Brezis' exercise 8.28.4: how to prove that $\langle Tf, f \rangle \ge 0$?

Question

Let $I$ be the open interval $(0, 1)$ and $H := L^2 (I)$ equipped with the usual inner product $\langle \cdot, \cdot \rangle$. Consider the linear map $T: H \to H$ defined by $$ (Tf) (x) = \int_0^x t f(t) dt + x \int_x^1 f(t)dt, \quad x \in I. $$

I am trying to solve (4.) in below problem from Brezis' Functional Analysis

Exercise 8.28

1. Check that $T$ is a bounded linear operator.

2. Check that $T$ is a compact operator.

3. Check that $T$ is self-adjoint.

4. Show that $\langle Tf, f \rangle \ge 0$ for all $f \in H$, and that $\langle Tf, f \rangle = 0$ implies $f =0$.

We have $$ \begin{align*} \langle Tf, f \rangle &= \int_0^1 \int_0^x t f(t) dt f(x) dx + \int_0^1 x \int_x^1 f(t) dt f (x) dx \\ &= \int_0^1 \int_0^1 t 1_{[0, x]} (t) f(t) f(x) dt dx + \int_0^1 \int_0^1 x 1_{[x, 1]} (t) f(t) f(x) dt dx \\ &= \int_0^1 \int_0^1 t 1_{[t, 1]} (x) f(t) f(x) dt dx + \int_0^1 \int_0^1 x 1_{[x, 1]} (t) f(t) f(x) dt dx \\ &= 2 \int_0^1 \int_0^1 t 1_{[t, 1]} (x) f(t) f(x) dt dx. \end{align*} $$

Could you provide some hints to prove $\langle Tf, f \rangle \ge 0$?

Answer 1

Notice that $$\langle Tf, f \rangle = \int_{[0,1]^2} (t\wedge x) f(t) f(x) dt dx$$ and write $t\wedge x$ as a simple integral.

Further hint:

$$t\wedge x=\int_0^1  \mathbf{1}_{[0,t]}(s) \mathbf{1}_{[0,x]}(s) ds.$$

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If you're familiar with Gaussian processes/Brownian motion, this is the same trick that's used to show that $K(t,x)= t\wedge x$ is a covariance function (i.e. positive semidefinite).

Answer 2

1. If $f$ is continuous, $Tf$ is derivable and $f$ has some primitive $F$. By integrating by parts, we find $\int^1_0 Tf f = [Tf F]^1_0 - \int^1_0 (Tf)' F$. This can be greatly simplified and proved to be nonnegative.

2. Using the density of continuous functions in $L^2$ and the previously proved fact that $T$ is bounded, it is a standard exercise to prove the inequality on all of $L^2$.

Answer 3

Because $T$ is continuous, we assume WLOG that $f \in C^\infty_c (I)$. We need to prove that $$ \beta := \int_0^1 \int_0^x t f(t) dt f(x) dx \ge 0. $$

We define $F (x) := \int_0^x t f(t) dt$. Then $F \in C^\infty (\bar I)$ with $F'(x) = xf(x)$ and $F'(0)=F(0)=0$. We need an auxiliary result (in the same book), i.e.,

Exercise 8.9

1. Let $u \in W^{2, p}(I)$ with $1<p<\infty$. Assume that $u(0)=u^{\prime}(0)=0$. Show that $\frac{u(x)}{x^2} \in L^p(I)$ and $\frac{u^{\prime}(x)}{x} \in L^p(I)$ with (1) $$ \left\|\frac{u(x)}{x^2}\right\|_{L^p(I)}+\left\|\frac{u^{\prime}(x)}{x}\right\|_{L^p(I)} \leq C_p\left\|u^{\prime \prime}\right\|_{L^p(I)} $$
2. Deduce that $v(x)=\frac{u(x)}{x} \in W^{1, p}(I)$ with $v(0)=0$.

Let $u(x) := \frac{F(x)}{x}$. By Exercise 8.9.2, $u \in H^1 (I)$ with $u(0)=0$. We have $$ u'(x) = \frac{F'(x)x- F(x)}{x^2} = \frac{F'(x)}{x} - \frac{F(x)}{x^2}. $$

Then $$ \begin{align*} \beta &= \int_0^1 F(x) f(x) dx \\ &= \int_0^1 F'(x) u(x) dx \\ &= F(x) u(x) |_0^1- \int_0^1 F(x) u'(x) dx \quad \text{by integration by parts} \\ &= F^2 (1) - \left ( \int_0^1 \frac{F(x) F'(x)}{x} dx - \int_0^1 \frac{F^2(x)}{x^2} dx \right ) \\ &= F^2 (1) - \beta + \int_0^1 \frac{F^2(x)}{x^2} dx. \end{align*} $$

It follows that $$ 2 \beta = F^2 (1) + \int_0^1 \frac{F^2(x)}{x^2} dx \ge 0. $$