Young's Inequality

Theorem

Young's inequality states that for a,b>0 and p,q>1 with 1p+1q=1

abapp+bqq.

Proof

The condition that 1p+1q=1 gives motivation for using the definition of convex functions. The fact that p,q>1 guarantee that 1p,1q(0,1) as required, and so we have that for any convex function f and real numbers x1,x2:

f(1px1+1qx2)1pf(x1)+1qf(x2).

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 1p and 1q to be powers on the right and end up with a product.

f(1px1+1qx2)1pf(x1)+1qf(x2)ln(1px1+1qx2)(1pln(x1)+1qln(x2))ln(1px1+1qx2)1pln(x1)+1qln(x2)1px1+1qx2exp(ln(x11p)+ln(x21q))1px1+1qx2x11px21q

Now all we must do is let x1=ap and x2=bq to get Young's inequality:

abapp+bqq.

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:R0R defined by

f(t)=1ptp+1qbqtb

where b0 and p and q satisfy the conditions of the theorem statement.

We will show that this function is non-negative for all values of t0. Firstly, notice that the derivative given by

f(t)=tp1b

is zero at exactly t=b1p1 noting that we are only considering positive values of t.

Furthermore, we know that

1p+1q=11+pq=p1p1=qp

so this point corresponds with t=bqp, at which point

f(bqp)=1pbq+(11p)bqbqp+1=0.

Furthermore, the function is convex everywhere it is defined since

f(t)=(p1)tp2>0,t>0.