# Equivalent conditions for a linear map between two normed spaces to be continuous everywhere

## Statement

Given two normed spaces [ilmath](X,\Vert\cdot\Vert_X)[/ilmath] and [ilmath](Y,\Vert\cdot\Vert_Y)[/ilmath] and a linear map [ilmath]L:X\rightarrow Y[/ilmath] between them, then the following are equivalent (meaning if you have 1 you have all the others):

1. If we have a sequence [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] with [ilmath]x_n\rightarrow 0[/ilmath] then [ilmath](\Vert L(x_n)\Vert_Y)_{n=1}^\infty[/ilmath] is a bounded sequence[1][Note 1]
2. [ilmath]L[/ilmath] is continuous at a particular point in [ilmath]X[/ilmath][1][2][Note 2]
3. [ilmath]L[/ilmath] is a bounded linear map, that is [ilmath]\exists A>0\ \forall x\in X[\Vert L(x)\Vert_Y\le A\Vert x\Vert_X][/ilmath][1][2]
4. [ilmath]L[/ilmath] is continuous everywhere[1][2]

## Proof

[ilmath]1)\implies 2)[/ilmath]: That [ilmath]L[/ilmath] maps all null sequences to bounded sequences [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] is continuous at [ilmath]p\in X[/ilmath]

This is a proof by contrapositive. That is we will show that if [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath] [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] takes a null sequence to one that isn't bounded (an unbounded one).

• Let the normed spaces [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be given, as well as a linear map [ilmath]L:X\rightarrow Y[/ilmath]
• Suppose that [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath], this means:
• [ilmath]\exists (x_n)_{n=1}^\infty\rightarrow p[/ilmath] such that $\lim_{n\rightarrow\infty}\left(L(x_n)\right)\ne L(p)$

Recall that continuity states that:

• [ilmath]L[/ilmath] is continuous at [ilmath]p[/ilmath] [ilmath]\iff[/ilmath] $\forall(x_n)_{n=1}^\infty\subseteq X \left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\implies\left(\lim_{n\rightarrow\infty}\left(L(x_n)\right)=L\left(\lim_{n\rightarrow\infty}(x_n)\right)=L(p)\right)\right]$

So it follows that to not be continuous at [ilmath]p[/ilmath]:

• $\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\neg\left(\lim_{n\rightarrow\infty}(L(x_n))=L(p)\right)\right]$, by negation of implies. Additionally we may negate the [ilmath]=[/ilmath] and thus we see this is the same as:
• $\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\left(\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)\right)\right]$

Which is exactly "there exists a sequence in [ilmath]X[/ilmath] whose limit is [ilmath]p[/ilmath] and where [ilmath]\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)[/ilmath]"

• Let us now take [ilmath]L(x_n)\not\rightarrow L(p)[/ilmath] and subtract [ilmath]L(p)[/ilmath] from both sides. We see:
• [ilmath]L(x_n)-L(p)\not\rightarrow L(p)-L(p)[/ilmath], using the fact that [ilmath]L[/ilmath] is linear we see that:
• [ilmath]L(x_n-p)\not\rightarrow L(0)[/ilmath] and [ilmath]L(0)=0\in Y[/ilmath] so:
• [ilmath]L(x_n-p)\not\rightarrow 0[/ilmath]
• Thus [ilmath]\Vert L(x_n-p)\Vert_Y\not\rightarrow 0[/ilmath] (as [ilmath]\Vert0\Vert_Y=0[/ilmath] by definition)
• So $\exists C>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge\Vert L(x_n-p)\Vert_Y>\epsilon]$

If a sequence converges to [ilmath]0[/ilmath] then we have:

• [ilmath]\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon][/ilmath] we know we don't have this, so we negate it:
• [ilmath]\exists\epsilon>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge d(x_n,x)>\epsilon][/ilmath].

That is there exists an [ilmath]\epsilon>0[/ilmath] such that for all [ilmath]N[/ilmath] there exists a bigger [ilmath]n[/ilmath] such that [ilmath]d(x_n,x)>\epsilon[/ilmath], we shall later call such an [ilmath]\epsilon[/ilmath] [ilmath]C[/ilmath] and construct a subsequence out of the [ilmath]n[/ilmath]s

• Thus it is possible to construct a subsequence, [ilmath](\Vert L(x_{n_k}-p)\Vert_Y)_{k=1}^\infty[/ilmath] of the image [ilmath](x_n)[/ilmath] where for every [ilmath]k[/ilmath] we have:
• [ilmath]\Vert L(x_{n_k}-p)\Vert_Y>C[/ilmath]

By the negation of convergent sequence we see that there exists a [ilmath]C[/ilmath] such that for all [ilmath]N\in\mathbb{N} [/ilmath] there exists another natural number, [ilmath]n[/ilmath] such that [ilmath]\Vert L(x_n-p)\Vert_Y>C[/ilmath], we construct a sequence, [ilmath](n_k)_{k=1}^\infty[/ilmath] of such [ilmath]n[/ilmath]-ns. That is we know there exists a [ilmath]C[/ilmath], so we pick [ilmath]N=1[/ilmath] and get the [ilmath]n[/ilmath] that works, this is our first term, we then set [ilmath]N=2[/ilmath] and the [ilmath]n[/ilmath] that works is our second term, and so forth.

Some of these [ilmath]n[/ilmath]s may be the same, but that doesn't matter.

[ilmath](x_{n_k})[/ilmath] is the subsequence of [ilmath](x_n)[/ilmath] which contains only the terms that satisfy [ilmath]\Vert L(x_n-p)\Vert_Y> C[/ilmath]

It would have been better if I used [ilmath]n[/ilmath] and [ilmath]m[/ilmath] as terms, to make which [ilmath]n[/ilmath] I am talking about clearer, but a reader able to attempt this proof should follow.

• We now have a sequence [ilmath](x_{n_k})[/ilmath] such that [ilmath]\Vert L(x_{n_k}-p)\Vert_Y>C[/ilmath]
• Define a new sequence $b_k:=\frac{1}{\sqrt{\Vert x_{n_k}-p\Vert} }$
• It is easy to see that [ilmath]b_k\rightarrow +\infty[/ilmath] (as [ilmath](x_{n_k}-p)\rightarrow 0[/ilmath])

TODO: Prove that this tends to [ilmath]+\infty[/ilmath]

• Define a new sequence $d_k:=b_k(x_{n_k}-p)$
• Clearly $d_k\rightarrow 0$

If [ilmath](d_k)_{k=1}^\infty\rightarrow 0[/ilmath] then [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies\Vert d_k-0\Vert_X<\epsilon][/ilmath].

• Notice $\Vert d_k\Vert_X=b_k\Vert x_{n_k}-p\Vert_X=\frac{\Vert x_{n_k}-p\Vert_X}{\sqrt{\Vert x_{n_k}-p\Vert_X} }$ $=\sqrt{\Vert x_{n_k}-p\Vert_X}$

TODO: Finish this proof

• But $\Vert L(d_k)\Vert_Y=\Vert L(b_k(x_{n_k}-p))\Vert_Y=b_k\Vert L(x_{n_k}-p)\Vert_Y\ge Cb_k\rightarrow +\infty$
• Thus we have shown if [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath] that the mapping of a null sequence is unbounded, the contrapositive of what we set out to claim

[ilmath]2)\implies 3)[/ilmath]: [ilmath]L[/ilmath] is continuous at [ilmath]p\in X[/ilmath] [ilmath]\implies[/ilmath] [ilmath]\exists A\ge 0\ \forall x\in X[\Vert L(x)\Vert_Y \le A\Vert x\Vert_X][/ilmath] (That is that [ilmath]L[/ilmath] is a bounded linear map)

The key to this proof is exploiting the linearity of [ilmath]L[/ilmath]. As will be explained in the blue box.

• Suppose that [ilmath]L:X\rightarrow Y[/ilmath] is continuous at [ilmath]p\in X[/ilmath]. Then:
• [ilmath]\forall\epsilon>0\ \exists\delta>0[\Vert x-p\Vert_X<\delta\implies\Vert L(x-p)\Vert_Y<\epsilon][/ilmath] (note that [ilmath]L(x-p)=L(x)-L(p)[/ilmath] due to linearity of [ilmath]L[/ilmath])
• Define [ilmath]u:=x-p[/ilmath], then:
• [ilmath]\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon][/ilmath] (writing [ilmath]Lu:=L(u)[/ilmath] as is common for linear maps)

We now know we may assume that [ilmath]\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon][/ilmath] - see this box for a guide on how to use this information and how I constructed the rest of the proof. The remainder of the proof follows this box.

The key to this proof is in the norm structure and the linearity of [ilmath]L[/ilmath].

• Pick (fix) some [ilmath]\epsilon>0[/ilmath] we now know:
• [ilmath]\exists\delta>0[/ilmath] such that if we have [ilmath]\Vert x\Vert_X<\delta[/ilmath] then we have [ilmath]\Vert Lx\Vert<\epsilon[/ilmath]

In the proof we will have to show at some point that: [ilmath]\forall x\in X[\Vert Lx\Vert_Y\le A\Vert x\Vert_X][/ilmath] this means:

• At some point we'll be given an arbitrary [ilmath]x\in X[/ilmath]

However we have a [ilmath]\delta[/ilmath] such that [ilmath]\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon[/ilmath]

• All we have to do is make [ilmath]\Vert x\Vert_X[/ilmath] so small that it is less than [ilmath]\delta[/ilmath] - we can do this using scalar multiplication.
• Note that if [ilmath]x=0[/ilmath] the result is trivial, so assume that arbitrary [ilmath]x[/ilmath] is [ilmath]\ne 0[/ilmath]
• With this in mind the task is clear: we need to multiply [ilmath]x[/ilmath] by something such that the actual vector part has [ilmath]\Vert\cdot\Vert_X<\delta[/ilmath]
• Then we can say (supposing [ilmath]x=\alpha p[/ilmath] for some positive [ilmath]\alpha[/ilmath] and [ilmath]\Vert p\Vert_X<\delta[/ilmath]) that [ilmath]\Vert L(\alpha p)\Vert_Y=\alpha\Vert Lp\Vert_Y[/ilmath]
• But [ilmath]\Vert p\Vert_X<\delta\implies\Vert Lp\Vert_Y<\epsilon[/ilmath]
• So [ilmath]\Vert L(\alpha p)\Vert_Y=\alpha\Vert Lp\Vert_Y<\alpha\epsilon[/ilmath], if we can get a [ilmath]\Vert x\Vert_X[/ilmath] involved in [ilmath]\alpha[/ilmath] the result will follow.

Getting an arbitrary [ilmath]\Vert x\Vert_X[/ilmath] to have magnitude [ilmath]<\delta[/ilmath]

1. Lets normalise [ilmath]x[/ilmath] and then multiply it by the magnitude again (so as to do nothing)
• So notice $x=\overbrace{\Vert x\Vert_X}^\text{scalar part}\cdot\overbrace{\frac{1}{\Vert x\Vert_X}x}^\text{vector part}$,
2. Now the vector part has magnitude [ilmath]1[/ilmath] we can get it to within [ilmath]\delta[/ilmath] by multiplying it by $\frac{\delta}{2}$
• So $x=\overbrace{\Vert x\Vert_X\cdot\frac{2}{\delta} }^\text{scalar part}\cdot\overbrace{\frac{\delta}{2\Vert x\Vert_X}x}^\text{vector part}$.

Now we have $\left\Vert\frac{\delta}{2\Vert x\Vert_X}x\right\Vert_Y<\delta$ which $\implies \left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon$

• Thus $\Vert Lx\Vert_Y=\frac{2\Vert x\Vert_X}{\delta}\cdot\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\frac{2\Vert x\Vert_X}{\delta}\cdot\epsilon=\frac{2\epsilon}{\delta}\Vert x\Vert_X$

But [ilmath]\epsilon[/ilmath] was fixed, and so was the [ilmath]\delta[/ilmath] we know to exist based off of this, so set [ilmath]A=\frac{2\epsilon}{\delta}[/ilmath] after fixing them and the result will follow!

• Fix some arbitrary [ilmath]\epsilon>0[/ilmath] (it doesn't matter what)
• We know there [ilmath]\exists\delta>0[\Vert x\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon][/ilmath] - take such a [ilmath]\delta[/ilmath] (which we know to exist by hypothesis) and fix it also.
• Define [ilmath]A:=\frac{2\epsilon}{\delta}[/ilmath] (if you are unsure of where this came from, see the blue box)
• Let [ilmath]x\in A[/ilmath] be given (this is the [ilmath]\forall x\in X[/ilmath] part of our proof, we have just claimed an [ilmath]A[/ilmath] exists on the above line)
• If [ilmath]x=0[/ilmath] then
• Trivially the result is true, as [ilmath]L(0_X)=0_Y[/ilmath], and [ilmath]\Vert 0_Y\Vert_Y=0[/ilmath] by definition, [ilmath]A\Vert x\Vert_X=0[/ilmath] as [ilmath]\Vert 0_X\Vert_X=0[/ilmath] so we have [ilmath]0\le 0[/ilmath] which is true. (This is more workings than the line is worth)
• Otherwise ([ilmath]x\ne 0[/ilmath])
• Notice that $x=\frac{2\Vert x\Vert_X}{\delta}\cdot\frac{\delta}{2\Vert x\Vert_X}x$ and $\left\Vert \frac{\delta}{2\Vert x\Vert_X}x\right\Vert_X<\delta$
• and $\left\Vert \frac{\delta}{2\Vert x\Vert_X}x\right\Vert_X<\delta\implies\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon$
• Thus we see $\Vert Lx\Vert_Y=\left\Vert L\left(\frac{2\Vert x\Vert_X}{\delta}\cdot\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y=\frac{2\Vert x\Vert_X}{\delta}\cdot\overbrace{\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y}^{\text{remember this is }<\epsilon}$ $<\frac{2\Vert x\Vert_X}{\delta}\cdot\epsilon=\frac{2\epsilon}{\delta}\Vert x\Vert_X=A\Vert x\Vert_X$
• Shortening the workings this states that: $\Vert Lx\Vert_Y< A\Vert x\Vert_X$
• So if [ilmath]x=0[/ilmath] we have equality, otherwise $\Vert Lx\Vert_Y< A\Vert x\Vert_X$

In either case, it is true that $\Vert Lx\Vert_Y\le A\Vert x\Vert_X$
This completes the proof

[ilmath]3)\implies 4)[/ilmath]: [ilmath]\exists A\ge 0\ \forall x\in X[\Vert L(x)\Vert_Y \le A\Vert x\Vert_X][/ilmath] (That is that [ilmath]L[/ilmath] is a bounded linear map) [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] is continuous at every point

This is a straight forward implication proof. We suppose by hypothesis that [ilmath]\exists A\ge 0\ \forall x\in X[\Vert L(x)\Vert_Y \le A\Vert x\Vert_X][/ilmath] is true.
We wish to show that [ilmath]L[/ilmath] is continuous everywhere, this means [ilmath]\forall x\in X[L\ \text{is continuous at }x][/ilmath], I will use sequential continuity in this proof (that is: [ilmath]\forall(x_n)_{n=1}^\infty\rightarrow x[\lim_{n\rightarrow\infty}(L(x_n))=L(x)][/ilmath], or the image of the sequence under [ilmath]L[/ilmath] converges to the image of [ilmath]x[/ilmath] under [ilmath]L[/ilmath])

• Let [ilmath]x\in X[/ilmath] be given (arbitrary point)
• Let [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] be given (an arbitrary sequence that converges to [ilmath]x[/ilmath])

Remember that to converge means [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon][/ilmath] and in our normed space:

• [ilmath]d(x_1,x_2):=\Vert x_1-x_2\Vert_X[/ilmath] (see metric induced by norm), so what we're really saying is:
• [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies \Vert x_n-x\Vert_X<\epsilon][/ilmath]
• Let [ilmath]\epsilon >0[/ilmath] be given (we will show that $\left\Vert L(x_n)-L(x)\right\Vert_Y<\epsilon$)
• Choose [ilmath]N[/ilmath] such that $\forall n\in\mathbb{N}\left[n>N\implies\Vert x_n-x\Vert_X<\frac{\epsilon}{A}\right]$ (Where [ilmath]A[/ilmath] is the [ilmath]A[/ilmath] in the hypothesis of [ilmath]L[/ilmath] being a bounded linear map)

We will want to show at some point that $\lim_{n\rightarrow\infty}\left(L(x_n)\right)=L(x)$ this means we'll need to do something along the lines of:

• [ilmath]\Vert L(x_n)-L(x)\Vert_Y=\Vert L(x_n-x)\Vert_Y\le A\Vert x_n-x\Vert_X[/ilmath]

If we pick [ilmath]N[/ilmath] such that $\Vert x_n-x\Vert_X<\frac{\epsilon}{A}$ then:

• [ilmath]\Vert L(x_n)-L(x)\Vert_Y=\Vert L(x_n-x)\Vert_Y\le A\Vert x_n-x\Vert_X[/ilmath]$<A\frac{\epsilon}{A}=\epsilon$

and that'll complete the proof.
Note: we know such an [ilmath]N[/ilmath] exists as by definition $\lim_{n\rightarrow\infty}(x_n)=x$

• Let [ilmath]n\in\mathbb{N} [/ilmath] be given
• If [ilmath]n\le N[/ilmath] then we're done. By the truth table of implies if the RHS ([ilmath]\Vert L(x_n)-L(x)\Vert_Y<\epsilon[/ilmath]) is true or false, the implication is true, so this literally doesn't matter.
• If [ilmath]n>N[/ilmath] then:
• $\Vert L(x_n)-L(x)\Vert_Y=\overbrace{\Vert L(x_n-x)\Vert_Y}^\text{by linearity}$$\le\overbrace{A\Vert x_n-x\Vert_X}^{\text{boundedness of }L}$$<A\frac{\epsilon}{A}=\epsilon$
• Thus we have shown $n>N\implies \Vert L(x_n)-L(x)\Vert_Y<\epsilon$ as required.

That completes the proof

[ilmath]4)\implies 1)[/ilmath]: [ilmath]L[/ilmath] is continuous at every point [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] maps all null sequences to bounded sequences

There is actually a slightly stronger result to be had here, I shall prove that, to which the above statement is a corollary.

• Let [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] be a sequence that converges - we must show that the image of this sequence under [ilmath]L[/ilmath] is bounded
• As [ilmath]L[/ilmath] is everywhere continuous we know that it is sequentially continuous at [ilmath]x[/ilmath]. This means that:
• $\forall\left((x_n)_{n=1}^\infty\rightarrow x\right)\left[\big(L(x_n)\big)_{n=1}^\infty\rightarrow L(x)\right]$
• So [ilmath]L[/ilmath] maps [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] to $\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(x)$
• Recall that if a sequence converges it is bounded, as $\left(L(x_n)\right)_{n=1}^\infty$ converges, it must therefore be bounded.

Corollary: the image of every null sequence under [ilmath]L[/ilmath] is bounded

This completes the proof.

## Notes

1. this may be better said (perhaps) as "[ilmath]L[/ilmath] maps all sequences convergent to [ilmath]0[/ilmath] to bounded sequences"
2. Gamelin and Green's Introduction to Topology only demonstrates continuity at the point [ilmath]0\in X[/ilmath]