Vrftsjtryk

The following question was part of an advanced calculus of variations class. A solution was presented, being different and way more simple then the proof being presented in the literature.

Let $Omega subset mathbb{R}^n$ be an open, bounded and connected set. Let $g: bar{Omega} times mathbb{R} to mathbb{R}$ be a function such that
$$
gin C^0(bar{Omega} times mathbb{R})
$$
as well as a growth estimate for all $x in bar{Omega}$
$$
|g(x,t)|leq C_1+C_2|t|^{r/p}
$$
Proof that the function $h: L^r(Omega)to L^p(Omega), phi to g(cdot,phi(cdot))$ is continuous.

The proof presented included the following steps:

For an arbitrary $phi_0 in L^r(Omega)$ and $z in L^r(Omega)$, consider the function $f:L^r(Omega)to L^p(Omega), z to g(cdot,phi_0+z)-g(cdot,phi_0)$ and study its continuity at $z(cdot)=0$. Note that $f(cdot,0)=0$. Using the growth estimate, one can derive a similar estimate for $f$ of the form:
$$
|f(x,z(x)|leq A_1+A_2 |phi_0(x)|^{r/p}+A_3|z(x)|^{r/p}
$$
So far, so good. Up until this point, I am confident the solution is correct.
In the last steps, I have my doubts if it is really that simple:
Now, taking any sequence $z_n xrightarrow{L^r}0$, we have that $z_n to 0$ pointwise a.e. and using the estimate for $f$ from above
$$
int_{Omega} |f(x,z_n(x))|^p leq int_{Omega}D_1+D_2 |phi_0(x)|^{r}+D_3|z_n(x)|^{r} leq infty
$$
the dominated convergence theorem can be applied. This should yield the result by using the continuity of $g$:
$$
lim_{n to infty} int_{Omega} |f(x,z_n(x))|^p=int_{Omega} lim_{n to infty}|f(x,z_n(x))|^p=int_{Omega} |f(x,0)|^p
$$

Can somebody please double check the last steps if the solution is actually correct?

The original solution suggested to split up the integral $int_{Omega} |f(x,z_n(x))|^p$ in 2 parts: One part of $Omega$, where the integrand is small due to $z_n$ being small in the $L^r$ sense, and the other part of $Omega$ ,where the grwoth estimates could be used, but the set where $|z_n| geq delta$ was small.

Appendix: As pointed out in the answer, the argument was flawed in the sense that we assumed $z_n to 0$ a.e pointwise. However, this is only correct for a subsequence. There seems to be no straight-forward way to proof that $int_{Omega} lim_{n to infty}|f(x,z_n(x))|^p=int_{Omega} |f(x,0)|^p$ without splitting up $Omega$ into different parts. Convergence in measure does require such a split.

in one part of your proof you have said:

$z_n to 0$ in $L^r$ then $z_n to 0$ pointwise.

That's not the case unfortunately, there is one standard counter example to this statement called the "Typewriter sequence":

consider the sequence of indicator functions defined on the interval $[0,1]$ $f_n=I[frac{n-2^k}{2^k},frac{n-2^k+1}{2^k}]$ whenever $k geq 0$ and $2^k leq n leq 2^{k+1}$.

You have that $f_n$ converges to $0$ in $L^1$ norm but $f_n$ doesn't converges to $0$ pointwise almost everywhere.

So maybe that's the why the author took that long way for the proof.

Best regards,

Andrea