How to construct a linear functional on the space of test functions $\mathcal{D}(Q)$ which is not continuous? In other words, how to find a linear map $T:\mathcal{D}(Q)\rightarrow\mathbb{C}$ such that there is $\varphi_k\rightarrow 0$ in $\mathcal{D}(Q)$ but $T(\varphi_k)\nrightarrow 0$ in $\mathbb{C}$? Here $Q=[-\pi, \pi]^n$ is the $n$-cube, $\mathcal{D}(Q)$ consists of smooth functions which are $2\pi$-periodic functions in all variables and convergence in $\mathcal{D}(Q)$ is defined to be uniform, i.e. $\lVert \partial^{\alpha }\varphi_k\rVert_{L^{\infty}}\rightarrow 0$ for all $\alpha\in\mathbb{N}^n$. Is there an explicit way to do this or does it require some form of axiom of choice etc.?

I tried to use the Fourier expansion $$ \varphi(x)=\sum_{k\in\mathbb{Z}^n}\hat{\varphi}(k)e^{ik\cdot x} $$ and to use the fact that $\{e^{ik\cdot x}\}$ is kind of orthonormal basis in $\mathcal{D}(Q)$. Therefore it would be enough to define $T$ only for basis vectors $e_k$ and extend $T$ by linearity to all $\varphi$. But I cannot find a good sequence $\varphi_k$ such that all its derivatives go uniformly to zero and at the same time, for example $|T(\varphi_k)|=1$.

Any help is appreciated.

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    $\begingroup$ @nicomezi thank you, edited it now $\endgroup$ – Infinitebig Feb 15 '19 at 9:42
  • $\begingroup$ I think I found an answer to this question. There is a theorem which says that all distributions, i.e. continuous linear functionals on $\mathcal{D}(Q)$ have finite order (I quess because $Q$ is compact). Therefore it is enough to find a linear functional which does not have finite order. For this we can use delta distribution and its derivatives, for example $T=\sum_{i=1}^{\infty}\delta^{(i)}_{\frac{1}{i}}$. Because $\delta^{(i)}$ has order $i$, $T$ cannot have finite order and thus it cannot be continuous. $\endgroup$ – peastick Feb 15 '19 at 19:12
  • $\begingroup$ @peastick but that series doesn't necessarily converge in $\mathbb{C}$ because we add up infinitely many terms $\endgroup$ – Infinitebig Feb 16 '19 at 9:18
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    $\begingroup$ Maybe math.stackexchange.com/questions/288075/… can help? $\endgroup$ – md2perpe Feb 16 '19 at 22:44
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    $\begingroup$ I know, but I think that the method there is not very dependent on the space having a norm. Therefore I thought that the method might give an idea, although it cannot be copied straight off. $\endgroup$ – md2perpe Feb 17 '19 at 8:45

For every infinite dimensional metrizable topological vector space there are discontinuous linear functionals: Fix a sequence $x_n$ of distinct elements belongin to some basis $B$. Using metrizability we find $a_n>0$ such that $a_nx_n\to 0$. Then we define $f(x_n)=1/a_n$, $f(b)=0$ for $b\in B\setminus\{x_n: n\in\mathbb N\}$ and extend it by linearity to the whole space.

  • $\begingroup$ what do you mean by "using metrizability we find $a_n>0$ such that $a_n x_n\rightarrow 0$"? and this clearly needs axiom of choice in the form of Zorn's lemma because we pick a basis $B$? $\endgroup$ – Infinitebig Feb 17 '19 at 19:11
  • $\begingroup$ Metrizability means that there is a metric inducing the topology. And yes, one needs choice. $\endgroup$ – Jochen Feb 17 '19 at 19:15
  • $\begingroup$ maybe my question was not clear, I meant that how do you find the sequence $a_n>0$ so that $a_n x_n\rightarrow 0$? $\endgroup$ – Infinitebig Feb 17 '19 at 19:19
  • $\begingroup$ If $U_n $ are a decreasing basis of $0$-neighbourhoods you need $a_n x_n \in U_n $. $\endgroup$ – Jochen Feb 17 '19 at 19:23

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