Existence of weak limits

Question

Background: Three months ago, I asked this question, which is a bit related to the following (if the answer to it was Yes, then answer to this one would be Yes too, but since that was a No, it still may be that this one has a positive answer).

Question: does existence of RHS imply existence of LHS in this formula: $$ w.\lim_{\lambda\to\lambda_{0}}f_\lambda(t)=w.\lim_{\Delta t\to 0}\frac{1}{2\Delta t}\left(w.\lim_{\lambda\to\lambda0}\int_{t-\Delta t}^{t+\Delta t}f_\lambda(\tau)d\tau\right) $$

where $w.\lim$ is meant to denote the limit in Schwartz distributions space (i.e. weak star limit on test function space), and $f_\lambda$ are generally distributions from the Schwartz space, parametrized with a real number $\lambda$ - family of Schwartz distributions. In case that the claim does not hold for all Schwartz distributions, it would be interesting to know whether it holds for locally integrable functions.

Proof that the existence of LHS implies existence of RHS is simple, since the integral of a test function is again a test function.

EDIT: As Pietro noted, integral represents the convolution of f with the characteristic function of an interval which is shrinking - so I believe the question can be reformulated in the spirit of my previous question, which was

It is provable that $f_\lambda\to f\Rightarrow f_\lambda*g\to f*g$ if $g$ has a compact support (shown in my textbook). In my particular case, $g=u(t+\triangle t)-u(t-\triangle t)$. Does for that particular case, $f_\lambda*g\to f*g\Rightarrow f_\lambda\to f$?

but now, we say that $\Delta t$ is not fixed (like in that previous question). Is that right?

Dec. 2011. EDIT: After a while, I returned to this question and felt the need to explain it a bit more. Zen Harper's answer shows that RHS is well-defined, but it doesn't seem to answer the question whether it's possible to have a family $f_\lambda$ which doesn't have a limit for $\lambda\to\lambda_0$ (i.e. LHS doesn't exist) but for which the iterated limit on the right has a value, i.e. RHS exists.

It appears to me it would be enough to show that if the inner limit on the right exists for all $\Delta t$ in some neighbourhood of $0$, then the LHS limit exists. Would it be enough?

Answer 1

For functions, the answer is NO, i.e. the limit need not be a function.

Let $I_h=(2h)^{-1} \chi_{[-h,h]}$, so that $I_h$ approximates the Dirac delta as $h \to 0^+$. You are considering $I_h * f_\lambda$.

A digression: similar things are common; e.g. $P_h(y) = \frac{1}{\pi} \frac{h}{h^2 + y^2}$ is the Poisson kernel, related to harmonic functions; $T_t(x) = \frac{1}{\sqrt{4 \pi t}}\exp(-x^2/4t)$ is the heat kernel, related to the heat equation. So questions like this have many connections with PDEs (fundamental solutions, etc.) I would guess that the specific functions $I_h$ don't make much difference here.

Anyway: choose $f_\lambda$ so that $\widehat{f_\lambda}(x) = \exp(- \lambda x^2)$ where $\widehat{f}$ denotes Fourier transform; so $f_\lambda$ is a Gaussian. Then, for each fixed test function $\varphi$, $$ \langle I_h * f_\lambda, \varphi \rangle = \langle \widehat{I_h} \exp(- \lambda x^2), \widehat{\varphi} \rangle \to \langle \widehat{I_h}, \widehat{\varphi} \rangle = \langle I_h, \varphi \rangle $$ as $\lambda \to 0^+$, depending on the constants in your Fourier transform and scalar product. Thus it is clear that $\langle I_h * f_\lambda, \varphi \rangle \to \langle I_h, \varphi \rangle$.

But now $\langle I_h, \varphi \rangle \to \varphi(0)$ as $h \to 0^+$, so $I_h \to \delta_0$ in the weak sense. So in this example, the RHS and LHS limits both exist; but the weak limit is $\delta_0$, the Dirac delta, not a function.

Actually this is not surprising, since functions (or just test functions) are dense in the space of all distributions, so you wouldn't expect distributional limits of functions to remain as functions.

For general distributions: assume that the inner limit of the RHS exists; so we can define a linear functional $U_h$ on test functions by $$ \langle U_h, \varphi \rangle = \lim_{\lambda \to \lambda_0} \langle I_h * f_\lambda, \varphi \rangle. $$ $U_h$ is well-defined by assumption. Now we can define another linear functional $F$ by $$ \langle F, \varphi \rangle = \lim_{h \to 0^+} \langle U_h, \varphi \rangle $$ Again this is well-defined by assumption. So $$ F = w.\lim_h U_h = w.\lim_h \left( w.\lim_\lambda I_h * f_\lambda \right) $$ So, the answer to your question would be yes, if we knew that $U_h, F$ were distributions.

So your question becomes: when can we show that a linear functional is a distribution?

I might be wrong, but isn't it consistent with ZF to assume that every linear functional defined on all test functions is a distribution? (But please correct me if I'm wrong; can anyone provide a reference for this result?) So you'll never be able to write down a counterexample without the Axiom of Choice or something similar. So, for all practical purposes, the answer should be yes.

(I myself do not believe full AC, so I would accept this as an answer, except that I haven't been able to find a reference for the ZF consistency result. But if you believe AC, then this probably won't satisfy you, sorry; we need some kind of horrible AC black magic.)