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Let $f: \mathbb R^2 \to \mathbb R$ be a function such that for every open set $U \subseteq \mathbb R$ and continuous function $h:U \to \mathbb R$ , $f|_{G(h)} : G(h) \to \mathbb R$ is continuous ($G(h):=\{(x,h(x))|x \in U\}$) , then is it true that $f$ is continuous ?

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    What happened when you tried to prove it? Use $\epsilon$,$\delta$ definition of continuity. – 2017-01-31
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    @GEdgar : I have pointed out the difficulty in the answer below by Surb ; my approach was the same and that was the difficulty – 2017-01-31

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Suppose $f$ is discontinuous at some $(x_0,y_0).$ Then there exists $\epsilon>0$ and a sequence $(x_n,y_n) \to (x_0,y_0)$ such that

$$\tag 1 |f(x_n,y_n) -f(x_0,y_0)| > \epsilon \text { for all } n.$$

By passing to a subsequence if necessary, we can assume $(x_n)$ is monotonic. WLOG, we can assume the sequence is decreasing, so that $x_1\ge x_2 \ge \cdots \ge x_0$ for all $n.$

Claim: We can choose a sequence $x_n'$ in $\mathbb R$ such that $x_1' > x_2' > \cdots \to x_0$ such $f(x_n',y_n) - f(x_n,y_n) \to 0.$

Suppose the claim is true. We then define a continuous $h$ on $\mathbb R$ as follows: $h$ is linear on each $[x_{n+1}',x_n'],$ with values $y_{n+1}, y_n$ at the endpoints. Then define $h(x) = y_0$ for $x\le x_0$ and $h(x) = y_1$ for $x\ge x_1.$ (Good to draw a picture here.)

By hypothesis, $f\circ h$ is continuous on $\mathbb R.$ Thus $f(x_n',h(x_n')) = f(x_n',y_n) \to f(x_0,y_0).$ The claim then forces $f(x_n,y_n) \to f(x_0,y_0)$ as well, contradiction.

Proof of the claim: The hypotheses imply that $f$ is continuous on every horizontal line. Thus

$$\lim_{x\to x_1^+} f(x,y_1) = f(x_1,y_1).$$

Therefore we can choose $x_1'>x_1$ such that

$$|x_1'-x_1| < 1, \text { and } |f(x_1',y_1) - f(x_1,y_1)| <1.$$

Because $x_2\le x_1$ and $\lim_{x\to x_2^+} f(x,y_2) = f(x_2,y_2),$ we can choose $x_2'\in (x_2,x_1')$ such that

$$|x_2'-x_2| < 1/2, \text { and } |f(x_2',y_2) - f(x_2,y_2)| <1/2.$$

Continiung by induction produces a sequence $x_n'$ such that $x_1' > x_2' > \cdots \to x_0$ and such that

$$|x_n'-x_n| < 1/n, |f(x_n',y_n) - f(x_n,y_n)| <1/n$$

for all $n.$ This proves the claim, and finishes the proof.

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Suppose it's not continuous at $(x,y)\in\mathbb R^2$. Then, $$\exists \varepsilon>0:\forall n\in\mathbb N, \exists (x_n,y_n)\in\mathbb R^2: \|(x_n,y_n)-(x,y)\|<\frac{1}{n}\quad \text{and}\quad |f(x_n,y_n)-f(x,y)|>\varepsilon.$$

Let $h$ a continuous function s.t. $y_n=h(x_n)$. In particular, by continuity of $h$, we have $y=h(x)$. Using the fact that $G(h)$ is closed, $(x,h(x))\in G(h)$, what contradict $f$ continuous on $G(h)$.

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    What if several $x_n$'s (or infinitely many) equal $x$? – 2017-01-31
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    @Surb : yes exactly , what if $x_n$ is ultimately constant ? – 2017-01-31
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    @SaunDev: I don't see in what it's a problem... But we can anyway take $x_n\neq x$ for all $n$. – 2017-01-31
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    @Surb Yes we can, but that might be considered the heart of the proof... After all, if we only assumed that $f$ were continuous along graphs of continuous functions containing $(x,y)$ (and not every graphs of continuous functions), we could construct a $f$ discontinuous along the vertical line $x$. So you need to complete your argument. – 2017-01-31
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    @Surb : No that would be a problem ... for example consider the sequence $(1,1/n)$ converging to $(1,0)$ . How would you construct your $h$ to consider the continuity of $f$ at $(1,0)$ along this sequence ? – 2017-01-31
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    @SaunDev: I agree, sorry to have not considered this case (to be honnest, I didn't think about it). But indeed, the problem is when there is $n$ s.t. $x_m=x$ **for all** $m\geq n$, other wise, it's not a problem. But you will be agree that either $x_n$ or $y_n$ must have a subsequence that is non trivial (other wise $(x_n,y_n)=(x,y)$ for all $n$). If $f|_{G(h)}$ is continuous, then it will be also continuous over $\{(h(x),x)\mid x\in U\}$, then you can suppose WLOG that there is a subsequence $x_{n_{k}}\neq x$ for all $k$. – 2017-01-31
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    @Surb : yes , I know that the problem is when $x_n$ is ultimately constant . And I also agree that WLOG can assume that either $x_n$ or $y_n$ consists of mutually distinct terms . But the problem is the " graph " of a function $h$ is defined as $\{(x,h(x))\}$ and not as $\{(h(x),x)\}$ ... and I am unable to see why $f|_{G(h)}$ is continuous for any $h$ implies $f|_{\{(h(x),x)|x \in U\}}$ is continuous ... – 2017-01-31
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    @SaunDev: Maybe I'm wrong to says that it will be continuous over $\{(h(x),x)\}$. I don't have time now, but if there is a counter example to find is in this way (something as $f(x,y)=??$ for $x\neq 0$ and $f(0,y)=1_{\mathbb Q}$... or something like that... maybe it can work. – 2017-01-31