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.
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.