x \rVert_E - \alpha}{\alpha - s} > 0 \,$. Thus $\ m=j-1 \in \mathbb{N} \$ and $\ j = m+1 \in \mathbb{N}^*$. By the minimality of $j$ we have either $\ m \notin J \ \Rightarrow \ m \leq \frac{2R \lVert x \rVert_E - \alpha}{\alpha - s} \,$. Therefore $\ \lVert x_m \rVert_E = \frac{\alpha + m(\alpha - s)}{2R} \leq \lVert x \rVert_E < \frac{\alpha + (m+1)(\alpha - s)}{2R} = \lVert x_{m+1} \rVert_E \$ and we are left with $\ \lVert x_m \rVert_E = \frac{\alpha + m(\alpha - s)}{2R} \leq \lVert x \rVert_E < \frac{\alpha + (m+1)(\alpha - s)}{2R} = \frac{\alpha + m(\alpha - s)}{2R} + \frac{\alpha - s}{2R} = \lVert x_{m} \rVert_E + \frac{\alpha - s}{2R} \$, that is, $\ \lVert x_m \rVert_E \leq \lVert x \rVert_E < \lVert x_{m} \rVert_E + \frac{\alpha - s}{2R} \$. Then $\ \lVert x - x_m \rVert_E \leq \big| \lVert x \rVert_E - \lVert x_m \rVert_E \big| < \frac{\alpha - s}{2R} \$, wich implies $\ x \in B_E \left( x_m , \frac{\alpha - s}{R} \right) \subset G \ \Rightarrow \ x \in G$.

As $x$ was arbitrarily chosen, we conclude that $\ G=E$.

Let $\ T \in \mathcal{C} \$ and $\ y \in im(T)$. There exists $\ x \in E \$ such that $\ y = T(x)$. But $\ x \in E = G = \bigcap_{A \in \mathcal{C}} A^{-1} \big[ B_F (0_F , \alpha) \big] \subset T^{-1} \big[ B_F (0_F , \alpha) \big] \$ and we have that $\ y = T(x) \in B_F(0_F, \alpha)$. Then $\ im(T) \subset B_F (0_F , \alpha) \$ and $\, im(T) \,$ is a bounded vector subspace of $F$. Hence $\ im(T) = \{ 0_F \} \ \Rightarrow \ T = 0_L \,$. Therefore we have that $\ \mathcal{C} = \{ 0_L \}$, a contradiction.

Is there anything wrong with this reasoning? Are my calculation steps correct? If it is right, then we cannot have $\ \alpha \neq s \$ and we are left with $\ s= \alpha \,$. That is exactly the case which I was not able to handle.

Like always, any hint is appreciated.

• Let $E = F = \mathbb{R}$ and $\mathcal{C} = \{1 - 1/n \mid 2 \ge 2\}$. Then, with $\alpha = 1$ you get $G = \bigcap_{n \ge 2} (-n/(n-1), n/(n-1)) = [-1,1]$. Or do I miss something? – gerw Jan 9 '17 at 8:44
• @gerw Do you have continuous linear transformations $\ T_n : E \to F \$ such that $$T_n^{-1} \big[ \ ] \! - \alpha , \alpha [ \ \big] = \ \bigg] \frac{-n}{n-1} , \frac{n}{n-1}\bigg[$$ for each $\ n \in \mathbb{N}^*$? – Gustavo Jan 9 '17 at 15:13
• @gerw Ok. Now I see. The set $\ \mathcal{C} = \{ T_n \}_{n \geq 2} \$ is the obvious counterexample, where $$T_n (x) = \frac{\alpha (n-1) x}{n}$$ Thanks – Gustavo Jan 9 '17 at 16:27