Let's consider $$f:\mathbb{R}\times [a,b]\to\mathbb{R}$$ a $C^1$ function such that, for every $y_0\in [a,b]$ we have that $f(x,y_0)\in H^1(\mathbb{R})$, would that mean that $$\sup_{(x,y)\in \mathbb{R}\times [a,b]}|f(x,y)|$$ is bounded?
I believe it's true, but I don't know how to prove it, any hint would be appreciated.
This is a (hopefully correct) counterexample: Take any positive function $\phi\in C_c^{\infty}(\mathbb{R})$ with $\sup_{x\in \mathbb{R}}\phi(x)=1$.
Let's define $f:\mathbb{R}\times [0,1]\to \mathbb{R}$ by
\begin{align} f(x,y)=\begin{cases} \frac{1}{y}\phi(x-1/y) \text{ for }y\neq0 \\ 0\text{ for }y=0. \end{cases} \end{align} I claim that $f\in C^{\infty}(\mathbb{R}\times[0,1])$. Clearly $f$ is smooth on $\mathbb{R}\times (0,1]$ so we only need to consider points of the form $(x_0,0)$. From the compact support of $\phi$ it follows that there is a neighbourhood $U$ of $(x_0,0)$ such that $f(x,y)=0$ for all $(x,y)\in U$ (Just take only small y's then the expression $\phi(x-1/y)$ will be zero) and therefore $f$ is smooth on $U$.
Clearly $f(\cdot,y)\in H^1(\mathbb{R})$ for every $y\in [0,1]$ and \begin{align} \sup_{x\in \mathbb{R}}f(x,y)=\frac{1}{y} \end{align} which implies \begin{align} \sup_{(x,y)\in\mathbb{R}\times [0,1]}f(x,y)=\infty \end{align}