I would like to to draw your attention on a problem that my class mate posted earlier, showing you our idea for a possible solution.
Let $p>1, s>0$ and $F$ be a function such that $$F(t)\le \frac{|t|^{p-p^{\prime}}}{p^{\prime}}e^{|t|^{p^{\prime}}}\quad\mbox{ for all } t\in\mathbb{R},$$ where $p^{\prime}$ denotes the conjugate exponent of $p$. We are wondering if the quantity $$\frac{|t|^p}{s+1} e^{|t|^{p^{\prime}}} -2N F(t)$$ is bounded from below. We guess the answer would be yes because we have $$\frac{|t|^p}{s+1} e^{|t|^{p^{\prime}}} -2N F(t)\ge |t|^p e^{|t|^{p^{\prime}}}(1/(s+1)-2(N-1))|t|^{-p^{\prime}}$$ which seems to me it is bounded from below.
Could someone please tell me if are we wrong or not? Thank you in advance.
The claim only holds for $N \geq 0$ and $p \geq 2$.
To see why you need $N \geq 0$, note that if $N < 0$ then $$\frac{|t|^p}{s+1}e^{|t|^{p'}} - 2NF(t) = \frac{|t|^p}{s+1}e^{|t|^{p'}} + CF(t)$$ for $C = -2N > 0$ and since there is no lower bound on $F(t)$ one cannot say whether or not this expression is bounded below. For example, let $F(t) = -1/|t|$ and choose $t$ close to 0.
Now lets look at the more subtle issue. If $p \in (1,2)$ then it's conjugate satisfies $$ p' = \frac{p}{p-1} \implies (p-1)p' = p \implies p' > p $$ where the last implication is from the fact that $p-1 \in (0,1)$. Now using this we have for $p \in 1,2$ that $p-p' < 0$. From here we have $$\lim_{t \rightarrow 0} \frac{|t|^{p-p'}}{p'} e^{|t|^{p'}} \geq \frac{1}{p'} \lim_{t\rightarrow 0} |t|^{p-p'} = \infty$$ since $e^{|t|^{p'}} \geq 1$ and $p-p' < 0$.
Now for $t \in [-1,1]$ we have $$\frac{|t|^p}{s+1} e^{|t|^p} \leq \frac{1}{s+1}e^1 = \frac{e}{s+1}.$$ Now consider the function when $F(t) = \frac{|t|^{p-p'}}{p'}e^{|t|^{p'}}$, which satisfies the upper bound in the assumption (just with equality everywhere). Then \begin{align} \lim_{t\in[-1,1], t\rightarrow 0} \frac{|t|^p}{s+1} e^{|t|^p} - 2NF(t) &= \lim_{t\in[-1,1], t\rightarrow 0} \frac{|t|^p}{s+1} e^{|t|^p} - 2N\frac{|t|^{p-p'}}{p'}e^{|t|^{p'}} \\ &\leq \lim_{t\in[-1,1], t\rightarrow 0} \frac{e}{s+1} - 2N\frac{|t|^{p-p'}}{p'}e^{|t|^{p'}} \\ &= \frac{e}{s+1} - 2N\lim_{t \in [-,1,1], t \rightarrow 0} \frac{|t|^{p-p'}}{p'} e^{|t|^{p'}} \\ &= -\infty \end{align} since the last limit was already shown to go to $+\infty$ and we have $N > 0$. This shows, even though the function isn't well-defined at $t = 0$, by choosing $t$ close enough to 0 it can be driven arbitrarily low. Or in other words, the quantity is not bounded below.
Now when $p \geq 2$ this problem doesn't happen since $p' \leq p$. In particular, by the assumptions $$\frac{|t|^p}{s+1}e^{|t|^{p'}} - 2NF(t) \geq \frac{|t|^p}{s+1}e^{|t|^{p'}} - 2N \frac{|t|^{p-p'}}{p'}e^{|t|^{p'}}.$$ This lower bound is a continuous function. Let's check where it can be negative \begin{align} \frac{|t|^p}{s+1}e^{|t|^{p'}} - 2N \frac{|t|^{p-p'}}{p'}e^{|t|^{p'}} &\leq 0 \\ \iff \frac{|t|^p}{s+1}e^{|t|^{p'}} &\leq 2N \frac{|t|^{p-p'}}{p'}e^{|t|^{p'}} \\ \iff \frac{|t|^p}{s+1} &\leq 2N \frac{|t|^{p-p'}}{p'} \\ \iff |t|^{p'} &\leq \frac{2N(s+1)}{p'} \\ \iff |t| &\leq \left ( \frac{2N(s+1)}{p'} \right )^{1/p'} \end{align} Outside the interval $\left [-\left ( \frac{2N(s+1)}{p'} \right )^{1/p'}, \left ( \frac{2N(s+1)}{p'} \right )^{1/p'} \right ]$ the lower bound is further bounded below by zero. On this interval, the lower bound is a continuous function on a compact interval which implies it achieves a finite minimum (this is a standard analysis fact). In particular let it's minimum on the interval be $M > -\infty$. Then we have the lower bound is further bounded by $M^* = \min(0,M)$. Remembering we are lower bounding a lower bound we have $$\frac{|t|^p}{s+1}e^{|t|^{p'}} - 2NF(t) \geq \frac{|t|^p}{s+1}e^{|t|^{p'}} - 2N \frac{|t|^{p-p'}}{p'}e^{|t|^{p'}} \geq M^*$$ which shows that the quantity of interest is indeed lower bounded.