Prob. 15, Chap. 6, in Baby Rudin: If $f$ is a real, continuously differentiable function on $[a, b]$, . . .

Question

Here is Prob. 15, Chap. 6, in the book Principles of Mathematical Analysis, by Walter Rudin, 3rd edition:

Suppose $f$ is a real, continuously differentiable function on $[a, b]$, $f(a) = f(b) = 0$, and $$ \int_a^b f^2 (x) \ \mathrm{d} x = 1. $$ Prove that $$ \int_a^b x f(x) f^\prime (x) \ \mathrm{d} x = - \frac{1}{2} $$ and that $$ \int_a^b \left[ f^\prime (x) \right]^2 \ \mathrm{d} x \cdot \int_a^b x^2 f^2 (x) \ \mathrm{d} x > \frac{1}{4}. $$

My Attempt:

As $f^\prime$ is continuous, so it is Riemann-integrable, by Theorem 6.8 in Rudin. Then by Theorem 6.22 in Rudin we can use integration by parts to obtain $$ \begin{align} \int_a^b x f(x) f^\prime (x) \ \mathrm{d} x &= \int_a^b x \frac{\mathrm{d}}{\mathrm{d} x} \left[ \frac{f^2(x)}{2} \right] \ \mathrm{d} x \\ &= b \frac{f^2(b)}{2} - a \frac{ f^2(a) }{2} - \int_a^b \frac{\mathrm{d} x}{\mathrm{d} x} \frac{ f^2(x) }{2} \ \mathrm{d} x \\ & \qquad \mbox{ [ using Theorem 6.22 in Rudin with $F(x) = x$ and $G(x) = \frac{f^2(x)}{2}$ ] } \\ &= 0 - 0 - \int_a^b \frac{f^2(x)}{2} \ \mathrm{d} x \\ &= -\frac{1}{2} \int_a^b f^2(x) \ \mathrm{d} x \\ &= -\frac{1}{2} \cdot 1 \\ &= -\frac{1}{2}. \tag{0} \end{align} $$

Here is the link to my Math SE post on the Holder's inequality for integrals:

Probs. 10 (a), (b), and (c), Chap. 6, in Baby Rudin: Holder's Inequality for Integrals

And, for $p=2$, the Holder's inequality is the Schwarz inequality for integrals.

Thus if $u$ and $v$ are complex Riemann-Stieltjes integrable functions with respect to a monotonically increasing function $\alpha$ on $[a, b]$, then $$ \left\lvert \int_a^b u v \ \mathrm{d} \alpha \right\rvert \leq \left( \int_a^b \lvert u \rvert^2 \ \mathrm{d} \alpha \right)^{1/2} \left( \int_a^b \lvert v \rvert^2 \ \mathrm{d} \alpha \right)^{1/2}. $$ Therefore, if $u$ and $v$ are real Riemann-Stieltjes integrable functions with respect to $\alpha$ on $[a, b]$, then we can rewrite the last inequality as $$ \left\lvert \int_a^b u v \ \mathrm{d} \alpha \right\rvert \leq \left( \int_a^b u^2 \ \mathrm{d} \alpha \right)^{1/2} \left( \int_a^b v^2 \ \mathrm{d} \alpha \right)^{1/2}. $$ If we put $\alpha = x$ in the last inequality, we obtain $$ \left\lvert \int_a^b u v \ \mathrm{d} x \right\rvert \leq \left( \int_a^b u^2 \ \mathrm{d} x \right)^{1/2} \left( \int_a^b v^2 \ \mathrm{d} x \right)^{1/2}, $$ and hence $$ \left\lvert \int_a^b u v \ \mathrm{d} x \right\rvert^2 \leq \int_a^b u^2 \ \mathrm{d} x \cdot \int_a^b v^2 \ \mathrm{d} x, $$ which is the same as $$ \left( \int_a^b u v \ \mathrm{d} x \right)^2 \leq \int_a^b u^2 \ \mathrm{d} x \cdot \int_a^b v^2 \ \mathrm{d} x, $$ which implies that $$ \int_a^b u^2 \ \mathrm{d} x \cdot \int_a^b v^2 \ \mathrm{d} x \geq \left( \int_a^b u v \ \mathrm{d} x \right)^2. \tag{1} $$

In (1) we put $u(x) = f^\prime(x)$ and $v(x) = x f(x)$ and obtain $$ \begin{align} \int_a^b \left[ f^\prime(x) \right]^2 \ \mathrm{d} x \cdot \int_a^b x^2 f^2(x) \ \mathrm{d} x &\geq \left( \int_a^b f^\prime(x) x f(x) \ \mathrm{d} x \right)^2 \\ &= \left( \int_a^b x f(x) f^\prime(x) \ \mathrm{d} x \right)^2 \\ &= \left( - \frac{1}{2} \right)^2 \qquad \mbox{ [ using (0) above ] } \\ &= \frac{1}{4}. \end{align} $$

Is what I've done so far correct? If so, then why is it that, unlike what Rudin has asserted, I've not obtained the strict inequality in the second calculation?

How to obtain the strict inequality from where I've left off?

Answer 1

If the equality was not strict, it is not to hard to prove that $$xf(x) = \lambda \cdot f'(x) \tag{1}$$ for some $\lambda \in \mathbb{R}$, i.e, the Cauchy-Schwarz inequality is an equality when the terms are linearly dependent.

Case 1 ($\lambda = 0 $)

Then $f = 0$ which leads to a contradiction with the fact that $$\int_a^b f^2 (x) \ \mathrm{d} x = 1.$$

Case 2 ($\lambda \neq 0$)

Since $\int_a^b f^2 (x) \ \mathrm{d} x = 1$, $\exists s \in [0,1] : f(s) \neq 0$. We can safely say that $s \in (0,1)$ due to the given condition that $f(0)=0=f(1)$.

Since $f$ is continuous, by the Extreme Value Theorem, $f$ has a local max/min on the interval [0,1], and due to the preceding observation at least one of these max/min values has to be non zero. In other words, $$\exists t \in (0,1) : f'(t) = 0 \text{ and } f(t) \neq 0$$ We arrived at this conclusion in this way because it is important that $f(t) \neq 0$ also, as we were already guaranteed a point where the derivative vanishes by Rolle's Theorem.

Evaluating what we have in (1) at $t$, $$f'(t) = 0 = \frac{1}{\lambda} \cdot t f(t)$$ This leads to a contradiction since $\frac{1}{\lambda},t,f(t)$ are all non zero.

In either case we have a contradiction. Thus the inequality is a strict one.