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I have doubts on how to find the kernel and image for this linear transformation $T:\mathbb{C(R)}\rightarrow\mathbb{C(R})$ defined by: $T(f(x))=\frac{f(x)+f(-x)}{2}$, where $\mathbb{C(R)}$ represents the set of continuous functions in $\mathbb{R}$. I'm lost. Any hints?

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    $\begingroup$ L ​ It looks like T(f(x)) should be replaced with (T(f))(x). ​ ​ ​ ​ $\endgroup$ – user57159 May 23 '16 at 16:26
  • $\begingroup$ Indeed. In fact the definition would be most explicitly written $T(f)=\left(x\mapsto\frac{f(x)+f(-x)}2\right)$. $\endgroup$ – Marc van Leeuwen May 24 '16 at 4:36
  • $\begingroup$ To get an idea of what the image is, try this for a few simple polynomials of degree 2, 3, 4, 5, 6. Then you might be able to guess which polynomials are in the kernel. After that it's a straight forward generalization. $\endgroup$ – Hans Engler May 25 '16 at 0:58
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For the kernel, you want $T(f(x))=0$, so $$\frac{f(x)+f(-x)}2=0$$ so $$f(x)=-f(-x).$$ This is the exact definition of an odd function.

The image is the set of even functions, because $T(f(-x))=T(f(x))$ by a simple calculation. And reciprocally, if you take an even function, you can always put it in this form.

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    $\begingroup$ Like in OP, the final formula should be written $T(f)(-x)=T(f)(x)$. One cannot apply $T$ to a real number, like $f(x)$. $\endgroup$ – Marc van Leeuwen May 25 '16 at 4:40
  • $\begingroup$ I couldn't agree more ! Though I used the notation of the question to avoid confusion. $\endgroup$ – E. Joseph May 25 '16 at 7:04
  • $\begingroup$ And how to make the image? $\endgroup$ – Tomáš Zato Dec 15 '16 at 1:34
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For example

$$f\in\ker T\iff f(x)=-f(-x)\iff f\;\;\text{is an odd function}$$

Now continue on this path and you'll find out the image is pretty simple, too.

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For $\ker(T)$ it suffices to apply the definition: the condition $T(f)=0$ means that $x\mapsto\frac{f(x)+f(-x)}2$ is the zero function, which means that $f(-x)=-f(x)$ for all $x\in\Bbb R$ (i.e., $f$ is an an odd function).

A direct computation shows that this linear transformation$~T$ satisfies $T^2=T$, so it is a projection. For a projection, being in the image is the same as being ones own image: if $f=T(g)$ for any function $g$ then one also has $f=T(T(g))=T(f)$. So the condition for $f$ being in the image of $T$ is that $0=T(f)-f=\left(x\mapsto\frac{f(x)+f(-x)}2-f(x)\right)=\left(x\mapsto\frac{-f(x)+f(-x)}2\right)$, in other words that $f(-x)=f(x)$ for all $x\in\Bbb R$ (i.e., $f$ is an an even function).

One may add, and this is true for any projection, that the whole space is the sum of the kernel and the image of $T$, since every $f$ satisfies $f=T(f)+(I-T)(f)$, where the first term is in the image of $T$ and the second term is (because of $T^2=T$) in the kernel of$~T$. Moreover this sum is direct (the decomposition of $f$ is unique), since a function $g$ cannot be simultaneously in the image of $T$ (so $T(g)=g$) and in the kernel of $T$ (so $T(g)=0$) unless $g=0$.

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