# Does Young's inequality hold only for conjugate exponents?

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.

Edit:

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.

• 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? – Asaf Shachar Jan 18 at 10:51
• 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) – Asaf Shachar Jan 18 at 10:55
• 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. – 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.

• 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. – Asaf Shachar Jan 16 at 13:19
• @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. – 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.