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Let $K(x)$ be a positive-valued function such that $$\int_{-\infty}^{\infty} K(x) \ dx = 1$$ and let $$K_{\lambda}(x) = \lambda K(\lambda x), \ \ \ (\lambda > 0);$$ that is to say, the family of functions $$\{K_{\lambda}(x)\}, \ \ \ \lambda \uparrow \infty $$ is an approximate identity generated by dilation.

Suppose $f(x)$ is a measurable function, but we do not know in advance whether or not $f(x) \in L^2$. Assume that for every $\lambda > 0$, the function $$K_{\lambda}(x)*f(x)$$ is finite valued at every $x$.

Further, we know that there exists a function $g(x) \in L^2$ such that $$ \lim_{\lambda \uparrow \infty} \int_{-\infty}^{\infty} [K_{\lambda}(x)*f(x) - g(x)]^{2} \ dx = 0.$$ Does it follow that $$g(x) = f(x) $$ almost everywhere?

If it makes the question easier to answer, one could add the assumption that $f(x)$ is locally integrable.

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    $\begingroup$ How do we define $K_{\lambda}*f$ if we only know that $f$ is measurable? $\endgroup$
    – Christian Remling
    Commented May 9, 2020 at 22:09
  • $\begingroup$ If $f$ is measurable, then the convolution function $K_{\lambda}*f$ is well-defined. I have added the condition that it is finite valued at every $x$. $\endgroup$
    – nickkatz2018
    Commented May 9, 2020 at 22:37
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    $\begingroup$ If $f$ corresponds to, say, a bounded distribution (a continuous linear functional on the class of test functions $\phi$ such that all derivatives of $\phi$ are absolutely integrable), then $K_\lambda * f$ converges to $f$ in the space of bounded distributions, and to $g$ in $L^2$. Thus, $f = g$ almost everywhere. That said, I do not know if it is sufficient to assume that the integrals in the definition of $K_\lambda * f$ are absolutely convergent. $\endgroup$
    – Mateusz Kwaśnicki
    Commented May 9, 2020 at 23:09

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