FFT result of windowed cosine function The frequency spectrum of cosine function is as below.

While adopting a square window function to the cosine function, it is same as implementing convolution of their frequency spectrum as below. The thick dash line is the sum of them. Is it correct to say the points on this dash line are the FFT result of corresponding frequency? I feel it is not true, but don't know where the problem is.
There is no problem in your plot if by fft you mean DTFT (or Discrete Time Fourier transform) and it is true to say &quot;the points on this dash line are the DTFT result of corresponding frequency&quot;:
If your function is a cosine:
$$
x[n] = \cos(\omega_0n) \Longrightarrow X(e^{j\omega}) = \pi\delta(\omega-\omega_0)+\pi\delta(\omega+\omega_0)\ ; \qquad|\omega|&lt;\pi
$$
and your window function is defined as follow:
$$
w[n] = \left\{
        \begin{array}{ll}
            1 &amp; \quad |n| \leq N \\
            0 &amp; \quad |n| &gt; N
        \end{array}
    \right.
\Longrightarrow W(e^{j\omega}) = \frac{\sin\left(\omega(\frac{N}{2}+1)\right)}{\sin\left(\frac{\omega}{2}\right)}
$$
as you mentioned if we multiply two functions in time domain, it's equal to periodic convolution in frequency domain:
$$
y[n] = x[n]w[n]\\ \Longrightarrow Y(e^{j\omega})=X(e^{j\omega})\circledast W(e^{j\omega})=\frac{1}{2\pi}\int_{2\pi}^{}X(e^{j\omega})W(e^{j(\omega-\theta)})  d\theta
$$
in this case (in one period):
$$
Y(e^{j\omega})=\frac{1}{2}\frac{\sin\left((\omega-\omega_0)\left(\frac{N}{2}+1\right)\right)}{\sin\left(\frac{\omega-\omega_0}{2}\right)}+\frac{1}{2}\frac{\sin\left((\omega+\omega_0)\left(\frac{N}{2}+1\right)\right)}{\sin\left(\frac{\omega+\omega_0}{2}\right)}
$$
if you plot this function in one period for $\omega_0=\frac{\pi}{5}$ and $N=2$ it look like this:

but if you change the window length to $N=4$ the result changes to:

and the cause for this, is that two main lobes overlap with one another.