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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?

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.

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 $f(x) \in L^2$.

However, 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?

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?

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 $f(x) \in L^2$.

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

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 $f(x) \in L^2$.

However, 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?