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In chapter 6 of Rudin's Principles of Mathematical Analysis, problem 15, he writes

Suppose $f$ is a real, continuously differentiable function on $[a,b], f(a) = f(b) = 0$, and

$$\int_a^b f^2(x)\ dx = 1.$$

Prove that

$$ \int_a^b x f(x) f'(x) \ dx = -\frac 1 2 $$

and that

$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx > \frac 1 4. $$

I've shown everything up to demonstrating

$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx \ge \frac 1 4 $$

However, I can't figure out how to eliminate the case where equality holds. I've seen several other resources say that equality would imply

$$ f'(x) = \lambda xf(x) $$

but I can't figure out why that would be implied.

Resources I've already looked at: How to show the inequality is strict?

Baby Rudin Chapter 6, Problem 15 : Strict inequality

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

Proving a strict inequality (Application of Hölder's Inequality)

Show that if $f(a)=f(b)=0$ and $\int_a^b [f(x)]^2dx=1$, then $\int_a^b [f'(x)]^2dx\cdot\int_a^b [xf(x)]^2dx\gt \frac14$

Addem
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  • If $f'(x)=\lambda xf(x)$ then after solving the DE we get $$|f(x)|=e^{\lambda x^2/2+C}\ge 1$$ hence since $\int f^2(x)=1$ we must have $$e^{\lambda x^2/2+C}=1 \forall x\in[a,b]$$ – Albus Dumbledore Feb 23 '21 at 01:52
  • @AlbusDumbledore That's not my question, though. I'm only asking how the integral equality would imply $f'(x) = \lambda xf(x)$. – Addem Feb 23 '21 at 01:53
  • since you may have used the cauchy schwarz inequality equality occurs there – Albus Dumbledore Feb 23 '21 at 01:54
  • @AlbusDumbledore Yes, but that involves an inclusive inequality. – Addem Feb 23 '21 at 01:55
  • There's a standard second part to Cauchy-Schwarz saying that equality occurs if and only if the two vectors are linearly dependent. The if is obvious, and to see the only if - suppose $v$ is non-zero, and $u = \lambda v + z$ for some non-zero $z$ orthogonal to $v$. Then $\lambda = \langle u, v\rangle/|v|^2$, and $|u|^2 = \lambda^2|v|^2 + |z|^2 > \lambda^2|v|^2 = \langle u, v \rangle^2/|v|^2$ (notice the strict inequality, which occurs since $|z| > 0$). – stochasticboy321 Feb 23 '21 at 02:25
  • In your case, you can apply this condition with $u = xf, v = f'$ in the inner product space $L_2(a,b)$. There's a small complication in that equality in $L_2$ is only almost everywhere, but this is handled by the fact that $f$ and $f'$ are both continuous. – stochasticboy321 Feb 23 '21 at 02:25
  • @stochasticboy321 I think this is beyond the scope of Rudin, at least up to this part of the text, since I don't believe we've formally proved that the integral forms a norm over the vector space of integrable functions. At the very least we don't have a notion of "almost everywhere" yet. – Addem Feb 23 '21 at 02:28
  • Ah, I'm not very familiar with Rudin. Sorry. I'm curious how you showed the inequality. Also I guess you only need to work over the vector space of continuous square integrable functions (so the whole almost everywhere stuff was unnecessary overkill). – stochasticboy321 Feb 23 '21 at 02:34
  • @stochasticboy321 I used Cauchy-Schwartz. It's just that in Rudin, this is given by an inclusive inequality. I'm not sure if you wanted more details than that about how I solved it. But it's really straight-forward once the first part of the problem is done, you just identify the parts of CS with the parts of that result. – Addem Feb 23 '21 at 02:40

2 Answers2

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I'll rederive Cauchy-Schwarz inequality (specifically the equality condition) for continuous functions on an interval. Suppose $f,g : [a,b] \to \mathbb{R}$ are continuous. Throughout the following, all integrals are over $[a,b]$. Since $f$ and $g$ are continuous on a compact set, they are bounded, and so $\int f^2, \int g^2, \int fg$ all exist. I'll assume that $\int f^2 > 0$. I'll also assume that you can show the following result - if $h$ is continuous on an interval $[a,b]$, then $\int_{[a,b]} h^2 = 0$ if and only if $h$ is identically zero on the interval. [This is why I can assume $\int f^2 > 0$ - otherwise $f$ is the $0$ function and there's nothing to show]

Consider the function $$ J(t) := \int (t f - g)^2 = t^2 \int f^2 - 2 t \int fg + \int g^2.$$

We can derive Cauchy-Schwarz by observing that $J(t) \ge 0$ since it is the integral of a non-negative function, and then minimising the quadratic. Notice that as long as $\int f^2 > 0,$ the quadratic has a unique minima at $t_* = \int fg/ \int f^2$. So we have that for all $t \in \mathbb{R},$ $$ J(t) \ge J(t_*) = \int g^2 - \frac{(\int fg)^2}{\int f^2} \ge 0.$$

Now, we want to argue that if $(\int fg)^2 = \int f^2 \int g^2,$ then there must exist a $t \in \mathbb{R}$ such that $tf = g.$ But notice that $(\int fg)^2 = \int f^2 \int g^2$ implies that for $t_* = \int fg/ \int f^2,$ $J(t_*) = 0$. Due to the definition of $J$, this means that $\int (t_*f - g)^2 = 0.$ But $t_* f - g$ is continuous, so this means that $t_* f = g$ everywhere on the interval.

Now you can apply this equality condition to the continuous functions $xf$ and $f'$.

stochasticboy321
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  • Definitely answers the question! And it also answers my suspicion, that everyone saying "It's a simple matter to prove $xf(x) = \lambda f'(x)$" were ... eh ... kinda full of it. – Addem Feb 23 '21 at 02:58
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    @Addem In a kinder reading, too often, people making such statements don't remember where they were back in the day and what they found simple or not. Anyway, the argument is really hinging on understanding the proof of the C-S inequality itself. In general, for most inequalities, figuring out when equality can hold is a useful exercise in understanding it, and can further be a useful fact in and of itself at times. I'd recommend the practice during your learning. – stochasticboy321 Feb 23 '21 at 03:00
  • Fair point, I'm probably a little too grumpy for being stressed and feeling dumb. Thank you for your help! – Addem Feb 23 '21 at 15:38
  • I'm just now poring over the details of this proof and it is really elegant. I've seen similar proofs of CS but this really distills it in an extremely economical and intelligible way. Did you come up with this or did you read it somewhere? – Addem Feb 23 '21 at 17:50
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    I certainly won't claim to be the originator of the proof, and I strongly suspect I've seen something like this before. Let me just tell you how I came up with it - I'll talk in terms of norms and inner products because that's how I think about this stuff, but you can replace $|f|^2 = \int f^2$ and $\langle f, g\rangle = \int fg$ in everything I say. You should also note that at the end of the day I knew the form of the result I wanted, which is a big guide in these sort of things. – stochasticboy321 Feb 23 '21 at 19:09
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    The starting point was the observation that $|f -g |^2 = |f |^2 + |g|^2 - 2\langle f, g \rangle$, because this is the simplest way to relate inner products and norms (and give the AM-GM type relation $\langle f, g\rangle \le (|f|^2 + |g|^2)/2$.) Then I used some back-filling: basically, I knew that the interesting behaviour is as you vary along the subspace of one of the functions, which is succinctly captured by replacing $f$ by $tf$ for $t \in R$ (I think this is very natural if you embrace the vector space structure of continuous functions. In any case, since the target is to – stochasticboy321 Feb 23 '21 at 19:09
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    show that equality means that the functions are equal up to a constant, it makes sense to chuck a free parameter into the simplest way to relate the norms and inner products, just to see what happens). But then I immediately noticed that this yields a quadratic function $J(t) = t^2 |f|^2 - 2t\langle f, g \rangle + |g|^2,$ which is non-negative. This means that the minimum value is non-negative, so it's natural to look at the value at the minimum, but this just gave C-S already. – stochasticboy321 Feb 23 '21 at 19:10
  • I find the answer you wrote and explanation of how you got it worthy of publication in a textbook--I personally find this very helpful for general skills in analysis. Thank you! – Addem Feb 23 '21 at 19:11
  • That's very kind, I'm glad to be helpful. BTW another useful picture is geometric, which directly suggests looking at $\min_t |tf -g|^2$ - basically, if you think of $f,g$ as vectors and $(\int f^2)^{1/2}$ as a norm, $\langle f,g \rangle/(\int f^2)^{1/2}$ is measuring the size of the projection of $g$ along $f$, and C-S is just saying that this is smaller than the length of $g$. Now, $\min_t |tf - g|^2$ is just measuring how much energy $g$ has outside of the subspace spanned by $f$, and so studying it seems like a good idea to show C-S. – stochasticboy321 Feb 23 '21 at 19:24
  • (contd.) (This is a long winded way of saying I should have seen that this way was natural sooner :P). Also, a useful exercise for you at this point might be to show that $\int fg $ is an inner product for continuous square integrable functions, which should solidify such pictures. – stochasticboy321 Feb 23 '21 at 19:25
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Use Cauchy Schwartz inequality, $p(x)$ and $q(x)$ are continuous in $[a,b]$, then $$\left(\int_{a}^{b} p(x) q(x) dx \right)^{2} \le \int_{a}^{b} p^2(x) dx \int_{a}^{b} q^2(x) dx,$$ by choosing $p(x)=f'(x), q(x)=x f(x).$ Equality holds when $p(x)=t q(x)$.

Here the equality will hold if $$txf(x)=f'(x) \implies f(x)=C e^{tx^2/2}.$$ Using $f(a)=0=f(b)$, we get $C=0$. so the equality will hold for the trivial case of $f(x)=0.$

Z Ahmed
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