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The textbook says that this function has to be continuous at least in the origin for it to be continuous everywhere. But how is it possible that a function is already linear but somehow not continuous?

For example, $E$ and $F$ are two normed vector spaces. $f:E\rightarrow F$ is a linear function. Obviously we know that $f(0) = 0$. Now, for a non-zero vector $a$ in $E$, as long as $f(a)$ has a definition, say $b=f(a)$ for some $b\in F$. Then for however small $\epsilon$, as long as $\lVert x\rVert<\lVert a\rVert\frac{\epsilon}{\lVert b\rVert}$, we have $\lVert f(x)\rVert<\epsilon$. So it seems that this function is continuous at the origin without stipulating it.

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I don't understand your argument. You make no use of $b$. – Qiaochu Yuan May 27 '12 at 4:23
Sorry, typos! I published it before I proofread it. Now I'm doing it. – Voldemort May 27 '12 at 4:25
Is it worth pointing out that for any linear $f: E \to F$, and any $v$ in $E$, the limit $\lim_{t \to 0} f(tv)$ will exist and be $0$? This is certainly true. A second and far more nontrivial true statement is that if $E$ is finite dimensional then every any linear map from $E$ to any normed space $F$ will be continuous. Probably, some blend of these two facts is where your intuition is coming from. But this intuition does not (and cannot) lead to a proof of a general statement, as examples like Qiaochu's show. – leslie townes May 27 '12 at 4:51
up vote 1 down vote accepted

It's a standard lemma that a linear operator $f : B \to C$ between two normed spaces is continuous if and only if it is bounded in the sense that the image of the unit ball in $B$ is bounded. It is easy to write down unbounded linear operators. For example, let $B = C$ be the subspace of compactly supported sequences in $\ell^1(\mathbb{Z})$ with basis $e_i, i \in \mathbb{Z}$ and consider the linear operator defined by $T(e_i) = i e_i$.

It is simply false that $||x|| < ||a|| \frac{\epsilon}{||b||}$ implies $||f(x)|| < \epsilon$. (Take $f = T, a = e_1, x = \frac{\epsilon}{2} e_3$.)

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@Voldemort: $\ell^1(\mathbb{Z})$ is the space of all functions $f_n : \mathbb{Z} \to \mathbb{C}$ such that $\sum |f_n|$ converges equipped with the norm $\sum |f_n|$. The compactly supported sequences in $\ell^1(\mathbb{Z})$ are the sequences with only finitely many nonzero terms; this is spanned as a vector space by the sequences $(e_i)_n = \delta_{in}$ (which are equal to $1$ if $i = n$ and equal to $0$ otherwise). – Qiaochu Yuan May 27 '12 at 4:57

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