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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.

Thanks in advance.

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  • $\begingroup$ 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? $\endgroup$ – gerw Jan 9 '17 at 8:44
  • $\begingroup$ Why should one expect an infinite intersection of open sets to be open ? $\endgroup$ – user42761 Jan 9 '17 at 9:05
  • $\begingroup$ @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}^*$? $\endgroup$ – Gustavo Jan 9 '17 at 15:13
  • $\begingroup$ @AndréS. I heard someone saying that this case was true, I wanted to be sure about it. But now I'm suspicious and I'm trying to find counterexamples. $\endgroup$ – Gustavo Jan 9 '17 at 15:18
  • $\begingroup$ @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 $\endgroup$ – Gustavo Jan 9 '17 at 16:27

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