Spectrum of operator : $Tf(x)=\dfrac{f(x)+f(-x)}{2}$

Question

Problem :

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Let :

$$T~:~L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$$ Such that :

$$Tf(x)=\dfrac{f(x)+f(-x)}{2}$$

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I'm going to find the spectrum of the above operator $T$

the operator $T$ are self -adjoins and $||T||$=1 We know that :

$$\sigma (\lambda)=\{\lambda\in\mathbb {C}~,~(T-\lambda f)\text{not invertible}\}$$ I'm start by :

$$Tf=\lambda f$$

So :

$$f(x)+f(-x)=2\lambda f(x)$$

So :

$$f(-x)=(2\lambda -1)f(x)$$

Also

$$f(x)=(2\lambda -1)f(-x)$$

This mean :

$$(2\lambda -1)^{2}f(x)=f(x)$$

So : $$2\lambda -1=±1 \implies \lambda =0,1$$

Finally: $$\sigma(\lambda)=\{0,1\}$$

Any other ideas?