Integral Triangle Inequality

From Absolute Values to Triangle Inequalities

The previous tutorial, Absolute Values and Integrability, showed how taking absolute values interacts with Riemann integrability. In particular, the Absolute-Value Inequality for the Riemann Integral theorem states that if $f$ is Riemann integrable on $[a,b]$, then $|f|$ is integrable and \[ \left|\int_a^b f(x)\,dx\right|\leq \int_a^b |f(x)|\,dx. \] We will use this established result without reproving it. The focus here is how triangle inequalities combine with linearity and order preservation to give useful estimates for sums of functions.

For real numbers $u$ and $v$, the ordinary triangle inequality says $|u+v|\leq |u|+|v|$. Applying it at each point gives a pointwise comparison of functions. Since order is preserved by the Riemann integral, this comparison can be integrated. The result controls the total magnitude of a sum, not just the signed value of its integral.

Theorem (Integral Triangle Inequality): If $f,g:[a,b]\to\mathbb{R}$ are Riemann integrable, then $f+g$ and $|f+g|$ are Riemann integrable, and $$ \int_a^b |f(x)+g(x)|\,dx \leq \int_a^b |f(x)|\,dx+\int_a^b |g(x)|\,dx. $$

Proof. By linearity, $f+g$ is Riemann integrable. The Absolute-Value Inequality for the Riemann Integral theorem then shows that $|f+g|$ is integrable; it also shows that $|f|$ and $|g|$ are integrable. For every $x\in[a,b]$, the real-number triangle inequality gives

|f(x)+g(x)|\leq |f(x)|+|g(x)|.

Both sides are integrable, so order preservation and linearity of the Riemann integral yield

\int_a^b |f(x)+g(x)|\,dx \leq \int_a^b \bigl(|f(x)|+|g(x)|\bigr)\,dx = \int_a^b |f(x)|\,dx+\int_a^b |g(x)|\,dx.

This proves the theorem. $\square$

The signed integral estimate follows as a consequence. Apply the Absolute-Value Inequality for the Riemann Integral theorem to $f+g$, and then apply the Integral Triangle Inequality:

\left|\int_a^b (f(x)+g(x))\,dx\right| \leq \int_a^b |f(x)+g(x)|\,dx \leq \int_a^b |f(x)|\,dx+\int_a^b |g(x)|\,dx.

Linearity also identifies the left side with $\left|\int_a^b f+\int_a^b g\right|$. Thus the familiar triangle inequality for two integrals is part of a stronger statement: the integral of the pointwise magnitude of a sum is bounded by the sum of the individual magnitudes.

Worked Example: A Strict Integral Triangle Inequality

On $[0,1]$, let $f(x)=1$ everywhere, and let $g(x)=-1$ for $0\leq x\leq 1/2$ and $g(x)=0$ for $1/2<x\leq1$. These are step functions, so they are Riemann integrable. The sum is zero on the first half and one on the second half. Therefore

\int_0^1 |f(x)+g(x)|\,dx =0\left(\frac12\right)+1\left(\frac12\right) =\frac12.

Meanwhile,

\int_0^1 |f(x)|\,dx=1, \qquad \int_0^1 |g(x)|\,dx=1\left(\frac12\right)+0\left(\frac12\right)=\frac12.

Hence the inequality reads $1/2\leq 3/2$, and is strict. On the first half of the interval, the positive values of $f$ and the negative values of $g$ cancel. Their separate magnitudes still contribute to the right side, but not to the magnitude of their sum.

Equality and Cancellation in the Signed Bound

The estimate $\left|\int f\right|\leq\int|f|$ compares signed accumulation with total magnitude. Its equality case can be described precisely using the positive and negative parts defined in Absolute Values and Integrability. Recall the Positive-Negative Decomposition theorem: for integrable $f$, both $f^+$ and $f^-$ are integrable, $f=f^+-f^-$, and $|f|=f^++f^-$.

Theorem (Equality Criterion for the Signed Integral Bound): Let $f:[a,b]\to\mathbb{R}$ be Riemann integrable. Then $$ \left|\int_a^b f(x)\,dx\right|=\int_a^b |f(x)|\,dx $$ if and only if at least one of $\int_a^b f^+(x)\,dx$ and $\int_a^b f^-(x)\,dx$ is zero.

Proof. Set $A=\int_a^b f^+(x)\,dx$ and $B=\int_a^b f^-(x)\,dx$. Since the positive and negative parts are nonnegative, order preservation gives $A\geq0$ and $B\geq0$. The Positive-Negative Decomposition theorem gives

\int_a^b f(x)\,dx=A-B, \qquad \int_a^b |f(x)|\,dx=A+B.

For nonnegative real numbers $A$ and $B$, $|A-B|\leq A+B$, and equality holds exactly when $A=0$ or $B=0$. Indeed, if $A\geq B$, equality means $A-B=A+B$, which is equivalent to $B=0$. If $B\geq A$, equality means $B-A=A+B$, which is equivalent to $A=0$. Substituting the integral expressions above proves both directions of the criterion. $\square$

The criterion is about the integrals of the two parts, not necessarily about the sign of $f$ at every point. A nonnegative integrable function can have integral zero without being zero at every point—for example, it may be nonzero at only finitely many points. Thus equality does not, in general, force $f$ to be nonnegative or nonpositive everywhere.

Worked Example: Equality Despite an Exceptional Sign

Define $f:[0,1]\to\mathbb{R}$ by $f(x)=-2$ for $x\ne 1/3$ and $f(1/3)=7$. This function differs from the constant function $-2$ at just one point, so the Finite Point Changes Do Not Affect the Integral theorem shows that $f$ is integrable and $\int_0^1 f(x)\,dx=-2$. Its absolute value differs from the constant function $2$ at just that point as well, so

\int_0^1 |f(x)|\,dx=2.

Consequently, $\left|\int_0^1 f(x)\,dx\right|=2=\int_0^1|f(x)|\,dx$, even though $f(1/3)=7$ is positive. Here $f^+$ is zero except at $1/3$, so its integral is zero; $f^-$ has integral $2$. This is exactly the equality case in the criterion.

For a function that has substantial positive and negative contributions, cancellation makes the signed bound strict. For example, let $h(x)=x-1/2$ on $[0,1]$. The function is negative on $[0,1/2)$ and positive on $(1/2,1]$, with equal-sized triangular contributions. Direct integration gives

\int_0^1 h(x)\,dx=0, \qquad \int_0^1 |h(x)|\,dx =2\int_0^{1/2}\left(\frac12-x\right)\,dx =2\left(\frac18\right) =\frac14.

The signed integral is zero because the two contributions cancel, while the integral of the absolute value records both. In the equality criterion, both positive-part and negative-part integrals equal $1/8$, so neither is zero.

A Reverse Estimate for Two Functions

The Integral Triangle Inequality also yields a useful way to compare the absolute integrals of two functions. If two functions are close in the sense that the integral of $|f-g|$ is small, then their integrals of absolute values cannot differ by much. This follows from the reverse triangle inequality for real numbers, $\bigl||u|-|v|\bigr|\leq|u-v|$.

Theorem (Reverse Triangle Inequality for Absolute Integrals): If $f,g:[a,b]\to\mathbb{R}$ are Riemann integrable, then $$ \left|\int_a^b |f(x)|\,dx-\int_a^b |g(x)|\,dx\right| \leq \int_a^b |f(x)-g(x)|\,dx. $$

Proof. The Absolute-Value Inequality for the Riemann Integral theorem ensures that $|f|$, $|g|$, and $|f-g|$ are integrable. Linearity makes $|f|-|g|$ integrable as well. Applying the same theorem to $|f|-|g|$, and then integrating the pointwise reverse triangle inequality, gives

\begin{aligned} \left|\int_a^b |f(x)|\,dx-\int_a^b |g(x)|\,dx\right| &=\left|\int_a^b (|f(x)|-|g(x)|)\,dx\right|\\ &\leq \int_a^b \bigl||f(x)|-|g(x)|\bigr|\,dx\\ &\leq \int_a^b |f(x)-g(x)|\,dx. \end{aligned}

The last inequality follows from $\bigl||u|-|v|\bigr|\leq|u-v|$ with $u=f(x)$ and $v=g(x)$, followed by order preservation. This proves the result. $\square$

Worked Example: Comparing Absolute Integrals

Take $f(x)=x$ and $g(x)=1/2$ on $[0,1]$. Both functions are nonnegative, so

\int_0^1 |f(x)|\,dx=\int_0^1 x\,dx=\frac12, \qquad \int_0^1 |g(x)|\,dx=\int_0^1 \frac12\,dx=\frac12.

The left side of the reverse estimate is therefore zero. For the right side, $|f(x)-g(x)|=|x-1/2|$, and symmetry about $1/2$ gives

\int_0^1 |x-\tfrac12|\,dx =2\int_0^{1/2}(\tfrac12-x)\,dx =2\left(\frac18\right) =\frac14.

Thus the estimate is $0\leq1/4$. The functions are not identical, but their absolute integrals agree; the estimate guarantees that the difference between those integrals is controlled by the integrated pointwise difference.

How to Use the Inequalities

The estimates here have different inputs and conclusions. The Integral Triangle Inequality bounds $\int|f+g|$ using the separate integrals of $|f|$ and $|g|$. The signed bound controls $\left|\int f\right|$ by $\int|f|$, with equality determined by the positive and negative contributions. The reverse estimate controls the difference between $\int|f|$ and $\int|g|$ by $\int|f-g|$. Keeping these quantities distinct prevents a common mistake: $\int|f|$ and $\left|\int f\right|$ are generally not the same.

A reliable proof strategy is to begin with a pointwise inequality, check that its functions are integrable using linearity and the Absolute-Value Inequality for the Riemann Integral theorem, and then apply order preservation. For equality questions, reduce the integral expressions to the nonnegative numbers given by the integrals of $f^+$ and $f^-$. This makes the role of cancellation explicit and avoids claiming pointwise sign conditions that the integral alone cannot establish.

Choose the quantity to estimate.
Decide whether the goal concerns the integral of a sum, the magnitude of a signed integral, or the difference between two absolute integrals.

Write the pointwise comparison.
Use $|f+g|\leq|f|+|g|$ or $\bigl||f|-|g|\bigr|\leq|f-g|$, as appropriate.

Check integrability and integrate.
Use the established absolute-value result, linearity, and order preservation to pass from the pointwise inequality to an integral estimate.

Inspect cancellation when equality matters.
For the signed bound, compare the integrals of the positive and negative parts rather than inferring a pointwise sign from equality.

Check Your Understanding

Use the integral inequalities and equality criterion developed here to answer the following questions.

State the Integral Triangle Inequality for two Riemann integrable functions.
How does the bound for $\left|\int_a^b f\right|$ differ from the bound for $\int_a^b |f|$?
In the equality criterion for the signed integral bound, which two nonnegative quantities determine whether equality holds?
Why does equality in $\left|\int_a^b f\right|\leq\int_a^b|f|$ not necessarily imply that $f$ has one sign at every point?
What does the reverse triangle inequality for absolute integrals say about $\int_a^b|f|-\int_a^b|g|$ when $\int_a^b|f-g|$ is small?

What You'll Learn