Linear transformation of a normed linear space to another normed linear space..

Question

Suppose a linear transformation $\Lambda$ of a normed linear space $X$ into a normed linear space $Y$. Im trying to prove that,,,.. If $\Lambda$ is continuous at one point of $X$ then $ \Lambda$ is bounded.

Proof: Suppose $\Lambda$ is continuous at $x_0$. To each $\epsilon>0$ there exist $\delta>0$ so that $\lVert x-x_0 \rVert<\delta$ implies $\lVert \Lambda x-\Lambda x_0 \rVert<\epsilon$. In other words, $\lVert x \rVert<\delta$ implies $\lVert\Lambda (x+x_0)-\Lambda x_0\rVert<\epsilon$. But then the linearity of the transformation $\Lambda$ shows that $\lVert\Lambda x\rVert<\epsilon$. Hence, $\lVert\Lambda\rVert\leq \epsilon/\delta$. This completes the proof.

I don t know how the last part of the proof was obtained.

Answer 1

If $x \neq 0$ then $y=(1-\frac 1 n) \delta \frac x { \|x\|}$ is a vector whose norm is $(1-\frac 1 n) \delta <\delta$. Hence $\|\Lambda y\|<\epsilon$. By linearity this gives $\|\Lambda x\|<\frac {\epsilon} {1-\frac 1 n} \frac 1 {\delta}\|x\|$. $\,\,\,$ (1)Let $n \to \infty$. [ $x=\|x\| (1-\frac 1 n)^{-1} \delta ^{-1} y$ so $\Lambda (x)=\|x\| (1-\frac 1 n)^{-1} \delta ^{-1} \Lambda ( y)$. Since $|\Lambda (y)|<\epsilon$ we get (1)].

Answer 2

One way to see this is looking at the definition using vectors of norm 1. For all $0<t<1$, we have that $$\|\Lambda\|=\text{sup}\{\|\Lambda x\|: x\in X, \|x\|=1\}=\text{sup}\{\|\Lambda\frac{x}{t\delta}\|:x\in X,\|x\|=t\delta\}=\text{sup}\{\frac{1}{t\delta}\|\Lambda x\|:x\in X,\|x\|=t\delta\}=\frac{1}{t\delta}\text{sup}\{\|\Lambda x\|:x\in X,\|x\|=t\delta\}<\frac{\epsilon}{t\delta}.$$

Note that this is enough to show that $\Lambda$ is bounded. To get the result at the end of the above proof, we may take a limit as $t\to 1^-$.