Show that there exists a linear functional $T:\mathcal{D}(Q)\rightarrow\mathbb{C}$ which is not continuous.

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.

• @nicomezi thank you, edited it now – Infinitebig Feb 15 at 9:42
• 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. – peastick Feb 15 at 19:12
• @peastick but that series doesn't necessarily converge in $\mathbb{C}$ because we add up infinitely many terms – Infinitebig Feb 16 at 9:18
• Maybe math.stackexchange.com/questions/288075/… can help? – md2perpe Feb 16 at 22:44
• 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. – md2perpe Feb 17 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.
• 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 $B$? – Infinitebig Feb 17 at 19:11
• maybe my question was not clear, I meant that how do you find the sequence $a_n>0$ so that $a_n x_n\rightarrow 0$? – Infinitebig Feb 17 at 19:19
• If $U_n$ are a decreasing basis of $0$-neighbourhoods you need $a_n x_n \in U_n$. – Jochen Feb 17 at 19:23