Prove or disprove the following. (Recall that $c_{00}$ is the space of sequences with only finitely many nonzero entries.)
Conjecture. Let $1<p<q<\infty$. Then there exists a function $\varphi:\mathbb{Z}^+\to\mathbb{R}^+$ such that
$\displaystyle\lim_{m\to\infty}\varphi(m)=\infty$,
and satisfying the following property. For each $m\in\mathbb{Z}^+$, there exists $\epsilon>0$ such that if $(\alpha_k)\in c_{00}$ and $\lVert(\alpha_k)\rVert_{\ell_q}\leq\epsilon\lVert(\alpha_k)\rVert_{\ell_p}$, then $\varphi(m)\lVert(\alpha_k)\rVert_{\ell_p}\leq\lVert(\alpha_k)\rVert_{\ell_1}$.
Discussion. I suspect the conjecture is false. However, I can't seem to find a counterexample.
If instead we try to prove the conjecture true, then without loss of generality we can assume $\varphi(m)=m$ for large $m$. We need to find $\epsilon>0$ with the desired property. Let $(\alpha_k)\in c_{00}$ satisfy $\lVert(\alpha_k)\rVert_{\ell_q}\leq\epsilon\lVert(\alpha_k)\rVert_{\ell_p}$, and without loss of generality assume $|\alpha_k|\geq 1$ whenever $\alpha_k\neq 0$. Then
$\lVert(\alpha_k)\rVert_{\ell_p}^p\leq\lVert(\alpha_k)\rVert_{\ell_q}^q\leq\epsilon^q\lVert(\alpha_k)\rVert_{\ell_p}^q\leq\epsilon^q\lVert(\alpha_k)\rVert_{\ell_p}^{p-1}\lVert(\alpha_k)\rVert_{\ell_1}^{1+q-p}$
and hence
$\epsilon^{-q}\lVert(\alpha_k)\rVert_{\ell_p}\leq\lVert(\alpha_k)\rVert_{\ell_1}^{1+q-p}$.
Of course, this is not good enough.
Any ideas would be much appreciated. Thanks!
This is true. Let $\lambda\in (0,1)$ be such that $$\frac{1}{p} = \frac{1-\lambda}{1} + \frac{\lambda}{q} \tag{1}$$ By Hölder's inequality, $$ \sum |\alpha_k|^p = \sum |\alpha_k|^{(1-\lambda)p} |\alpha_k|^{ \lambda p} \le \left(\sum |\alpha_k| \right)^{(1-\lambda)p} \left(\sum |\alpha_k|^q \right)^{ \lambda p/q} \tag{2}$$ which can be stated as $$\|\alpha\|_p\le \|\alpha\|_1^{1-\lambda} \|\alpha\|_q^{\lambda}\tag{3}$$ The inequality (3) is the standard $L^p$ interpolation inequality.
If $\|\alpha\|_q \le \epsilon \|\alpha\|_p$, then (3) implies $$\|\alpha\|_p\le \epsilon^{\lambda} \|\alpha\|_1^{1-\lambda} \|\alpha\|_p^{\lambda}$$ hence $$\|\alpha\|_p\le \epsilon^{\lambda/(1-\lambda)} \|\alpha\|_1 \tag{4}$$ which is the kind of inequality you want.