# $A$ is continuous in $0_X$ => $\exists c \in \mathbb{R}$, $c \geq 0$ : $||Ax||_Y \leq c ||x||_X$

the map $$A : X \rightarrow Y$$ is linear, where $$(X,||.||)$$ and $$(Y,||.||)$$ are normalized vector spaces.

I already have a solution, which is correct

but a friend of mine showed me his solution and asked me if his solution is correct. I couldn't tell him if his solution is right. So I have decided to ask here for help.

So this is his attempt:

proof by contradiction: assume $$\nexists c \in \mathbb{R}$$ , $$c \geq 0$$ with $$||Ax||_Y \leq c ||x||_X$$. Then we can contruct a sequence $$(x_n)_n$$ with $$||x_n|| = 1$$ $$\forall n \in \mathbb{N}$$ and $$||Ax_n|| > n \mbox{ } \forall n \in \mathbb{N}$$. It follows $$y_n := (\frac{x_n}{||Ax_n||})_n \rightarrow 0$$. ( because the denominator gets bigger and bigger but $$x_n$$ stagnates at $$1$$. ) Now use continuity in $$0$$ : It follows that $$||Ay_n|| \rightarrow 0 = ||A0||$$, but $$|| Ay_n|| = || A (\frac{x_n}{||Ax_n||}) || = \frac{1}{||Ax_n||}||Ax_n|| =1$$. Contradiction!

Is his proof correct? I need your feedback.

• What gave you the thoughts that it might be incorrect? – Stan Tendijck May 22 at 16:25
• Yes, it looks fine. – Mark May 22 at 16:26
• The solution looks fine to me (to answer your question). – Stan Tendijck May 22 at 16:27
• Thank you! Dear Stan I didn't think that his solution was wrong, but I was really, really unsure to be honest. Dear uniquesolution. I work with the $\epsilon$ $\delta$ definition. So no proof by contradiction. I choose $\epsilon = 1$ and showed that assumption. But I have to admit. My solution is way longer. – RukiaKuchiki May 22 at 16:34