Suppose that $ab \leq \frac{1}{p}a^p+\frac{1}{q}b^q$ holds for every real numbers $a,b\ge 0$. (where $p,q>0$ are some fixed numbers).

Is it true that $ \frac{1}{p}+\frac{1}{q}=1$?

I guess so, and I would like to find an easy proof of that fact. Plugging in $a=b=1$, we get $ \frac{1}{p}+\frac{1}{q}\ge1$. Is there an easy way to see that the converse inequality must hold?

To be clear, I am not looking for proofs of Young's inequality; only for a way to see why the relation between the conjugate exponents is the only possible option.


Here is a comment to help future readers (including future me) to see how can one come with Nicolás's nice idea to apply the inequality for $a=\lambda^{\frac{1}{p}}, b=\lambda^{\frac{1}{q}}$.

The idea is that it is inconvenient to compare a sum with another number; By using the specific choice of $a,b$ above, the sum is simplified, since the scales of the auxiliary parameter $\lambda$ in both summands are now identical.

Comment: I know that the relation $ \frac{1}{p}+\frac{1}{q}=1$ is necessary for Holder's inequality in general; this can be seen by scaling the measure. However, I don't think this approach is applicable here.


If you apply the inequality for $a=\lambda^{\frac{1}{p}}, b=\lambda^{\frac{1}{q}}$, you get something like $$ \lambda^{\frac{1}{p}+\frac{1}{q}}\leq \left( \frac{1}{p}+\frac{1}{q} \right)\lambda, \qquad \forall \lambda>0. $$ If you take $\lambda \to \infty$, it's clear that $\frac{1}{p}+\frac{1}{q}\leq 1$. Otherwise, after dividing by $\lambda$ you get $$ \lambda^{\frac{1}{p}+\frac{1}{q}-1} \leq \frac{1}{p}+\frac{1}{q}, $$ which is false for $\lambda$ large enough, as the right side is constant. Taking $\lambda \to 0$ gets the other inequality.

  • $\begingroup$ Thank you for this nice answer. Can you say something on "how did you came to think about this approach"? I thought of one possible motivation-the one I added to the question's body, which is the following: It is inconvenient to compare a sum with another number; By using your specific choice of $a$ and $b$, the sum is simplified, since the scales of the auxiliary parameter $\lambda$ in both summands are now identical. I wonder whether this was your line of thinking as well, or did you have a different "inspiration" on your mind? $\endgroup$ – Asaf Shachar Jan 18 at 10:51
  • $\begingroup$ What interests me is the fact that by scaling the measure, one can see that in general, Holder's inequality (whose proof crucially relies upon Young's inequality) can only hold when $p,q$ are conjugates. Is there a connection between your approach and that story? (For the Holder's story you can see here: math.stackexchange.com/a/2757556/104576) $\endgroup$ – Asaf Shachar Jan 18 at 10:55
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    $\begingroup$ Yeah, my idea was almost what you said. I didn't recall Hölder's proof, but I read your comment saying "you can show the necessity of $\frac{1}{p}+\frac{1}{q}=1$ by scalling the measure", so I tried to do the same. Here, "scalling the measure" was "multiply by an appropiate factor". Starting with Young's inequality for $a, b \in \mathbb{R}^+$, I tried to multiply by a factor that allowed my to factor at the right side (as $a \mapsto a\lambda^{1/p}, b \mapsto b\lambda^{1/q}$). Then $a$ and $b$ weren't important, so I set $a=b=1$ and we have the solution above. $\endgroup$ – Nicolás Vilches Jan 19 at 14:39

I think the following can help.

By AM-GM $$\frac{1}{p}a^p+\frac{1}{q}b^q\geq\left(\frac{1}{p}+\frac{1}{q}\right)\left(\left(a^p\right)^{\frac{1}{p}}\left(b^q\right)^{\frac{1}{q}}\right)^{\frac{1}{\frac{1}{p}+\frac{1}{q}}}=\left(\frac{1}{p}+\frac{1}{q}\right)(ab)^{\frac{1}{\frac{1}{p}+\frac{1}{q}}}.$$ The equality occurs, of course.

  • $\begingroup$ Thanks, but I do not understand; How exactly do you deduce from that something on the sum $\frac{1}{p}+\frac{1}{q}$? I tried putting $a^p=b^q$ (this is when we have equality in the AM-GM inequality) but I don't see how it helps. $\endgroup$ – Asaf Shachar Jan 16 at 13:19
  • $\begingroup$ @Asaf Shachar The equality, which I got is true for all positives $a$, $b$, $p$ and $q$. If we'll assume that $\frac{1}{p}+\frac{1}{q}\neq1$ then your inequality would wrong. $\endgroup$ – Michael Rozenberg Jan 16 at 13:27

I found an argument that works for $p,q\ge 1$, since I need convexity of $x\to x^p,x^q$. Assume by contradiction that $1/p+1/q>1$. Then, if $p'$ is the conjugate of $q$ we know that $p'>p$.

Now, for any $a>0$ we have $$ a^{p'}/p'=\max\limits_{b>0}\{ab-b^q/q\} $$ (just because $a\to a^{p'}/p'$ is the conjugate (Legendre transform) of $b\to b^q/q$). Then we can pick $a$ large enough such that $$ \frac{a^p}{p}<\frac{a^{p'}}{p'} $$ and for such $a$ we would have $$ a^{p}/p<\max\limits_{b>0}\{ab-b^q/q\} $$ which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.

Hope this helps.


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