Does the sequence of functions $k(x + \frac{1}{n}, y)$ converge uniformly to $k(x,y)$ if $k$ is continuous?

Question

Suppose I have a function $k(x,y): [0,1] \times [0,1] \rightarrow \mathbb{R}$ such that $k$ is continuous. Suppose further that for a certain application, I am fixing the $x$-coordinate and allowing $y$ to vary, and that I define $k_x: [0,1] \rightarrow \mathbb{R}$ so that $k_x(y) = k(x,y)$. Finally, for each $n \in \mathbb{N}$, I define $k_n: [0,1] \rightarrow \mathbb{R}$ so that $k_n(y) = (x + \frac{1}{n}, y)$. Then it is clear to me that $k_n \rightarrow k_x$ pointwise. I want to know if it is also the case that $k_n \rightarrow k_x$ uniformly. I think so, and it seems like it should be easy to prove. But I have just been totally stuck on this!

Answer 1

Since $[0,1]\times[0,1]$ is compact, $k$ is uniformly continuous. Take $\varepsilon>0$. Since $k$ is uniformly continuous, there is some $\delta>0$ such that$$\|(x_1,y_1)-(x_0,y_0)\|<\delta\implies\bigl|f(x_1,y_1)-f(x_,y_0)\bigr|<\varepsilon.$$Now, take $N\in\Bbb N$ such that $\frac1N<\delta$. Then\begin{align}n\geqslant N&\implies\left\|\left(x+\frac1n,y\right)-(x,y)\right\|<\delta\\&\implies\left|f\left(x+\frac1n,y\right)-f(x,y)\right|<\varepsilon.\end{align}