# Two weird proofs about continuity in normed vector spaces

I am reading a pair of "proofs" that a friend sent to me. I really don't understand some passages, so I hope someone could help me. The questions are the following

First Question. The result to be proved is: let $G$ and $H$ normed vector spaces and let $f\colon G \to H$ a linear map. If $f$ is continuous in a point $x_0$, then it is continuous everywhere.

The proof goes as $\forall \varepsilon >0, \exists \delta>0$ s.t. $\|(x_0+u)-x_0\|<\delta$ iff $\|u\|<\delta$, and that implies $\|f(x_0+u)-f(x_0)\|=\|f(u)\|<\varepsilon$...

I don't believe that this is a proof, I can't really see what has been proved. Should I assume that the proof is wrong? I know a proof of this theorem but it involves sequential continuity, any idea on how to prove it using an $\varepsilon-\delta$ argument?

Second Question. The result to be proved is: let $G$ and $H$ normed vector spaces and let $f\colon G \to H$ a linear continuous map. Then $f$ is bounded in norm.

The proof says: If $f$ is continuous in $x$, then it is continuous at every point, in particular in $u$. So, if $f$ is continuous in $u$, then for all $\varepsilon>0$ there exists $\delta$ s.t. $$\|u\|<\delta \implies \|f(u)\|<\varepsilon \iff \|f(\frac{d}{\|x\|}x)\|<\varepsilon$$

where $u=\frac{d}{\|x\|}x$. So linearity of $f$ implies $\|f(x)\|<\|x\|(\varepsilon/d)$. QED

Here I don't understand why it uses that $f$ is continuous in $x$: all I see is that it is really using the fact that it is continuous in $0$ (I know that these are equivalent, but it's the second that we are using in the proof).

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It looks like he is assuming that $f$ is continuous at $x=0$, and then showing it must be continuous at any $x_0$. But it's badly phrased.
Here's a correct version: Let $f$ be continuous at $0$. Let $\epsilon>0$. Choose $\delta>0$ such that if $\Vert x-0\Vert<\delta$, then $\Vert f(x)-f(0)\Vert<\epsilon$. Then if $\Vert x_0-x\Vert<\delta$, we have $\Vert f(x_0-x) - f(0)\Vert<\epsilon$. But $\Vert f(x_0-x)-f(0)\Vert =\Vert f(x_0)-f(x)\Vert$. From this, it follows that $f$ is continuous at $x_0$.
For your second question; yes, all you need is continuity at $0$ (and the "$d$" there, should be "$\delta$").