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Consider $u_{\epsilon} = \rho_{\epsilon} \ast \log$ in $B_{1/2} \subset \mathbb{R}^2$, where $\rho_{\epsilon}$ are standard mollifiers. Then $|u_{\epsilon}(0)|$ blows up as $\epsilon \rightarrow 0$, and $\Delta u_{\epsilon} = \rho_{\epsilon}$ (representation formula) has unit mass independent of $\epsilon$. Taking $V = \rho_{\epsilon}/u_{\epsilon}$ and $f = 0$, we see that the estimate doesn't hold.

Unbounded examples include $u = \log\log(r^{-1})$ in $B_{1/8} \subset \mathbb{R}^n$ for $n \geq 2$ (taking $V = \Delta u / u$ and $f = 0$). In the case $n = 2$ the Laplace is $(r|\log r|)^{-2}$ which is integrable near the origin, and in the case $n \geq 3$ the Laplace is order $r^{-2}|\log r|^{-1}$ which is $L^{n/2}$.

The underlying reason the estimate fails is that $\int |\Delta u|^{n/2}\,dx$ is invariant under the scaling $u \rightarrow u(Rx)$. This gives another approach to constructing examples: for $n = 1$ start with the building block $u_0$ that smoothly connects $0$ to the left and $x$ to the right, and take a weighted sum of translations, e.g. $\sum_{k > 0} k^{-3}u_0(k^3(x + 1/k))$ (a modification of $\log$).

Consider $u_{\epsilon} = \rho_{\epsilon} \ast \log$ in $B_{1/2} \subset \mathbb{R}^2$, where $\rho_{\epsilon}$ are standard mollifiers. Then $|u_{\epsilon}(0)|$ blows up as $\epsilon \rightarrow 0$, and $\Delta u_{\epsilon} = \rho_{\epsilon}$ (representation formula) has unit mass independent of $\epsilon$. Taking $V = \rho_{\epsilon}/u_{\epsilon}$ and $f = 0$, we see that the estimate doesn't hold.

Unbounded examples include $u = \log\log(r^{-1})$ in $B_{1/8} \subset \mathbb{R}^n$ for $n \geq 2$ (taking $V = \Delta u / u$ and $f = 0$). In the case $n = 2$ the Laplace is $(r|\log r|)^{-2}$ which is integrable near the origin, and in the case $n \geq 3$ the Laplace is order $r^{-2}|\log r|^{-1}$ which is $L^{n/2}$.

The underlying reason the estimate fails is that $\int |\Delta u|^{n/2}\,dx$ is invariant under the scaling $u \rightarrow u(Rx)$. This gives another approach to constructing examples: for $n = 1$ start with the building block $u_0$ that smoothly connects $0$ to the left and $x$ to the right, and take a weighted sum of translations, e.g. $u = 1 + \sum_{k > 0} k^{-3}u_0(k^3(x + 1/k))$ (a modification of $\log$). (In fact, by summing Lipschitz rescalings $a_k^{-1} u_0(a_k(x + 1/k))$ with $a_k$ as large as we like we can make $\int |u''|^{1/2}$ arbitrarily small).

Consider $u_{\epsilon} = \rho_{\epsilon} \ast \log$ in $B_{1/2} \subset \mathbb{R}^2$, where $\rho_{\epsilon}$ are standard mollifiers. Then $|u_{\epsilon}(0)|$ blows up as $\epsilon \rightarrow 0$, and $\Delta u_{\epsilon} = \rho_{\epsilon}$ (representation formula) has unit mass independent of $\epsilon$. Taking $V = \rho_{\epsilon}/u_{\epsilon}$ and $f = 0$, we see that the estimate doesn't hold.

Unbounded examples include $u = \log\log(r^{-1})$ in $B_{1/8} \subset \mathbb{R}^n$ for $n \geq 2$ (taking $V = \Delta u / u$ and $f = 0$). In the case $n = 2$ the Laplace is $(r|\log r|)^{-2}$ which is integrable near the origin, and in the case $n \geq 3$ the Laplace is order $r^{-2}|\log r|^{-1}$ which is $L^{n/2}$.

The underlying reason the estimate fails is that $\int |\Delta u|^{n/2}\,dx$ is invariant under the scaling $u \rightarrow u(Rx)$. This gives another approach to constructing examples: for $n = 1$ start with the building block $u_0$ that smoothly connects $0$ to the left and $x$ to the right, and take a weighted sum of translations, e.g. $\sum_{k > 0} k^{-3}u_0(k^3(x + 1/k))$.

Consider $u_{\epsilon} = \rho_{\epsilon} \ast \log$ in $B_{1/2} \subset \mathbb{R}^2$, where $\rho_{\epsilon}$ are standard mollifiers. Then $|u_{\epsilon}(0)|$ blows up as $\epsilon \rightarrow 0$, and $\Delta u_{\epsilon} = \rho_{\epsilon}$ (representation formula) has unit mass independent of $\epsilon$. Taking $V = \rho_{\epsilon}/u_{\epsilon}$ and $f = 0$, we see that the estimate doesn't hold.

Unbounded examples include $u = \log\log(r^{-1})$ in $B_{1/8} \subset \mathbb{R}^n$ for $n \geq 2$ (taking $V = \Delta u / u$ and $f = 0$). In the case $n = 2$ the Laplace is $(r|\log r|)^{-2}$ which is integrable near the origin, and in the case $n \geq 3$ the Laplace is order $r^{-2}|\log r|^{-1}$ which is $L^{n/2}$.

The underlying reason the estimate fails is that $\int |\Delta u|^{n/2}\,dx$ is invariant under the scaling $u \rightarrow u(Rx)$. This gives another approach to constructing examples: for $n = 1$ start with the building block $u_0$ that smoothly connects $0$ to the left and $x$ to the right, and take a weighted sum of translations, e.g. $\sum_{k > 0} k^{-3}u_0(k^3(x + 1/k))$ (a modification of $\log$).

Consider $u_{\epsilon} = \rho_{\epsilon} \ast \log$ in $B_{1/2} \subset \mathbb{R}^2$, where $\rho_{\epsilon}$ are standard mollifiers. Then $|u_{\epsilon}(0)|$ blows up as $\epsilon \rightarrow 0$, and $\Delta u_{\epsilon} = \rho_{\epsilon}$ (representation formula) has unit mass independent of $\epsilon$. Taking $V = \rho_{\epsilon}/u_{\epsilon}$ and $f = 0$, we see that the estimate doesn't hold.

Unbounded examples include $u = \log\log(r^{-1})$ in $B_{1/8} \subset \mathbb{R}^n$ for $n \geq 2$ (taking $V = \Delta u / u$ and $f = 0$). In the case $n = 2$ the Laplace is $(r|\log r|)^{-2}$ which is integrable near the origin, and in the case $n \geq 3$ the Laplace is order $r^{-2}|\log r|^{-1}$ which is $L^{n/2}$.

The underlying reason the estimate fails is that $\int |\Delta u|^{n/2}\,dx$ is invariant under the scaling $u \rightarrow u(Rx)$. This gives another approach to constructing examples.

Consider $u_{\epsilon} = \rho_{\epsilon} \ast \log$ in $B_{1/2} \subset \mathbb{R}^2$, where $\rho_{\epsilon}$ are standard mollifiers. Then $|u_{\epsilon}(0)|$ blows up as $\epsilon \rightarrow 0$, and $\Delta u_{\epsilon} = \rho_{\epsilon}$ (representation formula) has unit mass independent of $\epsilon$. Taking $V = \rho_{\epsilon}/u_{\epsilon}$ and $f = 0$, we see that the estimate doesn't hold.

Unbounded examples include $u = \log\log(r^{-1})$ in $B_{1/8} \subset \mathbb{R}^n$ for $n \geq 2$ (taking $V = \Delta u / u$ and $f = 0$). In the case $n = 2$ the Laplace is $(r|\log r|)^{-2}$ which is integrable near the origin, and in the case $n \geq 3$ the Laplace is order $r^{-2}|\log r|^{-1}$ which is $L^{n/2}$.

The underlying reason the estimate fails is that $\int |\Delta u|^{n/2}\,dx$ is invariant under the scaling $u \rightarrow u(Rx)$. This gives another approach to constructing examples: for $n = 1$ start with the building block $u_0$ that smoothly connects $0$ to the left and $x$ to the right, and take a weighted sum of translations, e.g. $\sum_{k > 0} k^{-3}u_0(k^3(x + 1/k))$.