is defined for all \(f \in C^0([a,b])\text{,}\) and indeed for somewhat more general functions. Here we recall some basic properties of this integral that we will need in the next section.
Theorem3.7.Properties of the Riemann integral.
Let \(f,g \in C^0([a,b])\text{,}\)\(\alpha,\beta \in \R\text{,}\) and \(c \in (a,b)\text{.}\) Then
The first inequality (3.3) follows from Theorem 3.7 and the fact that \(-\abs{f(t)} \le f(t) \le \abs{f(t)}\) for all \(t \in [a,b]\text{.}\) The second inequality (3.4) similarly follows from the inequalities
for all \(t \in [a,b]\) combined with the fact that \(\int_a^b 1\,dt =
b-a\text{.}\)
Lemma3.9.Inertia principle for integrals.
Let \(f \in C^0([a,b])\) and suppose that \(f(t) \ge 0\) for all \(t\in
[a,b]\text{.}\) If \(\int_a^b f(t)\, dt = 0\text{,}\) then \(f(t) = 0\) for all \(t
\in [a,b]\text{.}\)
Proof.
Suppose for the sake of contradiction that there exists some \(c \in
(a,b)\) such that \(f(c) \gt 0\text{.}\) Since \(f\) is continuous, we can find \(\delta \gt 0\) such that for any \(t \in [a,b]\) with \(\abs{t-c} \lt
\delta\) we have
where we have assumed that \(\delta \gt 0\) is chosen small enough that \(c \pm
\delta \in (a,b)\text{.}\) Since by assumption \(f(t) \ge 0\) for all \(t \in [a,b]\text{,}\) while \(f(t) \ge \frac 12 f(c)\) for \(t\) in the smaller interval \([c-\delta,c+\delta]\text{,}\) we can then estimate
contradicting the choice of \(c\text{.}\) The cases when \(c\) is one of the endpoints \(a,b\) are extremely similar, and so we omit them.
Finally, we remark that not all functions \(f \maps [a,b] \to \R\) have well-defined definite integrals, indeed, not even all bounded functions. This is discussed in great detail in the Measure theory & integration unit.
Argue that \(0 \le f(t) \le 1\) for all \(t \in [-1,1]\) and that \(f(t)=0\) for \(t \notin [0,d]\text{.}\)
Hint.
I strongly recommend sketching the graph of \(f\text{.}\)
Solution.
Like \(g_c\) and \(g_d\text{,}\) the function \(f\) is also piecewise linear, just with more pieces. Going through the various cases we find
\begin{equation*}
f(t) =
\begin{cases}
0 \amp \text{ if } t \in [-1,0] \\
t/c - t/d \amp \text{ if } t \in [0,c] \\
1 - t/d \amp \text{ if } t \in [c,d] \\
0 \amp \text{ if } t \in [d,1].
\end{cases}
\end{equation*}
In particular, \(f(t)=0\) for \(t \notin [0,d]\text{.}\) Since \(0 \lt c \le
d\text{,}\)\(f\) is increasing on \([0,c]\) and decreasing on \([0,d]\text{,}\) as shown in the following graph.
Graph of the function f
From the monotonicity, or just from looking at the graph, we see that
Do not just calculate the integrals explicitly. In addition to being tedious and slow, this will not help you learn the relevant analysis skills.
Hint1.
Seriously, do not explicitly calculate the integrals. Instead, take advantage of the inequalities you have already shown in the previous part.
Hint2.
Use Theorem 3.7 to split up the integrals into several pieces, and then estimate the different pieces using Corollary 3.8 and Part b.
Solution.
We have shown in Part b that \(f(t) \ge
0\) for all \(t \in [-1,1]\text{,}\) and so the inequalities on the left hand side follow from Theorem 3.7. To prove the inequalities on the right hand side, we split up the integral and use the upper bound in Part b,
While the question (and hint) both ask you to estimate these integrals rather than calculate them exactly, at least a few students still calculated them explicitly. I strongly recommend getting used to doing estimates as in the official solution. This can be much faster, works in many situations where the integral cannot be calculated exactly, and is less prone to calculation errors.
For what it’s worth, the explicit values of the integrals here are
