Power Functions and the Direction of Jensen’s Inequality
Jensen’s inequality turns convexity into a comparison between a function of an average and an average of function values. For power functions, the direction depends on the exponent: \(x^r\) is convex when \(r\ge1\), but concave when \(0<r<1\). That change reverses the inequality. Keeping track of the total weight or measure is equally important: the cleanest statement is for an average with total mass one, and finite-measure formulas follow by normalizing.
We use the integral form of Jensen’s inequality established in the earlier material on Integral Jensen Inequality With Proof. In the form needed here, if \((X,\mathcal{M},P)\) is a probability space, \(f:X\to I\) is measurable and integrable, and \(\phi:I\to\mathbb{R}\) is convex with \(\phi(f)\) integrable, then \(\phi(\int f\,dP)\le\int\phi(f)\,dP\). For concave \(\phi\), the inequality is reversed. We will arrange these integrability conditions explicitly. All functions in this tutorial are real-valued and nonnegative; integral inequalities are pointwise numerical inequalities between the resulting integrals.
Convexity and Concavity of \(x^r\)
Proof. For \(x>0\), the second derivative is \(r(r-1)x^{r-2}\). It is nonnegative if \(r\ge1\), and nonpositive if \(0<r\le1\). The second-derivative criterion therefore gives convexity or concavity, respectively, on \((0,\infty)\). For \(r>0\), \(x^r\) is continuous at \(0\), with \(0^r=0\). To extend either inequality defining convexity or concavity to endpoints that may equal zero, replace each such endpoint \(0\) by a positive number \(\delta\), apply the inequality on \((0,\infty)\), and let \(\delta\downarrow0\). Continuity passes the inequality to the limit. When \(r=1\), \(x^r=x\) is linear, so both inequalities hold with equality. \(\square\)
Proof. Put probability mass \(\lambda_i\) at \(x_i\), omitting any indices with \(\lambda_i=0\). The resulting finite probability space has expectation \(\sum_i\lambda_i x_i\). The power function is convex for \(r\ge1\) and concave for \(0<r<1\), by the proposition. Finite-valued functions on this space and their powers are integrable. Applying Jensen’s inequality gives the stated inequalities. If \(r=1\), both sides are equal by direct evaluation. \(\square\)
Worked Example: A Convex Power with Unequal Weights
Take weights \(\lambda_1=\tfrac14\), \(\lambda_2=\tfrac34\), values \(x_1=1\), \(x_2=3\), and exponent \(r=2\). The weights are nonnegative and sum to \(\tfrac14+\tfrac34=1\). The weighted average is $$ \frac14\cdot1+\frac34\cdot3=\frac14+\frac94=\frac{10}{4}=\frac52. $$ Its square is \((\tfrac52)^2=\tfrac{25}{4}\). The weighted average of the squares is $$ \frac14\cdot1^2+\frac34\cdot3^2=\frac14+\frac{27}{4}=\frac{28}{4}=7. $$ Thus \(\tfrac{25}{4}\le7\), as the convex-power inequality requires. Check: \(7-\tfrac{25}{4}=\tfrac{28-25}{4}=\tfrac34>0\).
Worked Example: A Concave Power Reverses the Comparison
Take equal weights \(\lambda_1=\lambda_2=\tfrac12\), values \(x_1=1\), \(x_2=9\), and exponent \(r=\tfrac12\). The weighted average is \(\tfrac12\cdot1+\tfrac12\cdot9=5\), so the square root of the average is \(\sqrt5\). The weighted average of the square roots is $$ \frac12\sqrt1+\frac12\sqrt9=\frac12+\frac32=2. $$ The concave-power inequality says \(2\le\sqrt5\). Both sides are nonnegative, and squaring gives \(4\le5\), verifying the comparison. Equality does not hold because the two input values differ.
Integral Inequalities on Probability Spaces
Proof. For \(r\ge1\), the hypotheses say that \(f\) and \(f^r\) are integrable, so the convex form of integral Jensen applies to \(x^r\) and gives the first inequality. For \(0<r<1\), \(f^r\) is integrable: for every \(t\ge0\), \(t^r\le1+t\), since this is immediate for \(0\le t\le1\), while \(t^r\le t\) for \(t\ge1\). Hence \(\int f^r\,dP\le\int(1+f)\,dP=1+\int f\,dP<\infty\). The concave form of integral Jensen now applies and gives the second inequality. If \(r=1\), the two expressions are identical. \(\square\)
For a finite measure space \((X,\mathcal{M},\mu)\), write \(M=\mu(X)\) and suppose \(0<M<\infty\). The probability measure \(P\) defined by \(P(E)=\mu(E)/M\) normalizes the total mass to one. Substituting \(dP=d\mu/M\) into the probability-space bounds gives the following formulas.
Proof. The measure \(P=\mu/M\) is a probability measure. For \(r\ge1\), the probability-space result gives \((\int f\,d\mu/M)^r\le\int f^r\,d\mu/M\). Multiplying by \(M^r\), which is positive, yields the first formula. For \(0<r<1\), it gives \(\int f^r\,d\mu/M\le(\int f\,d\mu/M)^r\); multiplying by \(M\) yields the second formula. If \(M=0\), every measurable nonnegative function has integral zero, so both integrals in question are zero; the normalized argument is not used. \(\square\)
Worked Example: A Convex-Power Bound on an Interval
On \(([0,1],\mathcal{L},m)\), let \(f(x)=2x\) and \(r=2\). The space has measure one, and \(f\) and \(f^2\) are integrable. Direct calculation gives $$ \int_0^1 2x\,dx=[x^2]_0^1=1,\qquad \int_0^1(2x)^2\,dx=4\left[\frac{x^3}{3}\right]_0^1=\frac43. $$ Therefore \((\int_0^1 f\,dm)^2=1\le\tfrac43=\int_0^1 f^2\,dm\). Check: the difference is \(\tfrac43-1=\tfrac13>0\). The same values also follow from the finite-measure formula with \(M=1\).
Worked Example: A Concave-Power Bound on an Interval
On \(([0,1],\mathcal{L},m)\), set \(f(x)=x\) and \(r=\tfrac12\). The function \(f\) is nonnegative and integrable, and \(f^{1/2}\) is integrable. We compute $$ \int_0^1 x\,dx=\left[\frac{x^2}{2}\right]_0^1=\frac12,\qquad \int_0^1 \sqrt{x}\,dx=\left[\frac23x^{3/2}\right]_0^1=\frac23. $$ The required inequality is \(\tfrac23\le(\tfrac12)^{1/2}=\tfrac1{\sqrt2}\). Both sides are nonnegative; squaring reduces it to \(\tfrac49\le\tfrac12\), which holds because \(8\le9\). This independently checks the exact comparison.
Normalization and the Most Common Reversal
The probability-space formulas compare an average with an average of powers. They do not generally say that an unnormalized integral can be substituted for an average. If \(M\ne1\), the factors \(M^{r-1}\) and \(M^{1-r}\) in the finite-measure formulas are essential. Also, changing from a convex power to a concave power changes the direction of Jensen’s inequality.
Worked Example: Correcting a Reversed Inequality
With equal weights on the values \(1\) and \(9\), the average is \(5\), and the average of the square roots is \(\tfrac12\cdot1+\tfrac12\cdot3=2\). A common incorrect calculation asserts \(\sqrt5\le2\), by using the convex-power direction for the concave function \(x^{1/2}\). But \(\sqrt5\le2\) would imply \(5\le4\) after squaring, which is false. Since \(0<\tfrac12<1\), the function is concave, so the correct inequality is \(2\le\sqrt5\); squaring confirms \(4\le5\).
These inequalities are useful whenever a power of an average must be controlled by an average of powers, or conversely. In particular, the exponent determines which quantity bounds the other; the total mass determines the scaling in the finite-measure case. The assumptions matter: for \(r\ge1\), if \(\int f^r\,dP=\infty\), the displayed bound has infinite right side and is automatic, while the finite-integrability hypothesis in the stated Jensen application is not available. For \(0<r<1\), integrability of \(f\) on a probability space itself ensures integrability of \(f^r\), as shown above.
Check Your Understanding
For each question, identify whether the power is convex or concave and check the relevant measure or weight normalization.
- For weights \(\tfrac13,\tfrac23\) and values \(0,6\), verify the weighted inequality for \(r=2\).
- On a probability space, state the inequality relating \(\int f^{1/3}\,dP\) to \((\int f\,dP)^{1/3}\) for nonnegative integrable \(f\).
- Derive the finite-measure bound for \(r=3\) from the probability-space inequality, and identify the power of \(M=\mu(X)\).
- Why does the finite-measure result require \(M>0\) for its normalization argument? What happens to the integrals when \(M=0\)?
- Give a concrete reason that the inequality direction for \(x^{1/2}\) cannot be the same as the direction for \(x^2\).