Young's Inequality

Theorem

Young's inequality states that for \(a, b > 0\) and \(p, q > 1\) with \(\frac{1}{p} + \frac{1}{q} = 1\)

\[ ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}.\]
Proof

The condition that \(\frac{1}{p} + \frac{1}{q} = 1\) gives motivation for using the definition of convex functions. The fact that \(p, q > 1\) guarantee that \(\frac{1}{p}, \frac{1}{q} \in (0, 1)\) as required, and so we have that for any convex function \(f\) and real numbers \(x_{1}, x_{2}\):

\[ f\left(\frac{1}{p}x_{1} + \frac{1}{q}x_{2}\right) \leq \frac{1}{p}f(x_{1}) + \frac{1}{q}f(x_{2}).\]

Then we choose \(f = -\ln\), since the left hand side looks kind of like what we want, and by choosing \(f\) as above and taking the exponential function, we can take \(\frac{1}{p}\) and \(\frac{1}{q}\) to be powers on the right and end up with a product.

\[\begin{align*} & f\left(\frac{1}{p}x_{1} + \frac{1}{q}x_{2}\right) \leq \frac{1}{p}f(x_{1}) + \frac{1}{q}f(x_{2}) \\ \implies & -\ln\left(\frac{1}{p}x_{1} + \frac{1}{q}x_{2}\right) \leq -\left(\frac{1}{p}\ln(x_{1}) + \frac{1}{q}\ln(x_{2})\right) \\ \implies & \ln\left(\frac{1}{p}x_{1} + \frac{1}{q}x_{2}\right) \geq \frac{1}{p}\ln(x_{1}) + \frac{1}{q}\ln(x_{2}) \\ \implies & \frac{1}{p}x_{1} + \frac{1}{q}x_{2} \geq \exp\left(\ln\left(x_{1}^{\frac{1}{p}}\right) + \ln\left(x_{2}^{\frac{1}{q}}\right)\right) \\ \implies & \frac{1}{p}x_{1} + \frac{1}{q}x_{2} \geq x_{1}^{\frac{1}{p}} x_{2}^{\frac{1}{q}} \\ \end{align*}\]

Now all we must do is let \(x_{1} = a^{p}\) and \(x_{2} = b^{q}\) to get Young's inequality:

\[ ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}.\]

The above proof is somewhat more motivated than the one we include below, however the proof below has the benefit of depending on less machinery.

Proof

Consider the function \(f : \mathbb{R}_{\geq 0} \to \mathbb{R}\) defined by

\[ f(t) = \frac{1}{p} t^p + \frac{1}{q} b^q - t b\]

where \(b \geq 0\) and \(p\) and \(q\) satisfy the conditions of the theorem statement.

We will show that this function is non-negative for all values of \(t \geq 0\). Firstly, notice that the derivative given by

\[ f'(t) = t^{p - 1} - b\]

is zero at exactly \(t = b^{\frac{1}{p - 1}}\) noting that we are only considering positive values of \(t\).

Furthermore, we know that

\[ \frac{1}{p} + \frac{1}{q} = 1 \implies 1 + \frac{p}{q} = p \implies \frac{1}{p - 1} = \frac{q}{p}\]

so this point corresponds with \(t = b^{\frac{q}{p}}\), at which point

\[ f(b^{\frac{q}{p}}) = \frac{1}{p} b^q + \left(1 - \frac{1}{p}\right) b^q - b^{\frac{q}{p} + 1} = 0.\]

Furthermore, the function is convex everywhere it is defined since

\[ f''(t) = (p - 1)t^{p - 2} > 0, \quad \forall t > 0.\]