Yes. If $K \subset \Omega$ is compact, let $\text{dist}(K, \Omega^c) > r > 0$. Then for any $p \in K$ and convex function $f$, $f(p)$ is bounded above by the average of $f$ over the ball of radius $r$ centred at $p$, and thus by a constant (depending only on $r$) times $\|f\|_{L^2(\Omega)}$.

For a lower bound, note that as $q$ varies over the ball $B_r(p)$ of radius $r$ centred at $p$, $(q+p)/2$ varies over $B_{r/2}(p)$, and $f(p) \ge 2 f((q+p)/2) - f(q)$, so $$f(p) \ge \dfrac{2}{m(B_{r/2}(p)} \int_{B_{r/2}(p)} f - \dfrac{1}{m(B_r(p))} \int_{B_r(p))} f \ge - C(r) \|f\|_{L^2(\Omega)}$$

Yes. If $K \subset \Omega$ is compact, let $\text{dist}(K, \Omega^c) > r > 0$. Then for any $p \in K$ and convex function $f$, $f(p)$ is bounded above by the average of $f$ over the ball of radius $r$ centred at $p$, and thus by a constant (depending only on $r$) times $\|f\|_{L^2(\Omega)}$.

For a lower bound, note that if $Q$ is a uniform random variable on the ball $B_r(p)$ of radius $r$ centred at $p$, $(Q+p)/2$ is uniform on $B_{r/2}(p)$, and $f(p) \ge 2 f((Q+p)/2) - f(Q)$, so $$f(p) \ge \dfrac{2}{m(B_{r/2}(p)} \int_{B_{r/2}(p)} f - \dfrac{1}{m(B_r(p))} \int_{B_r(p))} f \ge - C(r) \|f\|_{L^2(\Omega)}$$

Yes. If $K \subset \Omega$ is compact, let $\text{dist}(K, \Omega^c) > r > 0$. Then for any $p \in K$ and convex function $f$, $f(p)$ is bounded by the average of $f$ over the ball of radius $r$ centred at $p$, and thus by a constant (depending only on $r$) times $\|f\|_{L^2(\Omega)}$.

Yes. If $K \subset \Omega$ is compact, let $\text{dist}(K, \Omega^c) > r > 0$. Then for any $p \in K$ and convex function $f$, $f(p)$ is bounded above by the average of $f$ over the ball of radius $r$ centred at $p$, and thus by a constant (depending only on $r$) times $\|f\|_{L^2(\Omega)}$.

For a lower bound, note that as $q$ varies over the ball $B_r(p)$ of radius $r$ centred at $p$, $(q+p)/2$ varies over $B_{r/2}(p)$, and $f(p) \ge 2 f((q+p)/2) - f(q)$, so $$f(p) \ge \dfrac{2}{m(B_{r/2}(p)} \int_{B_{r/2}(p)} f - \dfrac{1}{m(B_r(p))} \int_{B_r(p))} f \ge - C(r) \|f\|_{L^2(\Omega)}$$