Simple question on linear mappimg $\phi : E \rightarrow F$ between two normed vector space $(E; \| . \|_E)$ and $(F; \| . \|_F)$

Question

Question

Let a linear mappimg $\phi : E \rightarrow F$ between two normed vector space $(E; \| . \|_E)$ and $(F; \| . \|_F)$.
a) Prove that in $ \mathbb{R}^+ \cup \infty $. We have:
$$ \| \phi \| =_{def} Sup \{ \frac{ || \phi(x) ||_E}{ \| x \|_F} : x \in E - \{ 0\} \} = Sup \{ || \phi(x) ||_F : \| x \|_E \leq 1 \} = Sup \{ || \phi(x) ||_F : \| x \|_E = 1 \} = min \{ C \in \mathbb{R} \; verifying \; \forall x \in E , \| \phi(x) \|_F \leq C \| x \|_E \}$$
b) And that $ \phi $ is continuous iff $ \| \phi \| < \infty$

Answer

a)
1- Using the fact that $\phi$ is a linear mapping $\frac{\| \phi(x) \|_E}{\| x \|_F} = \| \phi(\frac{x}{\| x\|_F}) \|_E $
2- By definition $\| \frac{x}{\| x\|_F} \|_E$ is a normalized vector of $F$ + the basic definition of $Sup$ we have that $ \| \phi(\frac{x}{\| x\|_F}) \|_E \leq Sup \{ \| \phi(y) \|_E : \| y \|_F = 1 \} \leq Sup \{ \| \phi(y) \|_E : \| y \|_F \leq 1 \} $
3- We continue and we get $ Sup \{ \| \phi(y) \|_E : \| y \|_F \leq 1 \} = Sup \{ \| \phi( \frac{y}{ \| y \|_F }) \|_E : \| y \|_F \leq 1 \} = Sup \{ \| \frac{\phi(y)}{ \| y \|_F } \|_E : \| y \|_F \leq 1 \} \leq Sup \{ \| \frac{\phi(y)}{ \| y \|_F } \|_E : \| y \|_F \in E - \{ 0\} \} $
4- We sum up everything and we finally get $$ \| \phi(\frac{x}{\| x\|_F}) \|_E \leq Sup \{ \| \phi(y) \|_E : \| y \|_F = 1 \} \leq Sup \{ \| \phi(y) \|_E : \| y \|_F \leq 1 \} = Sup \{ \| \phi( \frac{y}{ \| y \|_F }) \|_E : \| y \|_F \leq 1 \} = Sup \{ \| \frac{\phi(y)}{ \| y \|_F } \|_E : \| y \|_F \leq 1 \} \leq Sup \{ \| \frac{\phi(y)}{ \| y \|_F } \|_E : \| y \|_F \in E - \{ 0\} \} $$

5- And this give the two first equality. Now we want to prove the last equality $ = min \{ C \in \mathbb{R} \; verifying \; \forall x \in E , \| \phi(x) \|_F \leq C \| x \|_E \} $ but I do not succeed to do it. Anyone has an idea ? Or can help me?

Thank you for your help.

Answer 1

Let $C_0=\inf\{C \in \mathbb{R}:\forall x \in E, \|\phi x\| \leq C \|x\|\}$. This is equivalent to $$I:=\left \{C \in \mathbb{R}:\forall x \in E-\{0\}, \frac{\|\phi x\|}{\|x\|} \leq C \right\} \quad C_0=\inf I$$ since any $C \in \mathbb{R}$ satisfies the inequality in the first definition for $x=0$. I will prove it is equal to $$\|\phi\|=\sup \left \{ \frac{\|\phi x\|}{\|x\|} : \forall x \in E-\{0\} \right \}$$ We know that $I$ is non-empty since $\|\phi\| \in I$. $$\forall x \in E-\{0\},\forall C \in I \quad \frac{\|\phi x\|}{\|x\|} \leq C \implies \forall C \in I \quad \|\phi\| \leq C$$ which means $\|\phi\| \leq \inf I = C_0$. This establishes the first inequality. For the second inequality, note that for any $\epsilon>0$, $C_0 - \epsilon \notin I$ since otherwise $C_0-\epsilon \geq \inf I=C_0$. Thus $\exists x_0 \in E-\{0\}$ s.t. $$\frac{\| \phi x_0\|}{\|x_0\|} > C_0 - \epsilon \quad \implies \quad \|\phi\| > C_0 - \epsilon$$ which means $C_0 \leq \|\phi\|$ since $\epsilon$ was arbitrary.

Answer 2

Just to complete my answer regarding the part $b)$

Answer to b)

1- ($ \| \phi \| < \infty \Rightarrow \phi $ is continuous )
In such a case $ \phi $ is Lipschitz and thus continuous. Indeed by the linearity of $ \phi $ we can write according to the inequalities we wrote above that $ \forall x, y \in E \Rightarrow || \phi(x) - \phi(y) ||_F = || \phi(x-y) ||_F \leq || \phi || || x - y ||_E$

2- ( $ \phi$ is continuous $ \Rightarrow \| \phi \| < \infty $ )
If $ \phi $ is continuous she is too continuous at $0$ and so by definition $ \forall \epsilon >0 , \exists \delta>0 : \forall x \in B(0; \delta) \Rightarrow || \phi (x) || _F < \epsilon$ and if this is true for all $ \epsilon$ this is particularly true when $ \epsilon =1 $ and thus $\exists \delta_1>0: \| x \|_E < \delta \Rightarrow || \phi (x) || _F \leq 1 $
Now we continue and we note $ || \phi (x) || = || \phi(x \cdot\frac{\delta_1}{||x||} \cdot \frac{||x||}{\delta_1} ) || = \frac{||x||}{\delta_1}|| \phi( x \cdot\frac{\delta_1}{||x||} ) || \leq 1 \cdot \frac{||x||}{\delta_1} $.
In other words we get $|| \phi (x) || \leq \frac{||x||}{\delta_1}$ and this is exactly the definition of bounded for an operator.

Q.E.D.