Proving an infinite norm minimization problem has finite support (non-convex p-norms)

If $p=1$, $N=1$ and $a_1=(1/2,2/3,3/4,4/5,\ldots)$, the infimum equals 1 and is not achieved on a finitely supported vector (moreover, it is not achieved at all).

However if $0<p<1$ and the minimizer $x$ exists, it must have finite support (namely, of size at most $N$). To prove this, assume the contrary. Without loss of generality $x_1,\dots,x_{N+1}$ are positive. Choose a non-zero vector $b=(b_1,\dots,b_{N+1},0,0,\dots)$ orthogonal to all $a_i$'s. Choose small $t$ so that $x_i-t|b_i|>0>0$ for all $i=1,\dots,N+1$. Then by concavity of the function $x^p$ we have $$x_1^p+\ldots+x_{N+1}^p> \frac12\left((x_1+tb_1)^p+\ldots+(x_{N+1}+tb_{N+1})^p+\\+(x_1-tb_1)^p+\ldots+(x_{N+1}-tb_{N+1})^p\right).$$ Therefore one of the vectors $x\pm tb$ has smaller $p$-norm than $x$.

For $p>1$ the claim is completely false. Say, if $p=2$, $N=1$, the minimum is achieved on the vector proportional to $a_1$.

If one use Lagrange multipliers, there exist $\mu_1, \mu_2 , \cdots \mu_N$ such that $$ \begin{cases} p|x^*(i)|^{p-1}=\sum_{n=1}^N \mu_n a_n(i) \quad\text{ or}\\ |x^*(i)|=0\end{cases}$$ for all $i$. If $0<p<1$, and for every $n$, $\lim_{i\rightarrow \infty}a_n(i)= 0$ and then $$ \lim_{i\rightarrow \infty}\big(\frac{1}{p}\sum_n \mu_n a_n(i)\big)^{1/(p-1)}=\infty$$ but obviously $x^*(i)$ is bounded so $x^*(i)=0$ for large $i$.