There is a result of the form: a function $u \in H^1(0,T;L^2)$ if and only if $$\int_0^{T-h}\lVert u(t+h)-u(t) \rVert_{L^2}^2 \leq C|h|$$ holds for all $h \in [0,T]$.

I have only seen one place containing this result, which is book by Barbu. can someone cite another source for this? Also, does this inequality above give a bound on the time derivative $u'$ in terms of the constant $C$? Thanks

EDIT: please no Wloka since I can't access his books. EDIT: I ask because I saw this in a paper around page 10 and so: enter image description here enter image description here I assumed it was because estimate (33) gave a bound on the time derivatives in some way. But I should have read the cited paper..


This "result" is false. The rule of thumb is that spaces characterized by a supremum condition, such as $$\sup_{0<h<T} \frac{1}{h} \int_0^{T-h}\lVert u(t+h)-u(t) \rVert_{L^2}^2 <\infty \tag{1}$$ are non-separable spaces of Lipschitz/Hölder type, in which smooth functions are not norm-dense. This is quite unlike $H^1$.

As a concrete counterexample, consider $u(t,x)=t^{\alpha}$ for $0\le t\le 1$, where $\alpha\in (0,1/2)$ is fixed. (The function is constant in $x$.) Observe that $u\notin H^1$.

We have $\|u(t+h)-u(t)\|_{L^2} = C((t+h)^\alpha-t^\alpha)$ for some constant $C$. By the concavity of $t\mapsto t^\alpha$, $$(t+h)^\alpha-t^\alpha \le \alpha h t^{\alpha-1}$$ Hence, $$ \int_{h}^{1-h} ((t+h)^\alpha-t^\alpha)^2\,dt = O(h^2) \int_{h}^{1-h} t^{2\alpha-2} \,dt = O(h^2)O(h^{2\alpha-1}) = O(h^{2\alpha+1}) $$ Also, $\int_{0}^{h} ((t+h)^\alpha-t^\alpha)^2\,dt = O(h^{2\alpha+1})$ trivially. Therefore, (1) is satisfied.

  • $\begingroup$ Thanks. I think I misunderstood something. I assumed a bound on the differences like in my OP (which I've now edited) would give me a bound on the derivative which would give me a compactness result. However this compactness follows an entirely different result. $\endgroup$ – maximumtag Mar 15 '14 at 10:13

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