Integral operator and the kernel function Let $D$ be an open bounded set in $\mathbb{R}^{n}$. Let $p \in [1,\infty)$, and $q$ is the dual exponent to $p$.  Assume that $K : \mathbb{R} \times D \rightarrow \mathbb{R}$ is a bounded continuous function, and the function
\begin{align}
y \mapsto \int_{D} K(y,x) g(x) dx : \mathbb{R} \rightarrow \mathbb{R}, \hspace{1cm}(1)
\end{align}
is Lipschitz continuous for each $g \in L^{q}(D)$.
Prove that the vector valued map
\begin{align}
y \mapsto K(y,\cdot) : \mathbb{R} \mapsto L^{p}(D)
\end{align}
is Lipschitz continuous.
I was curious if someone could provide assistance on showing this is Lipschitz.  If we were asked to show the integral operator was Lipschitz we would start by applying Holder to (1), but I'm confused on showing the kernel is Lipschitz.
 A: Roughly speaking, this is a consequence of the closed graph theorem and the duality.
Let ${\rm Lip}(\mathbb R)$ be the Lipschitz space on $\mathbb R$ which defined by $${\rm Lip}(\mathbb R)=\{f\in C(\mathbb{R}):\|f\|_{Lip}:=|f(0)|+\sup_{y\neq y'} \frac{|f(y)-f(y')|}{|y-y'|} <\infty\}.$$
It will be convenient to write $K_y(\cdot)=K(y,\cdot)$.
Consider the integral operator $V: L^q(D)\to \rm{Lip}(\mathbb{R})$ $$Vg(y)=\int_{D} K_y(x)g(x)dx.$$ The condition in your question ensure that $V$ is will defined. By the closed graph theorem, one can show that $V$ is bounded.
The argument is standard, I just point out the following two fact which will be used:

${\rm Lip}(\mathbb{R})$ is a Banach space. Proof


If $f_n$ converges to $f$ in $\rm{Lip}(\mathbb{R})$, then $f_n$ convergens to $f$ pointswise.

The boundedness of $V$ gives a postive constant $C$, such that $$\sup_{y\neq y'}\frac{|Vg(y)-Vg(y')|}{|y-y'|}\leq |Vg(0)|+\sup_{y\neq y'}\frac{|Vg(y)-Vg(y')|}{|y-y'|}=\|Vg\|_{Lip}\leq C\|g\|_{L^q(D)}.$$ This deduces, for any $y$, $y'\in \mathbb R$ it holds that
\begin{align}
\bigg|\int_D \big(K_y-K_{y'}\big)g dx\bigg|=\big|Vg(y)-Vg(y')\big| \leq C|y-y'| \cdot \|g\|_{L^q(D)}.
\end{align}
The proof is finished by the duality:
\begin{align}
\|K_y-K_{y'}\|_{L^p}=\sup \big|\int_D \big(K_y-K_{y'}\big)g dx\big|,
\end{align}
where the supreme is taken over all $g\in L^q(D)$ with norm less that 1.
A: The conclusion follows directly from the Baire category theorem.
Indeed let $T: L^q(D)\to C(\mathbb{R})$ be defined
by $$(Tg)(x)=\int\limits_D K(x,t)g(t)\,dt$$
Let
$$A_N=\left \{g\in L^q\,:\, \|g\|_q\le 1, \ {|(Tg)(x)-(Tg)(x')|\over  |x-x'|}\le N,\, x\neq x'\right\}$$
Then $A_N$ is a convex set.
The set $A_N$ is  closed in $L^q(D).$ Indeed, assume $\|g_n-g\|_q\to 0,$ where $g_n\in A_N.$ Then $\|g_n\|_q\to \|g\|_q.$
Moreover by the Holder inequality we get
$$(Tg_n)(x)\to (Tg)(x),\qquad x\in D$$
Thus $$|(Tg)(x)-|(Tg)(x')|\le N|x-x'|$$ i.e. $g\in A_N.$
By assumptions we have
$$ L^q(D)=\bigcup_{k=1}^\infty \bigcup_{N=1}^\infty kA_N$$
Therefore, by the Baire category theorem $kA_N$ has nonempty interior for some $k$ and $N.$
By convexity the set  $kA_N$ contains an open neighborhood of $0,$ i.e.
$$\{g\in L^q\,:\, \|g\|_q\le m^{-1}\}\subset kA_N$$
Then $$\{g\in L^q\,:\, \|g\|_q\le 1\}\subset kmA_N$$
Let $L_x=K(x,\cdot).$ By duality between $L^p$ and $L^q$ we have
$$\displaylines{\|L_x-L_{x'}\|_p=\sup\left \{g\in L^q(D)\,:\, \left |\int\limits_D [L_x(t)-L_{x'}(t)]g(t)\,dt\right |,\ \|g\|_q\le 1\right \}\\
=\sup\left \{g\in L^q(D)\,:\, \left |(Tg)(x)-(Tg)(x')\right |,\ \|g\|_q\le 1\right \}\le kmN|x-x'|}$$
