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.

  • $\begingroup$ What gave you the thoughts that it might be incorrect? $\endgroup$ – Stan Tendijck May 22 at 16:25
  • $\begingroup$ Yes, it looks fine. $\endgroup$ – Mark May 22 at 16:26
  • $\begingroup$ The solution looks fine to me (to answer your question). $\endgroup$ – Stan Tendijck May 22 at 16:27
  • $\begingroup$ 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. $\endgroup$ – RukiaKuchiki May 22 at 16:34

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