Real analysis true/false

(1) Let $f$ be a continuous function on $[0,1]$. Then for every partition $P$ of $[0,1]$, the lower and upper Riemann sums of $f$ over $P$ satisfy $L(f,P)\neq U(f,P)$.

(2) Let $f:[0,\infty)\to\mathbb{R}$ be a continuous function such that $\lim_{x\to\infty}f(x)=0$. Then $f$ has a maximum value on $[0,\infty)$.

For (1), I think the answer is False because a continuous function on a closed interval is integrable, and the definition of integrability is that the upper and lower Riemann sums are within $\epsilon>0$ of each other, and thus can be equal. However, I do not know for sure what the correct answer is.

For (2), the correct answer is False, but I cannot understand why. I can see how it would be false if the domain of $f$ was $(0,\infty)$ instead, because then $f(x)=1/x$ is a counterexample, but since the interval is closed on the left, and $f$ is continuous, I don't see how $f$ could fail to have a maximum value and still be continuous. Even more confusing, according to my professor the statement would be True if $f$ were positive.

• Unless the meaning of "has a maximum value" really means "attains a maximum value" – Matt G Dec 15 '15 at 0:03
• Re (1): Integrability does not imply the Riemann sums are equal, but rather that their associated nets converge to the same thing. In other words, the upper and lower Riemann sums for a particular partition can definitely disagree. But nonetheless, to exhibit a counter example for (1) you need only find a continuous function and a partition where the the sums do agree. Consider any constant function. – Shawn O'Hare Dec 15 '15 at 0:10
• I assume maximum has been defined along the lines of "a point m that is larger than all other points" in which case the case of "f(x) = 0" would be continuous and have no maximum. Furthermore consider a function which starts out negative and heads to zero. Any f(x+1) > f(x) for all x in that case. Take -1/(x+1) as an example. – Kelechi Nze Dec 15 '15 at 0:11
• If you think it is true in general that for a continuous function and a given partition, the upper and lower Riemann sums are equal, then you surely do not understand what an upper and lower Riemann sum are. Before tackling a problem like this, you should work on understanding the definitions. A good way to start is to compute a few examples. – WillO Dec 15 '15 at 0:11
• For (2), how about a function with only negative values (and the other properties you list)? – John Dawkins Dec 15 '15 at 0:12

In regards to your response for 1, the definition of integrability does NOT mean that the upper and lower partition sums are equal. The definition states that for every $\epsilon > 0$ there is a $\delta > 0$ such that $U(f,P)-L(f,P) < \epsilon$ whenever the maximum width of an interval in $P$ is less than $\delta$.

For instance, if $f(x) = x$ and $P = \{0,\tfrac{1}{2},1\}$, then the upper sum would be $U(f,P) = \displaystyle\sum_{i = 1}^{2}\left[\max_{x_{i-1} \le x \le x_i}f(x)\right](x_i-x_{i-1}) = f(\tfrac{1}{2}) \cdot (\tfrac{1}{2}-0) + f(1) \cdot (1-\tfrac{1}{2}) = \tfrac{1}{2} \cdot \tfrac{1}{2} + 1 \cdot \tfrac{1}{2} = \tfrac{3}{4}$,

while the lower sum would be $L(f,P) = \displaystyle\sum_{i = 1}^{2}\left[\min_{x_{i-1} \le x \le x_i}f(x)\right](x_i-x_{i-1}) = f(0) \cdot (\tfrac{1}{2}-0) + f(\tfrac{1}{2}) \cdot (1-\tfrac{1}{2}) = 0 \cdot \tfrac{1}{2} + \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$

So $U(f,P)$ and $L(f,P)$ don't have to be equal.

However, the question asks if $U(f,P) \neq L(f,P)$ is always true. Can you exhibit an example where $L(f,P) = U(f,P)$?

Think about what happens if $f$ is constant.

For 2, consider a function such as $f(x) = -\dfrac{1}{x+1}$. Clearly, $\displaystyle\lim_{x \to \infty}f(x) = 0$. But does $f$ attain a maximum on $[0,\infty)$?

(1) is false because for example $f(x)=1$ does not meet it. If $P=\{0, x_1, x_2,\ldots,x_n,1\}$ is a partition ($0\leq x_1\leq x_2\leq \ldots$), the lower Riemann sums for $f(x)$ would be $1\cdot(x_1-0)+1\cdot(x_2-x_1)+\ldots$, whereas the upper sums would be $1\cdot(x_1-0)+1\cdot(x_2-x_1)+\ldots$, i.e. the same.

(2) is false because for example $f(x)=-1/e$ for $x\in [0,1]$, $f(x)=-e^{-x}$ for $x \geq 1$ does not meet it: it is always $< 0$, but its values approach $0$ as closely as one wishes. (By maximum, I understand a point $y\in[0,\infty)$ such that $f(x)\leq f(y)$ for all $x$.)