Bounded difference quotient implies weakly differentiable.

Question

Suppose $f\in L^2 (a,b)$ and $$\int_a^b \left(\frac{f(x+t)-f(x)}{t}\right)^2 dx \leq 1,$$ then we can show $f\in H^1(a,b)$

The proof we did in class was quite long, and we used the following argument:

From strongly bounded, there exists a subsequence $t_k$ such that the difference quotient converges weakly to $g$ in $L^2$. The weak limit $g$ is the distributional derivative of $f$, and the entire sequence of difference quotient actually converges to $g$ weakly from the uniqueness of limits in the sense distribution.

Is this a standard result? where could I find more about this type of result.

Answer 1

This is indeed a standard result. The proof is not too involved: let $g$ denote the weak limit in $L^2$ of the difference quotients with stepsizes $t_k$, i.e. $\frac1{t_k}(f(\cdot + t_k)-f(\cdot))\rightharpoonup g$.

Take a smooth test function $\phi\in C_0^\infty(a,b)$ with compact support. Then for $t_k$ small enough $$ \frac1{t_k}\int_a^b (f(x+t_k)-f(x))\phi(x)dx = \frac1{t_k}\int_a^b f(x) (\phi(x-t_k)-\phi(x))dx . $$ Since $\phi$ is smooth, we can pass to the limit on the right-hand side (dominated convergence) to obtain $$ \frac1{t_k}\int_a^b (f(x+t_k)-f(x))\phi(x)dx \to -\int_a^b f(x)\phi'(x)dx. $$ But the difference quotients of $f$ converge weakly to $g$, so we get the identity $$ \int_a^b g(x)\phi(x)dx = -\int_a^b f(x)\phi'(x)dx$$ for all smooth $\phi$. Hence, $g$ is the weak derivative of $f$, $f'=g\in L^2(a,b)$ and $f\in H^1(a,b)$.

Answer 2

I think this question and the corresponding answers here on Math.SE will be helpful to you:

Weak derivative as an $L^2$ limit of the difference quotient

Basically the idea is that the weak derivative can be written as a difference quotient if one takes the limit in $L^2$. Note that the integral you give above is the square of the difference quotient; this also points to the relevant of the $L^2$ limit. The fact that the above difference quotient is bounded then suggests that we can define limiting behavior for it without things blowing up, which is good, since if an appropriate limit exists, it is the weak derivative (if one believes that the weak derivative can be written as a difference quotient in $L^2$).

Answer 3

Seems to me it's much simpler than all that. Let's say $$\Delta_tf(x)=\frac{f(x+t)-f(x)}{t}.$$

First, it's easy to reduce to the case $f\in L^2(\Bbb T)$: We can certainly assume to start that $f\in L^2([0,\pi])$, and now $f$ extends to an even $2\pi$-periodic function.

Assume that $f$ is $2\pi$-periodic. Now we can use Fourier series to deduce everything very easily, without needing to know anything at all in advance:

Theorem. Suppose $f\in L^2(\Bbb T)$. TFAE:

1. There exists $g\in L^2$ with $\Delta_t f\to g$ weakly.

2. $||\Delta_t f||_2$ is bounded.

3. $\sum n^2|\hat f(n)|^2<\infty$.

4. There exists $g\in L^2$ with $||g-\Delta_tf||_2\to0$.

Proof (1) implies (2): Uniform Boundedness.

(2) implies (3): Note that $$\widehat{\Delta_t f}(n)=\frac{e^{int}-1}{t}\hat f(n)\to in\hat f(n).$$If $||\Delta_t f||\le c$ for all $t$ it follows that $\sum_{|n|\le N}n^2|\hat f(n)|^2\le c^2$ for all $N$, hence (3).

(3) implies (4): Since (3) holds we can define $g\in L^2$ by $g(x)=\sum in\hat f(n)e^{inx}$. Now Parseval shows that $||g-\Delta_t f||_2\to 0$. (I suppose we're going to need to apply dominated convergence in $\ell^2(\Bbb Z)$. For that we should note that $|e^{it}-1|\le|t|$.) QED