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Given a linear operator on normed vector spaces $T:\mathcal{D}(T)\subset X \rightarrow Y$, if it is continuous at one point $x_0 \in \mathcal{D}(T)$ then it is continuous for all points in $\mathcal{D}(T)$.

I am following Kreyszig intro to functional analysis, he uses boundedness iff continuous to prove this, which is a fine proof. I came up with another proof that I am curious to see if it is correct:

Since $T$ is continuous at $x_0$ for all $\epsilon > 0$ there exists $\delta>0$ such that $$||x_0-x||<\delta \implies ||Tx_0 - Tx|| < \epsilon.$$ By translation invariance of the norm (edit: here I mean the translation invariance of the induced distance under the norm), this implies $$||x_0+d-(x+d)||<\delta \implies ||T(x_0+d) - T(x+d)|| < \epsilon.$$ Since $d\in \mathcal{D}(T)$ is arbitrary, $x_0+d$ can be element in $\mathcal{D}(T)$ that we like. Hence, $T$ is continuous at $x_0+d$, whatever it may be.

I am worried about this proof, because it doesn't seem to use linearity?

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    @jcm your proof is fine. Ignore comments showing identical proofs with some variable substitution claiming there is something wrong with yours. –  Jun 03 '21 at 02:19
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    Your proof is fine, but you did use linearity. Assume that $|(x_0+d)-(x+d)|<\delta$. This is equivalent to $|x_0-x|<\delta$. Then this implies that $\epsilon>|Tx_0-Tx|=|Tx_0 +Td - Tx - Td|=|T(x_0+d)-T(x+d)|$. In the last equality you used a little less than linearity. You used additivity. – plop Jun 03 '21 at 02:27
  • That shows that rather than linearity what was important is that $T$ was a group homomorphism between the topological groups $(D(X),+)$ and $(Y,+)$. See here – plop Jun 03 '21 at 02:35
  • Right, @plop I see where I implicitly used linearity. Thanks. –  Jun 03 '21 at 02:59

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"By translation invariance of the norm, this implies..." No it doesn't. Your hypothesis is that $T$ is continuous at $x_0$. You cannot use a different $x_0$ in

If $\|x_0-x\|<\delta \implies \|Tx_0 - Tx\| < \epsilon$.

But it works if you use linearity. If $\|x_1-x\|<\delta$, then $\|x_0-(x_0-x_1+x)\|<\delta$. Then $$ \|Tx_0-T(x_0-x_1+x)\|<\epsilon, $$ which the linearity of $T$ lets you write as $$ \|Tx_1-Tx\|<\epsilon. $$

Martin Argerami
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