# Is there a linear function that is *not* continuous between two normed vector space?

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.

-

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

## 1 Answer

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 $f = T, a = e_1, x = \frac{\epsilon}{2} e_3$.)

-

@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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9584107398986816, "perplexity_flag": "head"}