Why this sequence of functions goes to zero?

Question

I have the function given by

$$_1\phi_k(x) = \frac{1}{k}e^{\left(\frac{-1}{1 - x^2}\right)} \operatorname{for } x\in (-1,1)$$ and $0$ elsewhere. This is a function in test-space and in that space this semi-norm is whell defined

$$\Vert _1\phi_k(x)\Vert_{m,q} = \operatorname{sup}\{(1 + \vert x \vert)^m\vert_1\phi_k^{(q)}(x)\vert,x \in \mathbb{R}\}$$

My question is that I have to prove that

$$\lim_{k \rightarrow \infty} \Vert _1\phi_k(x)\Vert_{m,q} = 0$$

My attempt was try to construct the $q$-dependence of the derivative and, after that, find the supremum. But I couldn't find any closed form, and also couldn't find the form of the polynomials in this form of the derivatives of the exponential. How can I compute the supremum? Is this necessary or I could use some inequality and the squeeze theorem?

Answer 1

hint: Per your question's title, $\dfrac{-1}{1-x^2} < 0$ for $-1 < x < 1$, thus $0 < \phi_k < \dfrac{1}{k}$.

Answer 2

Write $_1\phi_k(x) = \frac 1k\varphi(x)$, which gives you $$\|_1\phi_k\|_{m,q} = \frac 1k\|\varphi\|_{m,q}, $$ then notice that $\|\varphi\|_{m,q}$ is finite and does not depend on $k$.