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?