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Let $X=C([0,1])$ equipped with the norm $\Vert\cdot \Vert=\max_{x\in [0,1]}|f(x)|$. If $L:X\rightarrow X$ is linear, is $L$ continuous?

If not, what if $Lf\ge 0$ for $f\ge0$ for $\forall x \in [0,1]$ is assumed?

Edit: Rephrased the question.

Edit2: Attempt,

Let, $\Vert f_n \Vert \rightarrow 0$ when $n\rightarrow \infty$ and $f_n=\sum_{i=0}^{\infty}\alpha^{(n)}_i x^{i} $ then $\Vert Lf_n\Vert=\max_{x\in [0,1]}|\sum_{i=0}^{\infty}\alpha^{(n)}_i Lx^{i}| \rightarrow 0$ Because $\forall \alpha^{(n)}_i \rightarrow 0$ when $n\rightarrow \infty$.

But i don't know if this is allowed?

Stanley5664
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  • It's not clear what map you're talking about. Do you mean $$ L(f) = \sup_{x \in [0,1]} f(x)? $$ – Ben Grossmann Sep 23 '15 at 15:30
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    If the domain is infinite dimensional and incomplete, then there exist discontinuous linear maps. These can often be explicitly constructed; for example if we give both $C^1$ and $C^0$ the sup norm, then $f \mapsto f'$ is a discontinuous linear map. If the domain is infinite dimensional and complete, then we can use the axiom of choice to nonconstructively prove that there exist discontinuous linear maps. – Ian Sep 23 '15 at 15:31
  • @Omnomnomnom I think "the" was a non-native-speaker error. – Ian Sep 23 '15 at 15:31
  • @Ian but there's a question of whether we should have an absolute value in there. If we want the sup "norm", then we should have $\sup |f(x)|$ instead. – Ben Grossmann Sep 23 '15 at 15:32
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    @Omnomnomnom I think the question can be rephrased as "let $X=C([0,1])$ equipped with the sup norm. If $L : X \to X$ is linear, is $L$ continuous?" – Ian Sep 23 '15 at 15:33
  • @Ian thank you for that – Ben Grossmann Sep 23 '15 at 15:42
  • @Ian Thank you for the rephrase suggestion. – Stanley5664 Sep 23 '15 at 15:43

2 Answers2

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Let $X$ be a normed vector space and $Y$ be a normed vector space, $Y \neq \{ 0 \}$. If $X$ is infinite dimensional and incomplete, then there exist discontinuous linear maps $L : X \to Y$. Often, these can be explicitly constructed. For instance, if $X=C^1$ with the sup norm and $Y=C^0$ with the sup norm, then $L(f)=f'$ is a discontinuous linear map, since $f_n=\frac{1}{n} \sin(n^2x)$ converges in $C^1$ but $L(f_n)$ does not converge in $C^0$.

If $X$ is infinite dimensional and complete, then one can use the axiom of choice to nonconstructively prove that there exist discontinuous linear maps $L : X \to Y$. $C([0,1])$ with the sup norm is infinite dimensional and complete, so this case applies to your question.

Ian
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  • But can one use the same process to prove that there exists a discontinuous linear map satisfying $Lf \geq 0$ whenever $f \geq 0$? – Ben Grossmann Sep 23 '15 at 15:46
  • Is this the case also if the Linear operator has the property $Lf\ge0$ for $f\ge 0$ for $∀x∈[0,1]$? – Stanley5664 Sep 23 '15 at 15:47
  • In fact, I think that one may construct a basis consisting of non-negative operators, which would make such a map possible. I am not exactly sure how one would prove the existence of such a basis, however. – Ben Grossmann Sep 23 '15 at 15:48
  • What if expanding the functions in polynomials, Use the linearity and try to prove that the operator is continuous at f=0? – Stanley5664 Sep 23 '15 at 15:58
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Show that $\Vert Lf\Vert\le\alpha$ for all $\Vert f \Vert=1$.

Because

$1\ge \pm f, \forall x \in [0,1]$

we have that

$1\pm f\ge 0$,

thus,

$L(1\pm f)\ge 0$.

By linearity

$-L1\le Lf\le L1$,

that is

$|Lf|\le L1, \forall x \in [0,1]$,

hence

$\max_{x\in[0,1]}|Lf|= \Vert Lf \Vert \le \max_{x\in [0,1]}L1=\alpha$.

Because boundness gives continuity the conclusion follows.

Shervin
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