Show that f such that f(x+cy)=f(x)+cf(y) is continuous

[Bptrhp]

Homework Statement

Let f:R→R be a function such that f(x+cy)=f(x)+cf(y), ∀x,y∈R, ∀c∈R. Show that f is continuous.

Relevant Equations

f(x+cy)=f(x)+cf(y), ∀x,y∈R, ∀c∈R

We need to show that $\lim_{x \rightarrow a}f(x)=f(a), \forall a \in \mathbb{R}$ .
At first, I tried to show that f is continuous at 0 and from there I would show for all a∈R. But now, I think this may not even be true. I only got that f(0)=0. I'm very confused, I appreciate any help!

 

[fresh_42]

$f$ is continuous at $0$, and from there you can work with the translation $x\longmapsto x+a$. For continuity at $0$ consider $f(1/n)=1/n f(1).$

 

[Bptrhp]

f is continuous at 0, and from there you can work with the translation x⟼x+a. For continuity at 0 consider f(1/n)=1/nf(1).

I already managed to prove continuity for all $a\in\mathbb{R}$ assuming that $f$ is continuous at $0$, but I can't see how to prove continuity for 0 considering $f(1/n)=1/nf(1)$. Is it something related to sequences?

 

[PeroK]

I already managed to prove continuity for all $a\in\mathbb{R}$ assuming that $f$ is continuous at $0$, but I can't see how to prove continuity for 0 considering $f(1/n)=1/nf(1)$. Is it something related to sequences?

What about trying $x = a + h$, where $x \rightarrow a$ is equivalent to $h \rightarrow 0$?

 

[fresh_42]

I already managed to prove continuity for all $a\in\mathbb{R}$ assuming that $f$ is continuous at $0$, but I can't see how to prove continuity for 0 considering $f(1/n)=1/nf(1)$. Is it something related to sequences?

I would take the epsilon-delta definition, or just take an arbitrary sequence $x_n \longrightarrow 0$. The sequence $1/n \longrightarrow 0$ was meant as an example to help you find the clue. The key is the following equation:
$$
|f(x_n)-f(a)|=|x_n\cdot f(1)-a\cdot f(1)|=|x_n-a|\cdot |f(1)|
$$
where you can immediately read the epsilons and deltas from.

 

[fresh_42]

To remember the definitions, I use to think about the following picture, which isn't allow to happen on a continuous function:
\begin{align*}
\text{______________}&\\[14pt]
&\text{______________}
\end{align*}
Coming closer on the $x-$ axis doesn't allow a gap on the $y-$axis.

 

[willem2]

It's actually easier to compute a general solution for f, than to prove that f is continuous.
if you substitute x =0, y = u, c = 1/u, you get f(u) = (f(1)- f(0)) u.
These are obviously all continuous.

 

[Infrared]

@willem2 That argument only works when $u\neq 0$, so it doesn't get you around needing to prove continuity at $0.$

Edit: Although, this is easy to fix. Your expression shows that the limit is zero, and putting $x=y=0,c\neq 0$ shows that $f(0)=0.$

 

[Bptrhp]

I would take the epsilon-delta definition, or just take an arbitrary sequence $x_n \longrightarrow 0$. The sequence $1/n \longrightarrow 0$ was meant as an example to help you find the clue. The key is the following equation:
$$
|f(x_n)-f(a)|=|x_n\cdot f(1)-a\cdot f(1)|=|x_n-a|\cdot |f(1)|
$$
where you can immediately read the epsilons and deltas from.

Does that mean the continuity $\forall a \in \mathbb{R}$ can be shown by epsilon-delta definition without showing continuity at $0$ first?

My attempt:
Given $\epsilon >0$, take $\delta = \dfrac{\epsilon } {|f(1)|}>0$, so that for any $x\in\mathbb{R},|x-a|<\delta$, we have
$|f(x)-f(a)|=|x\cdot f(1)-a\cdot f(1)|=|x-a|\cdot |f(1)|<\delta\cdot |f(1)|=\epsilon$.

I feel like I'm missing something...

 

[fresh_42]

Does that mean the continuity $\forall a \in \mathbb{R}$ can be shown by epsilon-delta definition without showing continuity at $0$ first?

Yes.

My attempt:
Given $\epsilon >0$, take $\delta = \dfrac{\epsilon } {|f(1)|}>0$, so that for any $x\in\mathbb{R},|x-a|<\delta$, we have
$|f(x)-f(a)|=|x\cdot f(1)-a\cdot f(1)|=|x-a|\cdot |f(1)|<\delta\cdot |f(1)|=\epsilon$.

I feel like I'm missing something...

Why? Linear functions are always continuous (maybe some weird topologies apart).

 

[PeroK]

Does that mean the continuity $\forall a \in \mathbb{R}$ can be shown by epsilon-delta definition without showing continuity at $0$ first?

My attempt:
Given $\epsilon >0$, take $\delta = \dfrac{\epsilon } {|f(1)|}>0$, so that for any $x\in\mathbb{R},|x-a|<\delta$, we have
$|f(x)-f(a)|=|x\cdot f(1)-a\cdot f(1)|=|x-a|\cdot |f(1)|<\delta\cdot |f(1)|=\epsilon$.

I feel like I'm missing something...

... you need to take care of the case when $f(1) = 0$.

Alternatively, consider $f(a + h)$ as $h \rightarrow 0$.

 

[vela]

My initial thought was to use the approach @PeroK has suggested, yet the rest of you are taking more complicated approaches. Is there some subtlety I'm missing here?

 

[Bptrhp]

... you need to take care of the case when $f(1) = 0$.

I totally forgot that!

Alternatively, consider $f(a + h)$ as $h \rightarrow 0$.

I actually considered this when I proved the continuity in $\mathbb{R}$ assuming that $f$ is continuous at 0. I've obtained:

\begin{align*}
\lim_{x\rightarrow a}f(x)=\lim_{h\rightarrow 0}f(h+a)&=\lim_{h\rightarrow 0}(f(h)+f(a))\\
&=\lim_{h\rightarrow 0}f(h)+\lim_{h\rightarrow 0}f(a) \quad (because\, both\, limits\, exist)\\
&=0+f(a)\\
&=f(a).
\end{align*}
But how should I use $f(a+h)$ as $h\rightarrow 0$ to prove continuity at $0$? In this case, we can't assume the limit of the sum is the sum of the limits, right?

 

[vela]

Try $x = a+h = a+h \cdot 1$.

 

[PeroK]

I totally forgot that!I actually considered this when I proved the continuity in $\mathbb{R}$ assuming that $f$ is continuous at 0. I've obtained:

\begin{align*}
\lim_{x\rightarrow a}f(x)=\lim_{h\rightarrow 0}f(h+a)&=\lim_{h\rightarrow 0}(f(h)+f(a))\\
&=\lim_{h\rightarrow 0}f(h)+\lim_{h\rightarrow 0}f(a) \quad (because\, both\, limits\, exist)\\
&=0+f(a)\\
&=f(a).
\end{align*}
But how should I use $f(a+h)$ as $h\rightarrow 0$ to prove continuity at $0$? In this case, we can't assume the limit of the sum is the sum of the limits, right?

What about simply:
$$\lim_{h\rightarrow 0}f(a + h) = \lim_{h\rightarrow 0}(f(a)+hf(1))= f(a) + f(1)\lim_{h\rightarrow 0}(h) = f(a)$$

 

[Bptrhp]

Now I get it, I was really making things more complicated than they really are! Thanks a lot!

 

[epenguin]

Not my sort of thing but I ask doesn't f(x+cy)=f(x)+cf(y) if true for all y imply that if f(x) is discontinuous anywhere it is discontinuous everywhere?
Is this true and is it relevant?

 

[Bptrhp]

Not my sort of thing but I ask doesn't f(x+cy)=f(x)+cf(y) if true for all y imply that if f(x) is discontinuous anywhere it is discontinuous everywhere?
Is this true and is it relevant?

I think this is only true if $f(x+y)=f(x)+f(y)$. If I understood correctly, the fact that $f(cy)=cf(y)$ makes the function continuous, but I'm not sure...