**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?